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\title{Number Fields}

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\tableofcontents

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\section{Miscellaneous}

Book: Number Fields, Marcus

Course notes: www.dpmms.ac.uk/~jat58/nfl2018

\newpage

\section{Motivation}
\begin{thm}
If $p$ is an odd prime, then $p=a^2+b^2$ for $a,b \in \Z \iff p \equiv 1 \pmod 4$.
\begin{proof}
If $p=a^2+b^2$, then $p\equiv 0,1,2 \pmod 4$. So this condition on $p$ is necessary.\\
Suppose instead $p \equiv 1 \pmod 4$. Then $\left(\frac{-1}{p}\right) = 1$. Thus $\exists a \in \Z$ such that $a^2 \equiv -1 \pmod p$, or $p|a^2+1$. We can factor $a^2+1=(a+i)(a-i)$ in the ring $\Z[i]$. Here we introduce a notation: if $R \subseteq S$ are rings and $\alpha \in S$, then $$R[\alpha] = \{\sum_{i=0}^n a_i \alpha^i \in S | a_i \in R\}$$, the smallest subring of $S$ containing both $R$ and $\alpha$.

We know from IB GRM that $\Z[i]$ is a UFD. Now $p|(a+i)(a-i)$. If $p$ is irreducible in $\Z[i]$ then $p|a+i$ or $p|a-i$, contradiction. Thus $p$ is reducible in $\Z[i]$, hence $p=z_1z_2$ with $z_1,z_2 \in \Z[i]$. If $z_1 = A+Bi$, $A,B\in \Z$, then $A^2+B^2 = p$.
\end{proof}
\end{thm}

Another example is when $p$ is an odd prime. Does the equation $$x^p+y^p=z^p$$ have solutions with $x,y,z\in \Z$ and $xyz \neq 0$?

\begin{thm} (Kummer, 1850)\\
If $\Z[e^{2\pi i/p}]$ is a UFD, then there are no solutions.\\
Strategy: factor $x^p+y^p = \prod_{j=0}^{p-1} (x+e^{2\pi ij/p}y)$ in $\Z[e^{2\pi i/p}]$.
\end{thm}

However, we now know $\Z[e^{2\pi i/p}]$ is a UFD $\iff$ $p \leq 19$.

\begin{thm} (Kummer, 1850)\\
If $p$ is a \emph{regular} prime, then there are no solutions.\\
If $p<100$, then $p$ is regular $\iff$ $p \neq 37,59,67$.
\end{thm}

We have seen various examples such as $\Z \subseteq \Q$, $\Z[i] \subseteq \Q[i]$, $\Z[e^{2\pi i/p}] \subseteq \Q[e^{2\pi i/p}]$, or in general, $\mathcal{O}_L\subseteq L$, where a ring of "integers" lies in a number field.

\newpage

\section{Ring of integers}
Recall: A field extension $L/K$ is an inclusion $K \leq L$ of fields. The degree of $L/K$ is $[L:K] = \dim_K L$. We say $L/K$ is finite if $[L:K]<\infty$.

\begin{defi} (1.1)\\
A number field is a finite extension $L/\Q$. Here are two ways to construct number fields:\\
(1) Let $\alpha \in \C$ be an algebraic number. Then $L=\Q(\alpha)$ is a number field;\\
(2) Let $K$ be a number field, and let $f(X) \in K[X]$ be an irreducible polynomial. Then $L=K[X]/(f(X))$ is a number field.\\
(Recall Tower Law: $[L:Q] = [L:K][K:Q] < \infty$).
\end{defi}

\begin{defi} (1.2)\\
(1) Let $L/K$ be a field extension. Then we say $\alpha \in L$ is algebraic over $K$ if there exists a monic $f(X) \in K[X]$ such that $f(\alpha) = 0$;\\
(2) Let $L/\Q$ be a field extension. Then we say $\alpha \in L$ is an algebraic integer if there exists a monic $f(X) \in Z[X]$ such that $f(\alpha) = 0$.
\end{defi}

\begin{defi} (1.3)\\
Let $L/K$ be a field extension, and let $\alpha \in L$ be algebraic over $K$. We call the minimal polynomial of $\alpha$ over $K$ the monic polynomial $f_\alpha(X) \in K[X]$ of least degree such that $f_\alpha(\alpha)=0$.
\end{defi}

We recall why $f_\alpha(X)$ is well-defined: there exists some monic $f(X) \in K[X]$ with $f(\alpha)=0$ as $\alpha$ is algebraic. If $f_\alpha(\alpha),f'_\alpha(\alpha) \in K[X]$ both satisfy the definition of minimal polynomial, then we apply the polynomial division algorithm to write $$f_\alpha(X) = p(X) f'_\alpha(X) + r(X)$$ where $p(X),r(X) \in K[X]$, and $\deg r < \deg f'_\alpha$. Evaluate at $X=\alpha$, we have $0=f_\alpha(\alpha) = p(\alpha)f'_\alpha(\alpha)+r(\alpha)=r(\alpha)$. By minimality of $\deg f'_\alpha$, we must have $r=0$. Then $\deg f_\alpha = \deg f'_\alpha$, and $f_\alpha(X),f'(\alpha)$ are both monic, i.e. $p(X) = 1$ and $f_\alpha(X) = f'_\alpha(X)$.

\begin{lemma}(1.4)\\
Let $L/\Q$ be a field extension, and let $\alpha \in L$ be an algebraic integer. Then:\\
(1) The minimal polynomial $f_\alpha(X)$ of $\alpha$ over $\Q$ lies in $\Z[X]$;\\
(2) If $g(X) \in \Z[X]$ satisfies $g(\alpha) = 0$, then there exists $q(X) \in \Z[X]$ such that $g(X) = f_\alpha(X) q(X)$;\\
(3) The kernel of the ring homomorphism $\Z[X] \to L$ by $f(X) \to f(\alpha)$ equals $(f_\alpha(X))$, the ideal generated by $f_\alpha(X)$.
\begin{proof}
(1) Recall that if $f(X) = a_n X^n + ... + a_0 \in \Z[X]$, then we define from GRM, the content $c(f) = \gcd(a_n,...,a_0)$. Recall Gauss' Lemma: If $f(X), g(X) \in \Z[X]$, then $c(fg) = c(f)c(g)$. Since $\alpha \in L$ is an algebraic integer, there exists monic $f(X) \in \Z[X]$ such that $f(\alpha) = 0$, i.e. $c(f) = 1$. Apply polynomial division in $\Q[X]$ to get $f(X) = p(X) f_\alpha(X) +r(X)$, where $p(X),r(X) \in \Q[X]$, $\deg r < \deg f_\alpha$. The definition of $f_\alpha(X)$ implies that $r(X) = 0$, hence $f(X) = p(X) f_\alpha(X)$. Now choose integers $n,m \geq 1$ such that $np(X) \in \Z[X]$, $c(np) = 1$, and $mf_\alpha(X) \in \Z[x]$, $c(mf_\alpha) = 1$. Then $nmf(x) = (np(x))(mf_\alpha(x)) \implies c(nmf(x)) = nm = 1$. So $n=m=1$, hence $f_\alpha(x) \in \Z[X]$.\\
(2) Let $g(X) \in \Z[X]$ be such that $g(\alpha) =0 $. WLOG $g(x) \neq 0$ and $c(g) = 1$. Now apply polynomial division to write $g(x) = q(x) f_\alpha(x) + s(x)$ where $q(x),s(x) \in \Q[x]$, $\deg s < \deg f_\alpha$. Again by definition we have $s(x) = 0$. Choose an integer $k \geq 1$ such that $kq(x) \in Z[x]$ and $c(kq) =1$. Then $kg(x) = kq(x) f_\alpha(x) \implies k=c(kg) = c(kq) c(f_\alpha) = 1$. So $k=1$, hence $q(x) \in \Z[x]$.\\
(3) is a reformulation of (2).
\end{proof}
\end{lemma}

Let $L/\Q$ be a field extension. Last time we said $\alpha \in L$ is an algebraic integer if $\exists$ monic polynomial $f(x) \in \Z[x]$ such that $f(\alpha) = 0$. We proved that if $\alpha \in L$ is an algebraic integer and $f_\alpha(x) \in \Q[x]$ is the minimal polynomial of $\alpha$ over $\Q$, then $f_\alpha(x) \in \Z[x]$. However there is a small problem, so we'll prove again.
\begin{proof}
Choose $f(x) \in \Z[x]$ monic with $f(\alpha) = 0$, and write $$f(x) = q(x) f_\alpha(x) + r(x)$$ where $q(x),r(x) \in \Q[x]$, $\deg r < \deg f_\alpha$. Then $r(\alpha) = 0 \implies r(x) = 0$, by minimality of $\deg f_\alpha$. I said that we can find integer $n,m \geq 1$ s.t. $nf\alpha(x) \in \Z[x]$, $c(nf\alpha) = 1$, $mq(x) \in \Z[x]$, $c(mq) = 1$. However we need to explain why do they exist. Note $f_\alpha(x)$ and $q(x)$ are both monic. Choose integers $N,M \geq 1$ such that $Nf_\alpha(x) \in \Z[x]$, $Mq(x) \in \Z[x]$. Then $c(Nf_\alpha) | N$, $c(Mq)|M$ as those are the leading term of the polynomial. Now let $N/c(Nf\alpha) = n \in \Z$, $M/c(Mq) = m \in \Z$. Now $nmf(x) = (nf\alpha(x)) (mq(x))$, so $c(nmf(x)) = nm = 1 \implies n=m=1$.
\end{proof}

\begin{coro} (1.5)\\
If $\alpha \in \Q$, then $\alpha$ is an algebraic integer $\iff$ $\alpha \in \Z$.
\begin{proof}
By lemma 1.4, $\alpha$ is an algebraic integer $\iff f_\alpha(x) \in \Z[x]$. But if $\alpha \in \Q$, then $f_\alpha(x) = x-\alpha$, and the first needs to divide the second polynomial.
\end{proof}
\end{coro}

\begin{notation}
If $L/\Q$ is any field extension, we write $\mathcal{O}_L =\{\alpha \in L | \alpha$ is an algebraic integer$\}$.
\end{notation}

Now we proceed to the first non-trivial result of the course:

\begin{prop} (1.6)\\
If $L/\Q$ is a field extension, $\mathcal{O}_L$ is a ring.
\begin{proof}
Clearly $0,1 \in \mathcal{O}_L$. Now if $\alpha \in \mathcal{O}_L$, then $f_{-\alpha}(x) = (-1)^{\deg f_\alpha} f_\alpha (-x) \implies -\alpha \in \mathcal{O}_L$.\\
The hard part is to show that if $\alpha,\beta \in \mathcal{O}_L$, then $\alpha+\beta \in \mathcal{O}_L$ and $\alpha\beta \in \mathcal{O}_L$.\\
Observe that if $\alpha \in \mathcal{O}_L$, then $\Z[\alpha] \subseteq L$ is a finitely generated $\Z$-module. By definition, $\Z[\alpha]$ is generated by $1,\alpha,\alpha^2,\alpha^3,...$. Let $f_\alpha(x) = x^d + a_1 x^{d-1} + ... + ad$, $a_i \in \Z$. Then $\alpha^d = -(a_1 \alpha^{d-1} + ... + ad)$, so $\alpha^d \in \sum_{i=0}^{d-1} \Z \alpha^i$. By induction, we see that $\alpha^n \in \sum_{i=0}^{d-1} \Z \alpha^i$ for all $n \geq d$. Hence $\Z[\alpha] = \sum_{i=0}^{d-1} \Z \alpha^i$. Now take $\alpha,\beta \in \mathcal{O}_L$ and let $d = \deg f_\alpha$, $e = \deg f_\beta$.\\
By definition, $\Z[\alpha,\beta] =\Z[\alpha][\beta]$ is generated as a $\Z$-module by $\{\alpha^i\beta^j\}_{i,j \in \N}$. The same argument show that in fact this ring is generated as a $\Z$-module by $\{\alpha^i \beta^j\}$ for $0\leq i\leq d-1, 0 \leq j \leq e-1$. So $\Z[\alpha,\beta]$ is finitely generated. From GRM we know the classification of finitely generated $\Z$-modules implies that there's an isomorphism $\Z[\alpha,\beta] \cong \Z^r \oplus T$ for some $r \geq 1$ and finite abelian group $T$. In fact, $T=0$: if $\gamma \in T$, then $|T|\gamma = 0$, by Lagrange's theorem. But $\Z[\alpha,\beta] \subseteq L$, a $\Q$-vector space, so this forces $\gamma = 0$. Now we can therefore fix an isomorphism $\Z[\alpha,\beta] \cong \Z^r$ ($r \geq 1$. There's an endomorphism $m_{\alpha\beta}: \Z[\alpha,\beta] \to \Z[\alpha,\beta]$ by $\gamma \to \alpha\beta\gamma$ (as a $\Z$-module). $m_{\alpha\beta}$ corredponds to an $r \times r$ matrx $A_{\alpha\beta} \in M_{r \times r} (\Z)$.\\
Let $F_{\alpha\beta}(x) = \det (x\cdot 1_r -A_{\alpha\beta}) \in \Z[x]$, a monic polynomial. By the Cayley-Hamilton theorem, $F_{\alpha\beta}(m_{\alpha\beta}) = 0$ as endomorphisms of $\Z[\alpha,\beta]$. Write $F_{\alpha\beta}(x) = x^r + b_1x^{r-1} + ... + b_r$ for $b_i \in \Z$. Thus $m^r_{\alpha\beta} + b_1 m^{r-1}_{\alpha\beta} + ... + b_r \cdot 1_r = 0$ as endomorphisms of $\Z[\alpha,\beta]$.\\
Now the image of $1$ is $(\alpha\beta)^r+b_1(\alpha\beta)^{r-1} + ... + b_r = F_{\alpha\beta}(\alpha\beta) = 0$. So $\alpha\beta \in \mathcal{O}_L$.\\
The argument to show $\alpha+\beta \in \mathcal{O}_L$ is identical, replacing $m_{\alpha\beta}$ by $m_{\alpha+\beta}: \Z[\alpha,\beta] \to \Z[\alpha,\beta]$ by $\gamma \to (\alpha+\beta)\gamma$. The detail is omitted here.
\end{proof}
\end{prop}

We call $\mathcal{O}_L$ the ring of algebraic integers of $L$.

\begin{lemma} (1.7)\\
Let $L/\Q$ be a number field, and let $\alpha \in L$. Then $\exists n \geq 1$ an integer such that $n\alpha \in \mathcal{O}_L$.
\begin{proof}
Let $f(x) \in \Q[x]$ be a monic polynomial such that $f(\alpha) = 0$. Then $\exists n\in \Z, n \geq 1$ such that $g(x) = n^{\deg f} f(x/n) \in \Z[x]$ is monic. But then $g(n\alpha) = n^{\deg f} f(\alpha) = 0$. So $n\alpha \in \mathcal{O}_L$.
\end{proof}
\end{lemma}

\newpage

\section{Complex embeddings}
Let $L$ be a number field.
\begin{defi} (2.1)\\
A \emph{complex embedding} of $L$ is a field homomorphism $\sigma: L \to \C$. Note: in this case, $\sigma$ is injective, and $\sigma|_\Q$ is the usual embedding $\Q \to \C$.
\end{defi}

\begin{prop} (2.2)\\
Let $L/K$ be an extension of number fields, and let $\sigma_0:K \to \C$ be a complex embedding. Then there exist exactly $[L:K]$ embeddings $\sigma:L \to \C$ which extends $\sigma_0$ ($\sigma|_K = \sigma_0$).
\begin{proof}
Induction on $[L:K]$. If $[L:K] = 1$, then $L=K$, so $\sigma_0$ determines $\sigma$.\\
In general, choose $\alpha \in L-K$ and consider $L/K(\alpha)/K$. By the Tower law, $[L:K] = [L:K(\alpha)][K(\alpha):K]$ and $[K(\alpha):K]>1$. By induction, it's enough to show there are exactly $[K(\alpha):K]$ embeddings $\sigma:K(\alpha) \to \C$ extending $\sigma_0$.\\
Let $f_\alpha(x) \in K[x]$ be the minimal polynomial of $\alpha$ over $K$. Observe there's an isomorphism $K[x] / (f_\alpha(x)) \to K(\alpha)$ by sending $x \to \alpha$. To give a complex embedding $\sigma:K(\alpha) \to \C$ extending $\sigma_0$, it's equivalent to give a root $\beta$ of $(\sigma_0 f)(x)$ in $\C$ ($\sigma_0 f(x) \in \C[x]$ means apply $\sigma_0$ to the coefficients of $f(x)$). Dictionary: $\sigma \to \beta = \sigma (\alpha)$. We have $[K(\alpha):K] = \deg f_\alpha = \deg \sigma_0 f_\alpha$. It's enough to show $\sigma_0 f_\alpha$ has distinct roots in $\C$. The polynomial $f_\alpha(x) \in K[x]$ is irreducible, so is prime to its derivative $f'_\alpha(x)$ ($char\ K =0$). So $\alpha$ is separable over $K$.
\end{proof}
\end{prop}

Recall from last lecture, let $L$ be a number field, a complex embedding is a field homomorphism $\sigma:L \to \C$. The number of such embeddings is $[L:\Q$]. If $L = \Q(\alpha)$, and $f_\alpha(x) \in \Q[x]$ is the minimal polynomial, then there is a bijection $\{\sigma:L \to \C \} \leftrightarrow \{$ roots $\beta \in \C$ of $f_\alpha(x)\}$ by sending $\sigma \to \beta = \sigma(alpha)$.

Notation: if $\sigma:L \to \C$ is a complex embedding, then $\bar{\sigma} : L \to \C$ is also a complex embedding, where $\bar{\sigma}(\alpha) = \overline{\sigma(\alpha)}$ (complex conjugation). If $\sigma = \bar{\sigma}$, then $\sigma(L) \subseteq \R$. Otherwise $\sigma \neq \bar{\sigma}$ and $\sigma(L) \not\subseteq \R$.

We write $r$ for the number of complex embedding $\sigma$ such that $\sigma = \bar{\sigma}$, $s$ for the number of pairs of embeddings $\{\sigma,\bar{\sigma}\}$ where $\sigma \neq \bar{\sigma}$. Then $r+2s = [L:\Q]$.

\begin{eg}
Let $d \in \Z$ be square-free, $d \neq 0,1$. Let $\Q(\sqrt{d}) = \Q[x] / (x^2-d)$. If $d>0$, then $r=2,s=0$ (real quadratic field).\\
If $d<0$, then $r=0,s=1$ (imaginary quadratic field).
\end{eg}

\begin{eg}
Let $m \in \Z$ cube-free, $m \neq 0,1,-1$. Let $\Q(\sqrt[3]{m}) =\Q[x]/(x^3-m)$. Then $r=1,s=1$, since $x^3-m$ has one real and two complex roots.
\end{eg}

\begin{defi} (2.3)\\
Let $L/K$ be an extension of number fields, and let $\alpha \in L$. Let $m_\alpha:L\to L$ be the $K$-linear map defined by $m_\alpha(\beta) = \alpha\beta$. Then we define
\begin{equation*}
\begin{aligned}
\tr_{L/K}(\alpha) = \tr m_\alpha \in K\\
N_{L/K}(\alpha) = \det m_\alpha \in K
\end{aligned}
\end{equation*}
the trace and norm of $\alpha$ respectively.
\end{defi}

\begin{lemma} (2.4)\\
If $L/K$ is an extension of number fields and $\alpha \in L$, then
\begin{equation*}
\begin{aligned}
\tr_{L/K}(\alpha) = [L:K(\alpha)] \tr_{K(\alpha)/K}(\alpha)\\
N_{L/K}(\alpha) = N_{K(\alpha)/K} (\alpha)^{[L:K(\alpha)]}
\end{aligned}
\end{equation*}
\begin{proof}
There's an isomorphism $L \cong K(\alpha)^{[L:K(\alpha)]}$ of $K(\alpha)$-vector spaces(?).
\end{proof}
\end{lemma}

\begin{lemma} (2.5)\\
Let $L/K$ be an extension of number fields and let $\alpha \in L$. Let $\sigma_0:K \to \C$ be a complex embedding, and let $\sigma_1,...,\sigma_n:L \to \C$ be the embeddings of $L$ extending $\sigma_0$.\\
Then
\begin{equation*}
\begin{aligned}
\sigma_0(\tr_{L/K} (\alpha)) = \sigma_1(\alpha) + ... + \sigma_n(\alpha)\\
\sigma_0(N_{L/K}(\alpha)) = \sigma_1(\alpha)...\sigma_n(\alpha).
\end{aligned}
\end{equation*}
\begin{proof}
WLOG let $L=K(\alpha)$. Let $f_\alpha(x) \in K[x]$ be the minimal polynomial of $\alpha$ over $K$. Then $$(\sigma_0 f_\alpha)(x) = (x-\sigma_1(\alpha))(x-\sigma_2(\alpha))...(x-\sigma_n(\alpha))$$
If $f(\alpha) = x^n+a_1x^{n-1}+...+a_n$, then $\sigma_0(a_1) = -(\sigma_1(\alpha)+...+\sigma_n(\alpha))$, $\sigma_0(a_n) = (-1)^n \sigma_1(\alpha)...\sigma_n(\alpha)$.\\
Let $g(x) \in K[x]$ be the characteristic polynomial of $m_\alpha$. If $g(x) = x^n+b_1x^{n-1}+...+b_n$, then $b_1 = -\tr m_\alpha = -\tr_{L/K}(\alpha)$, $b_n = (-1)^n \det m_\alpha = (-1)^n N_{L/K}(\alpha)$.By Cayley-Hamilton, $g(m_\alpha) = 0 \implies g(\alpha) = 0 \implies f_\alpha(x) = g(x)$.
\end{proof}
\end{lemma}

\begin{coro} (2.6)\\
If $\alpha \in \mathcal{O}_L$, then $\tr_{L/K}(\alpha)$, $N_{L/K}(\alpha) \in \mathcal{O}_K$.
\begin{proof}
If $\beta \in K$ then $\beta \in \mathcal{O}_K$ $\iff$ $\sigma_0(\beta) \in \mathcal{O}_\C$ (as $\forall f(x) \in \Z[x], f(\beta) = 0 \iff f(\sigma_0(\beta)) = 0$).\\
By the lemma, $\sigma_0 \tr_{L/K} (\alpha) = \sigma_1(\alpha)+...+\sigma_n(\alpha)$. If $\alpha \in \mathcal{O}_L$, then $\sigma_1(\alpha),...,\sigma_n(\alpha) \in \mathcal{O}_\C \implies \sigma_1(\alpha)+...+\sigma_n(\alpha) \in \mathcal{O}_\C$ $\implies \sigma_0 \tr_{L/K}(\alpha) \in \mathcal{O}_\C$ $\implies \tr_{L/K}(\alpha) \in \mathcal{O}_K$.

The same argument works for the norm.
\end{proof}
\end{coro}

\begin{prop} (2.7)\\
Let $d \in \Z$ be squarefree, $d \neq 0,1$, and let $L = \Q(\sqrt{d})$. Then
\begin{equation*}
\begin{aligned}
\mathcal{O}_L = \left\{\begin{array}{ll}
\Z[\sqrt{d}] & d \equiv 2,3 \pmod 4\\
\Z[\frac{1+\sqrt{d}}{2}] & d \equiv 1 \pmod 4
\end{array}
\right.
\end{aligned}
\end{equation*}
\begin{proof}
If $\alpha \in L$, then $\alpha \in \mathcal{O}_L$ if and only if both trace and norm (over $L/\Q$) of $\alpha$ is in $\Z$. Why? Forward direction is the previous corollary; if $\alpha \in L$, then $f(\alpha) = 0$, where $f(x) = (x-\sigma_1(\alpha))(x-\sigma_2(\alpha)) = x^2 - \tr_{L/\Q}(\alpha)x+N_{L/\Q}(\alpha) \in \Q[x]$, where $\sigma_1,\sigma_2$ are complex embeddings of $L$. So backward holds too.

Let $\alpha \in L$. Write $\alpha = \frac{u}{2} + \frac{v}{2}\sqrt{d}$ where $u,v \in \Q$. If $\alpha \in \mathcal{O}_L$, then $\tr_{L/\Q}(\alpha) = u \in \Z$, and $N_{L/\Q}(\alpha) = \frac{1}{4} (u+\sqrt{d} v) (u-\sqrt{d} v) = \frac{1}{4} (u^2-dv^2) \in \Z$ $\implies u^2-dv^2 \in 4\Z$ $\implies dv^2\in\Z$.\\
Write $v=\frac{r}{s}$ where $r,s \in \Z, s \neq 0, (r,s) = 1$. Then we get $dr^2 \in s^2\Z$ $\implies s^2 | dr^2$.\\
If $p$ is a prime and $p|s$ then $p^2|d$. But we assumed $d$ is square-free. So $s=1$, so $v \in \Z$.

We've shown if $\alpha \in \mathcal{O}_L$, then $\alpha = \frac{u}{2} + \frac{v}{2}\sqrt{d}$ where $u,v \in \Z$ and $u^2 \equiv d^2 \pmod 4$.\\

Case 1: $d \equiv 2,3 \pmod 4$. Then $u^2,v^2 \equiv 0,1 \pmod 4$. Considering the congruence $u^2\equiv dv^2 \pmod 4$ shows that both $u,v \in 2\Z$. Hence $\alpha \in \Z[\sqrt{d}] = \{a+b\sqrt{d} | a,b \in \Z\}$, and $\mathcal{O}_L = \Z[\sqrt{d}]$.

Case 2: $d \equiv 1 \pmod 4$. Hence $u^2\equiv v^2 \pmod 4$, so $u \equiv v \pmod 2$. Hence $\mathcal{O}_L \subseteq \{\frac{u}{2}+\frac{v}{2}\sqrt{d} | u,v \in \Z, u \equiv 1 \pmod 2\} =\Z \oplus \Z(\frac{1+\sqrt{d}}{2})$. It remains to show that $\frac{1+\sqrt{d}}{2}$ is an algebraic integer.\\
We have $\tr_{L/\Q}(\frac{1+\sqrt{d}}{2}) = 1$, $N_{L/\Q}(\frac{1+\sqrt{d}}{2}) = \frac{1-d}{4} \in \Z$.
\end{proof}
\end{prop}

Recall that if $R$ is a ring, then a unit in $R$ is an element $u \in R$ such that there exists $v \in R$ such that $uv = 1$.

The set $\R^* = \{u \in R | u $ is a unit$\}$ forms a group under multiplication.

\begin{lemma} (2.8)\\
If $L$ is a number field, then the units in $\mathcal{O}_L$ are $\mathcal{O}_L^* = \{\alpha \in \mathcal{O}_L| N_{L/\Q}(\alpha) = \pm 1\}$.
\begin{proof}
next time.

It's next time now! Let's prove this lemma.\\
$N_{L/\Q}(\alpha\beta)=N_{L/\Q}(\alpha)N_{L/\Q}(\beta)$ for any $\alpha,\beta \in L$.\\
If $\alpha \in \mathcal{O}_L^*$, then $\exists \beta \in \mathcal{O}_L$ such that $\alpha\beta = 1 \implies N_{L/\Q}(\alpha) N_{L/\Q}(\beta) = 1$. Since $N_{L/\Q}(\alpha),N_{L/\Q}(\beta) \in \Z$, we get $N_{L/\Q}(\alpha) \in \{\pm 1\}$.\\
Conversely, suppose $\alpha \in \mathcal{O}_L$ and $N_{L/\Q} (\alpha) = \pm 1$. Then $\alpha^{-1} \in L$. Let $\sigma_1,...,\sigma_n:L \to \C$ be the distinct complex embeddings of $L$. Then
\begin{equation*}
\begin{aligned}
N_{L/\Q}(\alpha) = \sigma_1(\alpha)...\sigma_n(\alpha) = \pm 1\\
\implies \sigma_1(\alpha^{-1}) = \pm \sigma_2(\alpha)...\sigma_n(\alpha) \in \mathcal{O}_\C\\
\implies \alpha^{-1} \in \mathcal{O}_L
\end{aligned}
\end{equation*}
\end{proof}
\end{lemma}

\begin{rem}
We'll prove later in the course that $\mathcal{O}_L^*$ is a finite group $\iff$ either $L=\Q$ or $L$ is an imaginary quadratic field.
\end{rem}

\newpage

\section{Discriminants and integral bases}
Let $L$ be a number field, $n=[L:\Q]$, $\sigma_1,...,\sigma_n:L \to \C$ be distinct complex embeddings.

\begin{defi} (3.1)\\
Let $\alpha_1,...,\alpha_n \in L$. Then their discriminant is $disc(\alpha_1,...,\alpha_n) = \det(D)^2$, where $D=M_{n\times n}(F)$ is $D_{ij} = \sigma_i (\alpha_j)$. Note: this is independent of the choice of ordering of $\sigma_1,...,\sigma_n$ and $\alpha_1,...,\alpha_n$, as that's just permuting the rows or columns, hence changing only possibly signs; but we took a square in the definition.
\end{defi}

\begin{lemma} (3.2)\\
Let $\alpha_1,...,\alpha_n \in L$. Then $disc(\alpha_1,...,\alpha_n) = \det(T)$, where $T \in M_{n \times n}(\Q)$ is $T_{ij} = \tr_{L/\Q}(\alpha_i\alpha_j)$.
\begin{proof}
$T_{ij} = \sum_{k=1}^n \sigma_k (\alpha_i\alpha_j) = \sum_{k=1}^n D_{ki} D_{kj} = (D^T D)_{ij}$.
\end{proof}
\end{lemma}

\begin{coro} (3.3)\\
$disc(\alpha_1,...,\alpha_n) \in \Q$. If $\alpha_1,...,\alpha_n \in \mathcal{O}_L$, then $disc(\alpha_1,...,\alpha_n) \in \Z$.
\begin{proof}
$disc(\alpha_1,...,\alpha_n) = \det(T)$, and entries of $T$ is trace of some elements of $L$ (over $\Q$) so is in the base field $\Q$ (think a bit). So this must be rational. If $\alpha_1,...,\alpha_n \in \mathcal{O}_L$, then $\forall i,j$, $D_{ij} \in \mathcal{O}_\C \implies disc(\alpha_1,...,\alpha_n) \in \mathcal{O}_\C \cap \Q =\Z$.
\end{proof}
\end{coro}

\begin{prop} (3.4)\\
Let $\alpha_1,...,\alpha_n \in L$. Then $disc(\alpha_1,...,\alpha_n) \neq 0 \iff \alpha_1,...,\alpha_n$ form a basis of $L$ as $\Q$-vector space.\\
\begin{proof}
First suppose $\alpha_1,...,\alpha_n$ are linearly dependent. Then the columns of the matrix $D_{ij} = \sigma_i(\alpha_j)$ are linearly depnedent $\implies disc(\alpha_1,...,\alpha_n) = 0$ (determinant is 0).\\
Now suppose $\alpha_1,...,\alpha_n$ are linearly independent. Then $disc(\alpha_1,...,\alpha_n) \neq 0$ $\iff \det (T) \neq 0$ $\iff$ the symmetric bilinear form $\phi:L \times L \to \Q$ by $\phi(\alpha,\beta) = \tr_{L/\Q}(\alpha\beta)$ is non-degenerate, i.e. $\forall \alpha \in L^*, \exists \beta \in L$ such that $\phi(\alpha,\beta) \neq 0$.\\
If $\alpha \in L^*$, then $\phi(\alpha,\alpha^{-1}) = \tr_{L/\Q}(1) = n \neq 0$.
\end{proof}
\end{prop}

\begin{defi} (3.5)\\
We say elements $\alpha_1,...,\alpha_n \in L$ form an \emph{integral basis for $\mathcal{O}_L$}, if:\\
(i) $\alpha_1,...,\alpha_n \in \mathcal{O}_L$;\\
(ii) $\alpha_1,...,\alpha_n$ generate $\mathcal{O}_L$ as a $\Z$-module.
\end{defi}

\begin{lemma} (3.6)\\
If $\alpha_1,...,\alpha_n$ form an integral basis for $\mathcal{O}_L$, then the function
\begin{equation*}
\begin{aligned}
f: \Z^n &\to \mathcal{O}_L\\
(m_1,...,m_n) &\to \sum_{i=1}^n m_i\alpha_i
\end{aligned}
\end{equation*}
is an isomorphism of $\Z$-module.
\begin{proof}
$f$ is a homomorphism, we must show it's bijective. Observe that $\alpha_1,...,\alpha_n$ form a basis of $L$ as $\Q$-vector space. We know that if $\beta \in L$, then $\exists N \in \Z^+$ such that $N\beta \in \mathcal{O}_L$ (I think (1.7)). So we can write $N\beta = \sum_{i=1}^n m_i \alpha_i$ for some $m_1 \in \Z$ $\implies \beta = \sum_{i=1}^n \frac{m_i}{N}\alpha_i$. Hence $\alpha_1,...,\alpha_n$ span $L$, so they form a basis of $L$.\\
If $f(m_1,...,m_n) = 0$, then $\sum_{i=1}^n m_i \alpha_i = 0 \implies (m_1,...,m_n) = (0,...,0)$, as $\alpha_1,...,\alpha_n$ are independent over $\Q$. This shows $f$ is injective. It's surjecitve by definition.
\end{proof}
\end{lemma}

\begin{lemma} (3.7, sandwich lemma)\\
(i) If $H \leq G$ are groups and $G \cong \Z^a$ for some $a \geq 0$, then $H \cong \Z^b$ for some $b \leq a$.\\
(ii) If $K \leq H \leq G$ are groups and $K \cong \Z^a$, $G \cong \Z^a$ for some $a \geq 0$, then $H \cong \Z^a$.\\
(iii) If $H \leq G$ are groups and $H \cong \Z^a$, $G \cong \Z^a$ for some $a \geq 0$, then $G/H$ is finite.
\begin{proof}
(i) $H \leq G$, $G \cong \Z^a$. Then $G/H$ is f.g abelian group. By the classification, there's an isomorphism $G/H \cong \Z^N \oplus A$, $A$ finite abelian group. Choose $p$ prime, $p \not\mid |A|$. Then the map $f:G/H \to G/H$ by $x+H \to px + H$ is injective, so $f':H/pH \to G/pG$ by $x+pH \to x+pG$ is injecitve -- why? If $x \in H, x \in pG$, then $x = py$ for some $y \in G$; then $y+H \in \ker(f) =H$. Hence $x \in pH$. So indeed $f'$ is injective. By the classification, $H \cong \Z^b$. $f'$ injective $\implies |H/pH| \leq |G/pG|$, i.e. $p^b \leq p^a$ so $b\leq a$.\\
(ii) Apply (i) to $K \leq H$ and $H \leq G$ to get $H \cong \Z^b$ where $a\leq b\leq a$.\\
(iii) $H \leq G$, $H \cong \Z^a, G \cong \Z^a$. Again $G/H$ is finitely generated, so by the classification $G/H \cong \Z^N \oplus A$ where $A$ is a finite abelian group.\\
Let $p$ be a prime, $p \not\mid |A|$. same proof as in (i) shows that $f':H/pH \to G/pG$ is injecitve. Since $|H/pH| = |G/pG| = p^a$, $f'$ is a group isomorphism $G/H + pG \cong (\Z/p\Z)^N$. There's a surjective homomorphism $G/pG \to G/H+pG$ which has kernel containing the image of $f'$. Hence $G/pG \to G/H + pG$ is surjective with kernel $G/pG$. This forces $N=0$.
\end{proof}
\end{lemma}

Let $L$ be a number field, $n = [L:\Q]$, $\sigma_1,...,\sigma_n:L \to \C$ be distinct complex embeddings; $\alpha_1,...,\alpha_n \in L$, we defined $disc(\alpha_1,...,\alpha_n) = \det(\sigma_i(\alpha_j))^2$. An alternative notation is $\Delta(\alpha_1,...,\alpha_n)$. We also said $\alpha_1,...,\alpha_n$ form an integral basis for $\mathcal{O}_L$ if they generate $\mathcal{O}_L$ as a $\Z$-module.

\begin{prop} (3.8)\\
There exists an integral basis for $\mathcal{O}_L$.
\begin{proof}
Let $\beta_1,...,\beta_n \in L$ be a basis for $L$ as $\Q$-vector space. WLOG, $\beta_1,...,\beta_n \in \mathcal{O}_L$. Then $\mathcal{O}_L \supset \oplus_{i=1}^n \Z \beta_i$.\\
Recall $\phi:L \times L \to \Q$ by sending $(\alpha,\beta) \to \tr_{L/\Q} (\alpha\beta)$ is a non-degenerate symmetric bilinear form (we showed that last time). Let $\beta_1^*,...,\beta_n^*$ be the dual basis. Then $\tr_{L/\Q(\beta_i \beta_j^*)} = \delta_{ij}$ (why?).\\
If $\alpha \in \mathcal{O}_L$, then we can write $\alpha = \sum_{i=1}^n a_i \beta_i^*$ where $a_i \in \Q$. We know $\alpha\beta_i \in \mathcal{O}_L$, hence $\tr_{L/\Q} (\alpha\beta) \in \Z$. However LHS = $\sum_{j=1}^n \tr_{L/\Q} (a_j \beta_j^* \beta_i) = \sum_{j=1}^n a_j \tr_{L/\Q} (\beta_j^* \beta_i) = a_j$. So $\mathcal{O}_L \subseteq \oplus_{i=1}^n \Z \beta_i^*$. By sandwich lemma there is an isomorphism between $\Z^n$ and $\mathcal{O}_L$.
\end{proof}
\end{prop}

If $\alpha_1,...,\alpha_n$, $\beta_1,...,\beta_n$ are both integral bases for $\mathcal{O}_L$, then there exists $A \in M_{n \times n} (\Z)$ such that $\beta_j = \sum_{i=1}^n A_{ij}\alpha_i$ for each $j=1,...,n$. Moreover, we must have $\det(A) \in \{\pm 1\}$, and $A \in GL_n(\Z)$. Then $disc(\beta_1,...,\beta_n) = \det(D')^2$, where $D'_{ij} = \sigma_i(\beta_j), D_{ij} = \sigma_i(\alpha_j)$. We have $D'_{ij} = \sum_{k=1}^n \sigma_i (A_{kj} \alpha_k) = \sum_{k=1}^n \sigma_i (\alpha_k) A_{kj} = (DA)_{ij}$.

We find $disc(\beta_1,...,\beta_n) = \det(D')^2 = \det(DA)^2 = \det(D)^2 = disc(\alpha_1,...,\alpha_n)$. Therefore we could define:

\begin{defi} (3.9)\\
The discriminant $D_L$ of the number field $L$ is $disc(\alpha_1,...,\alpha_n)$, where $\alpha_1,...,\alpha_n$ is any integral basis for $\mathcal{O}_L$.
\end{defi}

\begin{prop} (3.10)\\
Let $L=\Q(\alpha)$, and let $f(x) \in \Q[x]$ be the minimal polynomial of $\alpha$ over $\Q$. Then
\begin{equation*}
\begin{aligned}
disc (1,\alpha,\alpha^2,...,\alpha^{n-1}) = \prod_{i<j} (\sigma_i(\alpha)-\sigma_j(\alpha))^2 = (-1)^{n(n-1)/2} N_{L/\Q}(f'(\alpha))
\end{aligned}
\end{equation*}
In part II Galois theory, we defined the discrimant of a polynomial, $disc f = \prod_{i<j} (\sigma_i(\alpha) - \sigma_j(\alpha))^2$ where $\alpha_i$'s are the roots of $f$.
\begin{proof}
If $D_{ij} = \sigma_i(\alpha^{j-1})$, $D \in M_{n\times n} (\C)$, then $disc(1,\alpha,...,\alpha^{n-1}) = \det(D)^2$. $D$ is a Vandermonde matrix, so we know $\det(D) = \prod_{i<j} (\sigma_j(\alpha) - \sigma_i(\alpha))$.\\
On the other hand, $N_{L/\Q} (f'(\alpha)) = \prod_{i=1}^n \sigma_i (f'(\alpha)) = \prod_{i=1}^n f'(\sigma_i(\alpha))$.\\
Using $f(x) = \prod_{j=1}^n (x-\sigma_j(\alpha))$, we get RHS = $\prod_{i=1}^n \prod_{j \neq i} (\sigma_i(\alpha) - \sigma_j(\alpha)) = (-1)^{n \choose 2} \prod_{i < j} (\sigma_i(\alpha) - \sigma_j(\alpha))^2$.
\end{proof}
\end{prop}

Note: if $\alpha \in \mathcal{O}_L$ and $\Z[\alpha] = \mathcal{O}_L$, then $1,\alpha,...,\alpha^{n-1}$ is an integral basi for $\mathcal{O}_L$. We can then use proposition to calculate $D_L$.

\begin{eg}
Let $d\in \Z$ square-free, $d \neq 0,1$, $L = \Q(\sqrt{d})$. Then
\begin{equation*}
\begin{aligned}
D_L = \left\{\begin{array}{ll}
4d & d \equiv 2,3 \pmod 4\\
d & d \equiv 1 \pmod 4
\end{array}
\right.
\end{aligned}
\end{equation*}
To see this, if $d \equiv 2,3 \pmod 4$, then $\mathcal{O}_L = \Z[\sqrt{d}]$ (shown previously). Apply proposition to $x^2-d = f(x)$, we get $D_L = disc(1,\sqrt{d}) = -N_{L/\Q}(2\sqrt{d}) = 4d$.\\
On the other hand, if $d \equiv 1 \pmod 4$, then $\mathcal{O}_L = \Z[\frac{1+\sqrt{d}}{2}]$. Apply proposition to the minimal polynomial of this element, $f(x) = x^2-x+\frac{1-d}{4}$, so $f'(x) = 2x-1$, so $f'(\alpha) = \sqrt{d}$. Therefore $D_L = -N_{L/\Q}(\sqrt{d}) = \sqrt{d}$.
\end{eg}

\begin{prop}
If $\alpha_1,...,\alpha_n \in \mathcal{O}_L$ are such that $disc(\alpha_1,...,\alpha_n)$ is a non-zero square-free integer, then $\alpha_1,...,\alpha_n$ form an integral basis for $\mathcal{O}_L$.\\
Note: this is a sufficient condition, but is not necessary (the previous example).\\
\begin{proof}
Let $\beta_1,...,\beta_n$ be an integral basis for $\mathcal{O}_L$. There exists $A \in M_{n \times n} (\Z)$ such that $\alpha_j = \sum_{i=1}^n A_{ij} \beta_i$ $\forall j = 1,...,n$. Then $disc(\alpha_1,...,\alpha_n) = \det(A)^2 disc(\beta_1,...,\beta_n)$ (we proved this in the beginning of lecture: $D'=DA$). In particular, if this is square-free and non-zero, then $\det(A)$ must be $\{\pm 1\}$. So $A \in GL_n(\Z)$. Hence $\alpha_1,...,\alpha_n$ generate $\mathcal{O}_L$ (as they can generate $\beta_i$) and form an integral basis.
\end{proof}
\end{prop}

This could save a lot of calculation if we are lucky.
\begin{eg}
Let $f(x) = x^3-x-1$. Then $disc f = -4a^3 - 27b^2 = -23$. This is square-free! If $L=\Q(\alpha)$, $\alpha$ a root of $f(x)$, then $\mathcal{O}_L = \Z[\alpha]$.
\end{eg}

\begin{defi} (3.12)\\
Let $I \subseteq \mathcal{O}_L$ be a no-zero ideal. Then elements $\alpha_1,...,\alpha_n \in L$ form an integral basis for $I$ if:\\
(i) $\alpha_1,...,\alpha_n \in I$;\\
(ii) $\alpha_1,...,\alpha_n$ generate $I$ as a $\Z$-module.
\end{defi}

\begin{prop} (3.13)\\
Let $I \subseteq \mathcal{O}_L$ be a non-zero ideal. Then there exists an integral basis for $I$.
\begin{defi}
By definition, $I \subseteq \mathcal{O}_L \cong \Z^n$. Let $\alpha_1,...,\alpha_n \in \mathcal{O}_L$ be an integral basis for $\mathcal{O}_L$. Let $\alpha \in I$ be non-zero. Then $(\alpha)\subseteq I$, hence $\oplus_{i=1}^n \Z \alpha \alpha_i \subseteq I \subseteq \mathcal{O}_L$. So by sandwich lemma, there is an isomorphism between $I$ and $\Z^n$ as $\Z$-module. Hence there exists an integral basis for $I$.
\end{defi}
\end{prop}

An interesting consequence of the proof:
\begin{defi} (3.14)\\
If $I \subseteq \mathcal{O}_L$ is a non-zero ideal, then we define its norm
\begin{equation*}
\begin{aligned}
N(I) = [\mathcal{O}_L:I]
\end{aligned}
\end{equation*}
which is finite by the sandwich lemma.
\end{defi}

\begin{defi} (3.15)\\
If $I \subset \mathcal{O}_L$ is a non-zero ideal then we define $disc(I) = disc(\alpha_1,...,\alpha_n)$ where $\alpha_1,...,\alpha_n$ is an integral basis for $I$. (same argument shows $disc(I)$ depends only on $I$).
\end{defi}

\begin{lemma} (3.16)\\
If $I \subseteq \mathcal{O}_L$ is a non-zero ideal, then $disc(I) = disc(\mathcal{O}_L) N(I)^2$.
\begin{proof}
Let $\alpha_1,...,\alpha_n$, $\beta_1,...,\beta_n$ be integral bases for $\mathcal{O}_L$ and $I$ respectively. Then $\exists A \in M_{n \times n} (\Z)$ such that $\beta_j = \sum_{i=1}^n A_{ij} \alpha_i$ $\forall j = 1,...n$, and $disc (\alpha_1,...,\alpha_n) \det(A)^2 = disc(\beta_1,...,\beta_n)$. We must show $\det(A)^2 = [\mathcal{O}_L:I]^2$.

In fact, we'll show if $B \in M_{n \times n} (\Z)$ and $\det(B) \neq 0$, then $|\Z^n / B\Z^n| = |\det(B)|$. This suffices after identify $\mathcal{O}_L \cong \Z^n$.

Recall: $\exists P,Q \in GL_n(\Z)$ such that $PBQ = D = Diag(d_1,...,d_n)$, $d_i \in \Z$ (Smith normal form). Hence we have $\Z^n/B\Z^n \cong \Z^n/D\Z^n \cong \oplus_{i=1}^n \Z/d_i \Z \implies |\Z^n / B\Z^n| = |\Z^n / D\Z^n| = \prod_{i=1}^n |d_i|$.\\
On the other hand, $|\det(B)| = |\det(D)| = \prod_{i=1}^n |d_i|$.
\end{proof}
\end{lemma}

Remember we have $L$ a number field, $n=[L:\Q]$, $\sigma_1,...,\sigma_n:L \to \C$ are distinct complex embeddings of $L$.

\begin{lemma} (3.17)\\
Let $\alpha \in \mathcal{O}_L \setminus\{0\}$. Then $N((\alpha)) = |N_{L/\Q}(\alpha)|$ (Note that's an ideal).
\begin{proof}
Let $\alpha_1,...,\alpha_n$ be an integral basis for $\mathcal{O}_L$. Then $\alpha\alpha_1,...,\alpha\alpha_n$ is an integral basis for $I=(\alpha)$. So
\begin{equation*}
\begin{aligned}
disc(I) &= disc(\alpha\alpha_1,...,\alpha\alpha_n)\\
&= \det(\sigma_i(\alpha\alpha_j))^2\\
&= \det(\sigma_i(\alpha)\sigma_i(\alpha_j))^2\\
&= (\prod_{i=1}^n \sigma_i(\alpha))^2 \det(\sigma_i(\alpha_j))^2\\
&= N_{L/\Q}(\alpha)^2 disc(\mathcal{O}_L)
\end{aligned}
\end{equation*}
And we showed last time that for any non-zero ideal $J \subseteq \mathcal{O}_L$, $disc(J) = N(J)^2 disc(\mathcal{O}_L)$.
\end{proof}
\end{lemma}

Notation: If $\alpha \in \mathcal{L}-\{0\}$, we let $N(\alpha) = N((\alpha)) N(0) = 0$.\\
Then $\forall \alpha,\beta \in \mathcal{O}_L$, $N(\alpha\beta) = N(\alpha)N(\beta)$.

\newpage
\section{Unique factorisation in $\mathcal{O}_L$}
Recall: we say a ring $R$ is a unique factorisation domain (UFD) if\\
(i) $R$ is an integral domain;\\
(ii) if $x \in R$ is non-zero and not a unit, then there exists an expression $x=p_1...p_r$ where $p_i \in R$ are irreducible elements. This expression is unique in the sense that if $x = q_1...q_s$ is another such expression, then $r=s$ and after re-ordering, each $q_i$ is an associate of $p_i$ (i.e. $q_i \in R^* p_i$, where $R^*$ is the field of units).

After 2 years of Cambridge Maths we certainly know $\Z$ is a UFD. However, if $L$ is a number field, $\mathcal{O}_L$ need not be a UFD.

In fact, any non-zero $x \in \mathcal{O}_L$ which is not a unit can be expressed as a product of irreducible elements.

If $x \in \mathcal{O}_L$, then $x$ is a no-zero non-unit $\iff N(x)>1$. Suppose $x \in \mathcal{O}_L$ is a non-zero non-unit which cannot be written as a product of irreducible elements, and with $N(x)$ minimal among elements with this property. Then $x=yz$ with $N(y) >1$, $N(z)>1$, hence $N(y)<N(x)$, $N(z)<N(x)$. By minimality of $N(x)$, both $y,z$ can be written as products of irreducible; contradiction.

\begin{eg}
Consider $L=\Q(\sqrt{-5}$, $\mathcal{O}_L = \Z[\sqrt{-5}]$, and $\mathcal{O}_L^* = \{\pm 1\}$. In $\mathcal{O}_L$ we have $6 = 2 \times 3 = (1+\sqrt{-5})(1-\sqrt{-5})$, and all of the four are irreducibles, and no two are associates (norms). So $\mathcal{O}_L$ is not a UFD (famous example).
\end{eg}

Idea: introduce ideal multiplication in order to reduce elements further.

Recall that if $R$ is a ring and $I,J$ are ideals of $R$, then we define
\begin{equation*}
\begin{aligned}
IJ = \{\sum_{i=1}^k a_ib_i | a_i \in I, b_i \in J\},\\
I+J = \{a+b|a\in I, b \in J\}
\end{aligned}
\end{equation*}
We can define an ideal $I \subsetneq R$ to be irreducible if it does not admit an expression $I=JK$ where $J,K$ are proper ideals of $R$.

Key point: even if $\alpha \in \mathcal{O}_L$ is irreducible, the ideal $(\alpha)$ need not be irreducible. For example in $\Z[\sqrt{-5}]$, we have $(2) = (2,1+\sqrt{-5})^2$, $(3) = (3,1+\sqrt{-5})(3,1-\sqrt{-5})$.

\begin{defi} (4.1)\\
If $R$ is a ring, we say that an ideal $P \subsetneq R$ is prime if $\forall x,y \in R$, $xy \in P$ $\implies x \in P$ or $y \in P$.
\end{defi}

\begin{lemma} (4.2)\\
Let $R$ be a ring, and let $I,J,P \subseteq R$ be ideals, and suppose $P$ is prime and $IJ \subseteq P$. Then $I \subseteq P$ or $J \subseteq P$.
\begin{proof}
WLOG $I \not\subseteq P$. Choose some $x \in I \setminus P$. If $y \in J$, is any element, then $xy \in IJ \subseteq P$. So $y \in P$. So $J \subseteq P$.
\end{proof}
\end{lemma}

From now on, $L$ is a number field.

\begin{lemma} (4.3)\\
Any non-zero prime ideal $P \subseteq \mathcal{O}_L$ is a maximal ideal.
\begin{proof}
Recall: if $R$ is a ring and $I \subsetneq R$ is an ideal, then $I$ is prime $\iff R/I$ is an integral domain, and $I$ is maximal $\iff R/I$ is a field. If you don't remember these statements then I strongly encourage you to review GRM. If $p \subseteq \mathcal{O}_L$ is a non-zero prime ideal, then $\mathcal{O}_L/P$ is a finite integral domain (of cardinality $N(P)$); any such ring is a field, so $P$ is also maximal.
\end{proof}
\end{lemma}

\begin{lemma} (4.4)\\
If $I \subsetneq \mathcal{O}_L$ is a non-zero ideal, then there exist non-zero prime ideals $P_1,...,P_r \subseteq \mathcal{O}_L$ such that $P_1...P_r \subseteq I$.
\begin{proof}
For contradiction, let $I \subsetneq \mathcal{O}_L$ be an ideal whicih does not have this property, and such that $N(I)$ is minimal among ideals not having this property. Then $I$ is not prime, so there exist elements $x,y \in \mathcal{O}_L$ such that $xy \in I$ but $x \not\in I$, $y \not\in I$. But then it follows that $I \subsetneq I+(x)$ and $I \subsetneq I+(y)$. So $N(I+(x)), N(I+(y)) < N(I)$. By minimality of $N(I)$, we can find non-zero prime ideals $P_1...P_r \subseteq I+(x)$ and $Q_1...Q_r \subseteq I+(y)$. Then $P_1 ... P_rQ_1...Q_r \subseteq (I+(x))(I+(y)) \subseteq I^2 +xI+yI+(xy) \subseteq I$. Contradiction.
\end{proof}
\end{lemma}

\begin{lemma} (4.5)\\
If $I \subsetneq \mathcal{O}_L$ is a non-zero ideal, then there exists $\gamma \in L\setminus \mathcal{O}_L$ such that $\gamma I \subseteq \mathcal{O}_L$.
\begin{proof}
Let $\alpha \in I \setminus \{0\}$. Let $P_1,...,P_r \subseteq \mathcal{O}_L$ be non-zero prime ideals such that $P_1...P_r \subseteq (\alpha)$. WLOG $r$ is minimal with this property. Let $P$ be a minimal ideal containing $I$. Then $P \supseteq I \supseteq (\alpha) \supseteq P_1...P_r$, hence $P \supset P_i$ for some $i$. After relabelling assume $P \supset P_1$. Since non-zero prime ideals are maximla, we have $P=P_1$. Since $r$ is minimal, we have $P_2...P_r \not\subseteq(\alpha)$. Choose $\beta \in P_2...P_r \setminus (\alpha)$.\\
Claim: the element $\gamma = \beta/\alpha$ has the desired property.\\
If $\gamma \in \mathcal{O}_L$, then $\beta = \alpha\gamma \in (\alpha)$, contradiction;\\
$\gamma I = \frac{\beta}{\alpha} I \subseteq \frac{1}{\alpha} P_2...P_r \cdot I \subseteq \frac{1}{\alpha} P_1P_2...P_r \subseteq \mathcal{O}_L$.
\end{proof}
\end{lemma}

Let $L$ be a number field. Last lecture we proved that if $I \subsetneq \mathcal{O}_L$ is a non-zero ideal, then there exist $\gamma \in L\setminus \mathcal{O}_L$ such that $\gamma I \subseteq \mathcal{O}_L$.

\begin{prop} (4.6)\\
If $I \subseteq \mathcal{O}_L$ is a non-zero ideal, there exists a non-zero ideal $J \subseteq \mathcal{O}_L$, such that $IJ$ is principal.
\begin{proof}
Choose $\alpha \in I \setminus \{0\}$. Define $J = \{\beta \in \mathcal{O}_L | \beta I \subseteq (\alpha)\}$. $J$ is a non-zero ideal, as $\alpha \in J$. We have $IJ \subseteq (\alpha)$. We will show $IJ = (\alpha$.\\
Let $K = \frac{1}{\alpha}IJ \subseteq \mathcal{O}_L$. We will show in fact that $K=\mathcal{O}_L$. Suppose otherwise, that $K \neq \mathcal{O}_L$, then $\exists \gamma \in L \setminus \mathcal{O}_L$ such that $\gamma K \subseteq \mathcal{O}_L$.\\
We have $(\alpha) \subseteq I$, hence $\frac{1}{\alpha} I \supseteq \mathcal{O}_L$, hence $underbrace{\frac{1}{\alpha} IJ}_{K} \supset J$. Hence $\gamma J \subseteq \gamma K \subseteq \mathcal{O}_L$.\\
Another observation is that, we also have $\gamma IJ = \gamma \alpha K \subseteq (\alpha)$.\\
If we have $\beta \in \gamma J$, on one hand $\beta \in \mathcal{O}_L$; on the other hand, $\beta I \subseteq (\alpha)$. So $\beta \in J$, hence $\gamma J \subseteq J$.\\
Recall that $J$ admits an integral basis, so ther's an isomorphism $J \cong \Z^n$. If $A \in M_{n \times n} (\Z)$ is the matrix representing multiplication by $\gamma$, and if $f(x) \in \Z[x]$ is the characteristic polynomial of $A$, then $f(\gamma) = 0$.\\
Hence $\gamma \in \mathcal{O}_L$. Contradiction. So $K = \mathcal{O}_L$.
\end{proof}
\end{prop}

\begin{coro} (4.7)\\
If $I,J,K \subseteq \mathcal{O}_L$ are non-zero ideals and $IJ = IK$, then $J=K$.
\begin{proof}
Choose a non-zero ideal $A \subseteq \mathcal{O}_L$ such that $AI = (\alpha)$ is principal. Then $AIJ = \alpha J = AIK = \alpha K \implies J=K$.
\end{proof}
\end{coro}

If $I,J \subseteq \mathcal{O}_L$ are non-zero ideals, say $I$ divides $J$ (or $I | J$) if there exists an ideal $K \subseteq \mathcal{O}_L$ such that $IK = J$.

\begin{coro} (4.8)\\
If $I,J \subseteq \mathcal{O}_L$ are non-zero ideals, then $I|J \iff I \supseteq J$.
\begin{proof}
If $IK=J$, then $J \subseteq I$.\\
Suppose instead that $I \supseteq J$. Choose a non-zero ideal $A \mathcal{O}_L$ such that $AI = (\alpha)$ is principal (by 4.6). Then $AI =(\alpha) \supseteq AJ$, hence $\mathcal{O}_L \supseteq \frac{1}{\alpha} AJ$. So $K=\frac{1}{\alpha} AJ$ is a non-zero ideal of $\mathcal{O}_L$, and $IK = \frac{1}{\alpha} AIJ = J$.
\end{proof}
\end{coro}

\begin{thm} (4.9)\\
If $I \subseteq \mathcal{O}_L$ is a non-zero ideal, then there exist prime ideals $P_1,...,P_r \subseteq \mathcal{O}_L$ such that $I = P_1P_2...P_r$. Moreover, this expression is unique up to re-ordering of terms.
\begin{proof}
We show existence by contradiction. Suppose $I$ is an ideal which cannot be written as product of primes, and with $N(I)$ minimal subject to this condition. We can find a maximal ideal $P \supset I$. $P$ is also prime. Then $P|I$, so we can write $I=PJ$ for some ideal $J\subseteq \mathcal{O}_L$. Then $J|I$, hence $J \supset I$. If $J=I$, then we get $I=IP$, hence $\mathcal{O}_L = P$ as we can cancel, but that's a contradiction as prime ideals by definition cannot be $\mathcal{O}_L$.\\
Therefore $J \supsetneq I$, hence $N(J) < N(I)$. By minimality, we can write $J$ as $J=P_2...P_r$ where each $P_i\subseteq \mathcal{O}_L$ are prime ideals. Then we have $I=PJ$. Contradiction. This shows existence.

For uniqueness, suppose $P_1,...,P_r$, $Q_1,...,Q_s$ are non-zero prime ideals in $\mathcal{O}_L$ such that $P_1...P_r = Q_1...Q_s$. Then $P_1 | Q_1...Q_r$, so $P_1 \supseteq Q_i$ for some $i=1,...,s$. WLOG $P_1 \supset Q_1$. Since both $P_1,Q_1$ are maximal, $P_1 = Q_1$. Then we cancel to obtain $P_2...P_r = Q_2...Q_s$; continue this to get $r=s$ and $P_i = Q_i$ after re-ordering.
\end{proof}
\end{thm}

\begin{defi} (4.10)\\
The ideal class group $Cl(\mathcal{O}_L) = \{I\subseteq \mathcal{O}_L$ non-zero ideal$\}$. $I \sim J$ if $\exists \alpha \in L^*$ such that $\alpha I = J$.\\
We write $[I]$ for the equivalence class containing $I$.
\end{defi}

\begin{lemma} (4.11)\\
$Cl(\mathcal{O}_L$ is a group under the operation
\begin{equation*}
\begin{aligned}
[I][J] = [IJ]
\end{aligned}
\end{equation*}
with identity $[\mathcal{O}_L]$.
\begin{proof}
If $I,J \subseteq \mathcal{O}_L$ are non-zero ideals and $\alpha,\beta \in L^*$ are such that $\alpha I \subseteq \mathcal{O}_L$ and $\beta J \subseteq \mathcal{O}_L$. Then
\begin{equation*}
\begin{aligned}
(\alpha I)(\beta J) = \alpha\beta IJ
\end{aligned}
\end{equation*}
so ideal multiplication is well-defined on equivalent classes.\\
For any $I \subseteq \mathcal{O}_L$, $\mathcal{O}_L I = I$, so $[\mathcal{O}_L]$ is an identity.\\
We showed that if $I \subseteq \mathcal{O}_L$ is any non-zero ideal, then there exists a non-zero ideal $J \subseteq \mathcal{O}_L$ such that $IJ = (\alpha)$ is principal. Then $[I][J] = [IJ] = [(\alpha)] = [\mathcal{O}_L]$. Hence $[I]^{-1} = [J]$.
\end{proof}
\end{lemma}

\begin{prop} (4.12)\\
The following are equivalent:\\
(i) $\mathcal{O}_L$ is a PID;\\
(ii) $\mathcal{O}_L$ is a UFD;\\
(iii) The ideal class group, $Cl(\mathcal{O}_L)$, is trivial.
\begin{proof}
(i) implies (ii): In IB GRM.\\
(ii) implies (iii): We must show any ideal $I \subseteq \mathcal{O}_L$ is principal. We know that we can write $I=P_1...P_r$ as a product of prime ideals.\\
It's therefore enough to show that every prime ideal of $\mathcal{O}_L$ is principal. Let $P \subseteq \mathcal{O}_L$ be a non-zero prime ideal, let $\alpha \in P$ be non-zero, and let $\alpha=\alpha_1...\alpha_r$ be an expression of $\alpha$ as a product of irreducibles.\\
Recall: if $R$ is a ring, then we say $x \in R$ is prime if $\forall y,z \in R, x|yz \implies x|y$ or $x|z$. Also we learned from GRM that if $R$ is a UFD then irreducible elements of $R$ are prime.\\
We find $P \supset \alpha = (\alpha_1)...(\alpha_r) \implies P | P_1...P_r$ where $P_i = (\alpha_i)$. Since $\alpha_i$ is prime, $P_i$ is a prime ideal. Hence we must have $P=P_i = (\alpha_i)$ for some $i$, and hence $P$ is principal.\\
(iii) implies (i): Let $I \subseteq \mathcal{O}_L$ be a non-zero ideal. Since $Cl(\mathcal{O}_L$ is trivial, we have $[I] = [\mathcal{O}_L]$, so there exists $\alpha \in L^*$ such that $\alpha \mathcal{O}_L = I$. We have $\alpha \cdot 1 = \alpha \in I \subseteq \mathcal{O}_L$, so $\alpha \in \mathcal{O}_L$, hence $I=(\alpha)$ is principal.
\end{proof}
\end{prop}

\begin{lemma} (4.13)\\
If $I,J \subseteq \mathcal{O}_L$ are non-zero ideals, then $N(IJ) = N(I)N(J)$.
\begin{proof}
Example sheet 2.
\end{proof}
\end{lemma}

Example sheet 2 now available!

Last time we learned that, if $L$ is a number field, then we know any non-zero ideal $I \subseteq \mathcal{O}_L$ canbe written uniquely as $I = \prod_{i=1}^r P_i^{e_i}$, wher the $p_i$ are distinct prime ideals, and $e_i \geq 1$. We also defined $Cl(\mathcal{O}_L)$ as the obstruction to $\mathcal{O}_L$ being a UFD.

\newpage
\section{Dedekind's criteion}
If $P\subseteq \mathcal{O}_L$ is a non-zero prime ideal, then there's a unique prime number $p \in \Z_{\geq 0}$ such that $p \in P$. $(p) = \ker(\Z \to \mathcal{O}_L/P)$. Then $P | p \mathcal{O}_L$, and $N(P) = p^f$ for some $f \geq 1$.

\begin{lemma} (5.1)\\
Let $p$ be a prime number, and factor $p\mathcal{O}_L = \prod_{i=1}^r P_i^{e_i}$ where $P_1,...,P_r$ are distinct prime ideals of $\mathcal{O}_L$, $e_i \geq 1$. Define $f_i \geq 1$ by $N(P_i) = p^{f_i}$. Then $\sum_{i=1}^r e_i f_i = [L:\Q]$. In particular, $r \leq [L:\Q]$.
\begin{proof}
Apply norm to get $N(p\mathcal{O}_L) (=p^{[L:\Q]}) = \prod_{i=1}^r N(P_i)^{e_i} (=p^{\prod_{i=1}^r e_if_i})$.
\end{proof}
\end{lemma}

\begin{defi} (5.2)\\
Let $p$ be a prime number, and let $p\mathcal{O}_L = \prod_{i=1}^r P_i^{e_i}$ be the factorization as above.\\
(i) We say $p$ \emph{ramifies} in $L$ if $e_i > 1$ for some $i$. We say $p$ is totally \emph{ramified} if $r=1$ and $e_1 = [L:\Q]$. In other words, $p\mathcal{O}_L = P_i^{[L:\Q]}$.\\
(ii) We say $p$ is \emph{inert} in $L$ if $r=1$ and $e_1 = 1$, i.e. $p\mathcal{O}_L$ is prime.\\
(iii) We say $p$ \emph{splits completely} in $L$ if $r=[L:\Q]$ and $e_i=f_i=1$ for all $i$.

Note that these don't cover all the possible cases.
\end{defi}

\begin{thm} (5.3, Dedekind's criterion)\\
Let $\alpha \in \mathcal{O}_L$ be such that $L=\Q(\alpha)$. Let $f(x) \in \Z[x]$ be its minimal polynomial and let $p$ be a prime such that $p \nmid [\mathcal{O}_L:\Z[\alpha]]$.\\
Let $\bar{f}(x) = f(x) \pmod p$, and factor $\bar{f}(x) = \prod_{i=1}^r \bar{g}_i(x)^{e_i}$ in $F_p[x]$, where $\bar{g}_1(x),...,\bar{g}_r(x) \in F_p[x]$ are distinct monic irreducible polynomials. Let $g_i(x) \in \Z[x]$ be any polynomial with $g_i(x) \pmod p = \bar{g}_i(x)$, and define $Q_i = (p,g_i(\alpha)) \subseteq \mathcal{O}_L$, an ideal of $\mathcal{O}_L$. Let $f_i = \deg \bar{g}_i(x)$.\\
Then $Q_1,...,Q_r$ are distinct prime ideals of $\mathcal{O}_L$, and $p\mathcal{O}_L = \prod_{i=1}^r Q-i^{e_i}$, and $N(Q_i) = p^{f_i}$.
\end{thm}

For example, let's take $L=\Q(\sqrt{-11})$, $p=5$. We see $-11 \equiv 1 \pmod 4$, so $\mathcal{O}_L = \Z [ \frac{1+\sqrt{-11}}{2}]$. Thus $\Z[\sqrt{-11}] \subseteq \mathcal{O}_L$ has index 2 as an additive subgroup. Therefore we can apply Dedekind's criterion to $\alpha = \sqrt{-11}$, with $f(x) = x^2+11$ in order to factorize $5\mathcal{O}_L$. We see $\bar{f}(x) = f(x) \pmod 5 = x^2+1 = (x+2)(x+3)$ in $F_5[x]$. So $t\mathcal{O}_L = PQ$ where $P = (5,\sqrt{-11}+2), Q = (5,\sqrt{-11},3)$, and hence $P,Q$ are the same prime ideals (of $\mathcal{O}_L$). Thus $5\mathcal{O}_L$ splits completely in $\mathcal{Q}\sqrt{-11}$.

\begin{proof} (of 5.3)\\
Recall: if $R$ is a ring and $I \subseteq R$ is an ideal, then there's a bijection between ideals containing $I$ and idealks of $R/I$. 3rd isomorphism theorem gives $R/J \cong (R/I)/(J/I)$. We have $\Z[\alpha] \subseteq \mathcal{O}_L$ of finite index. Let $A = \Z[\alpha]$, $\phi:A \to \mathcal{O}_L$. By reduction mod $p$, we get another ring homomorphism $\bar{\phi}: A/pA \to \mathcal{O}_L / p\mathcal{O}_L$ by $\bar{\phi}(\beta+pA) = \beta + p\mathcal{O}_L$.\\
We claim that this is actually an isomorphism. Both source and targe have cardinality $p^{[L:\Q]}$, so it's enough to show $\bar{\phi}$ is surjective. Let $N=[\mathcal{O}_L:\Z[\alpha]]$. We can find $a,b \in \Z$ such that $aN+bp = 1$. If $\beta \in \mathcal{O}_L$, then $N\beta \in \Z[\alpha]$ (by Lagrance), and $\beta = aN\beta + bp\beta \implies \bar{\phi} (aN\beta + pA) = \beta + p\mathcal{O}_L$. Therefore there is a bijection between ideals in $\mathcal{O}_L$ containing $p$ and ideals of $A/pA$.\\
We have $A=\Z[\alpha] \cong \Z[x] / (f(x))$ by sending $\alpha$ to $x$. Reduction mod $p$ gives an isomorphism $A/pA \cong \Z[x] / (p,f(x)) \cong F_p[x]/ (\bar{f}(x))$. We have $\bar{f}(x) = \prod_{i=1}^r \bar{g}_i(x)^{e_i}$, so there are homomorphisms $F_p[x] / (\bar{f}(x)) \to \F_p[x]/(\bar{g}_i(x))$, given by quotient by the ideal $(\bar{g}_i (x)) \supseteq (\bar{f}(x))$. Define $\Q_i \subseteq \mathcal{O}_L$ to be the ideal containing $p$ such that $\Q_i/(p)$ is the kernel of the ring homomorphism $\mathcal{O}_L/p\mathcal{O}_L \xrightarrow{\bar{\phi}^{-1}} A/pA \xrightarrow{\cong} F_p[x]/(\bar{f}(x)) \to F_p[x]/(\bar{g}_i(x))$. This ring homomorphism is surjective, and its image is a field of cardinality $p^{f_i}$. Hence $\mathcal{O}_L/\Q_i$ is a finite field of cardinality $p^{f_i}$, hence $\Q_i$ is a prime ideal of norm $N(\Q_i) = p^{f_i}$.\\
Also, the $\Q_i$ are distinct, because their images in $\mathcal{O}_L/p\mathcal{O}_L$ are distinct, as if $i \neq j$ then $(\bar{g}_i(x),\bar{g}_j(x))$ is the unit ideal of $F_p[x]$. To show $\Q_i = (p,g_i(\alpha))$, it's enough to show $\Q_i/(p) \subseteq \mathcal{O}_L/p\mathcal{O}_L$ is generated by $\bar{g}_i(\alpha)$. This is equivalent to showing that $\ker(F_p[x] / (\bar{f}(x)) \to F_p[x]/(\bar{g}_i(x)))$ is generated by $\bar{g}_i(x)$. This is true by definition.\\
It remains to show $Q_1^{e_1}...Q_r^{e_r} = p\mathcal{O}_L$. We have
\begin{equation*}
\begin{aligned}
Q_1^{e_1}...Q_r^{e_r} &= (p_1g_1(\alpha))^{e_1}...(p_rg_r(\alpha))^{e_r}\\
&=(p_1g_1(\alpha)^{e_1}) ... (p_1g_r(\alpha)^{e_r})\\
&\leq (p,g_1(\alpha)^{e_1})...(g_r(\alpha)^{e_r}) = (p,f(\alpha))=(p)
\end{aligned}
\end{equation*}
Take norms, $N(LHS) = \prod_{i=1}^r N(Q_i)^{e_i} = p^{\sum_{i=1}^r e_i f_i} =p^{\deg f} = p^{[L:\Q]} = N(p) = N(RHS)$. This forces $Q_1^{e_1}...Q_r^{e_r} = p\mathcal{O}_L$.
\end{proof}

Let $L$ be a number field. Last time we had that if $\alpha \in \mathcal{O}_L$, $\Q(\alpha) = L$, $p \nmid [\mathcal{O}_L:\Z[\alpha]]$. Dedekind's criterion: can factor $p\mathcal{O}_L$ by factoring $f_\alpha(x) \pmod p$.

\begin{prop} (5.4)\\
Let $d$ be a square-free integer, $d \neq 0,1$, $L = \Q(\sqrt{d})$, and let $p$ be a prime number. Then\\
(1) If $p$ is odd, then:\\
$\bullet$ if $p|d$, then $(p) = P^2$, so $p$ ramifies in $L$;\\
$\bullet$ if $p \nmid d$ and $(\frac{d}{p}) = 1$, then $(p) = PQ$, so $p$ splits completely in $L$;\\
$\bullet$ if $p \nmid d$ and $(\frac{d}{p} = -1$, then $(p)$ is prime and $p$ is inert in $L$.\\
(2) If $p=2$, then:\\
$\bullet$ if $d \equiv 2,3 \pmod 4$, then 2 ramifies in $L$;\\
$\bullet$ if $d \equiv 1 \pmod 8$, then 2 splits completely in $L$;\\
$\bullet$ if $d \equiv 5 \pmod 8$, then $2$ is inert in $L$.
\begin{proof}
We just do the case where $p=2$. If $d \equiv 2,3 \pmod 4$, then $\mathcal{O}_L = \Z[\sqrt{d}]$, so by Dedekind's criterion, we must factor $x^2 - d \pmod 2$. But $x^2-d \equiv (x-d)^2 \pmod 2$. If $d \equiv 1 \pmod 4,$ then $\mathcal{O}_L = \Z[\frac{1+\sqrt{d}}{2}]$, so we must factor $x^2+x+\frac{1-d}{4} \pmod 2$. If $d \equiv 1 \pmod 8$, this is $x^2+x = x(x+1) \pmod 2$. If $d \equiv 5 \pmod 8$, this is $x^2+x+1 \pmod 2$ which is irreducible.
\end{proof}
\end{prop}

\newpage
\section{Geometry of numbers}

\begin{defi} (6.1)\\
If $V$ is a finite dimensional $\R$-vector space, then a lattice in $V$ is a subgroup of the form $\Lambda = \oplus_{i=1}^m \Z v_i$, where $v_1,...,v_n$ is a basis of $V$ as $\R$-vector space (for example, $\Z^n \subseteq \R^n$).
\end{defi}

\begin{defi} (6.2)\\
If $V$ is a finite-dimensional inner product space over $\R$, and $\Lambda \subseteq V$ is a lattice, then the covolume of $\Lambda$ is
\begin{equation*}
\begin{aligned}
A(\Lambda) = vol(\{\sum_{i=1}^n t_i v_i | t_i \in [0,1)\})
\end{aligned}
\end{equation*}
where $\Lambda = \oplus_{i=1}^n \Z v_i$.\\
Check: this is independent of the choice of basis $v_1,...,v_n$.
\end{defi}

For today, let's consider only a fixed imaginary quadratic field $L = \Q(\sqrt{d})$ where $d<0$ is a square-free integer. Let's take $\sigma:L \to \Q$ be a complex embedding. Then $\sigma(\mathcal{O}_L)$ is a lattice in $\phi$. If $d \equiv 2,3 \pmod 4$, then $\sigma(\mathcal{O}_L) = \Z \oplus \Z [\sqrt{d}]$; if $d \equiv 1 \pmod 4$ then $\sigma(\mathcal{O}_L) = \Z \oplus \Z (\frac{1+\sqrt{d}}{2})$\\
If $I \leq \mathcal{O}_L$ is a non-zero ideal, then $\sigma(I)$ is a lattice in $\C$.

\begin{lemma} (6.3)\\
If $I \subseteq \mathcal{O}_L$ is a non-zero ideal, then $A(I) = \frac{1}{2} \sqrt{|disc(I)|} = \frac{N(I)}{2} \sqrt{|D_L|}$.
\begin{proof}
Let $\alpha_1,\alpha_2$ be an integral basis for $I$. Then $\sigma(I) = \Z \sigma(\alpha_1) \oplus \Z \sigma (\alpha_2)$. Write $\alpha_1 = x_1+iy_1,\alpha_2=x_2+iy_2$, then $A(\sigma(I)) = |\det{{x_1 \ x_2} \choose {y_1 \ y_2}}|$ (area of a parallelogram).\\
Then
\begin{equation*}
\begin{aligned}
disc(I) = \det
\begin{pmatrix}
x_1+iy_1 & x_2 + iy_2\\
x_1 - iy_1 & x_2 - iy^2
\end{pmatrix}
=
(2i)^2 \det
\begin{pmatrix}
y_1 & y_2\\
x_1 & x_2
\end{pmatrix}
\end{aligned}
\end{equation*}
\end{proof}
\end{lemma}

\begin{thm} (6.4, special case of Minkovski's theorem)\\
Let $\Lambda \subseteq \R^2$ be a lattice, and let $S=D(0,r) \subseteq \R^2$ be the closed disk of radius $r$. Then if $area(S) \geq 4A(\Lambda)$, then $\exists \lambda \in \Lambda - \{0\}$ such that $\lambda \in S$.\\
In particular, there exists $\lambda \in \Lambda - \{0\}$ such that $|\lambda|^2 \leq \frac{4}{\pi} A(\Lambda)$.
\end{thm}

\begin{coro} (6.5)\\
If $I \subseteq \mathcal{O}_L$ is a non-zero ideal, then there exists $\alpha \in I-\{0\}$ s.t. $N(\alpha) \leq c_LN(I)$, where $c_L := \frac{2}{\pi} \sqrt{|D_L|}$.
\begin{proof}
We apply the theorem to $\sigma(I) \subseteq \C$ to get $\lambda \in \sigma(I) - \{0\}$, such that $|\lambda|^2 \leq \frac{4}{\pi} \cdot \frac{N(I)}{2} \sqrt{|D_L|} = c_l N(I)$. If $\alpha \in I$ is such that $\sigma(\alpha) = \lambda$, then $N(\alpha) = \sigma(\alpha) \overline{\sigma(\alpha)} = |\sigma(\alpha)|^2 = |\lambda|^2$.
\end{proof}
\end{coro}

\begin{coro} (6.6)\\
If $[I] \in Cl(\mathcal{O}_L)$, then there exists $J \in [I]$ such that $N(J) \leq c_L$.
\begin{proof}
Choose $k \in [I]^{-1}$ so that $IK$ is principal. Apply the corollary to find $\alpha \in K-\{0\}$, such that $N(\alpha) \leq c_L N(K)$. Then $(\alpha) \subseteq K \implies K | (\alpha) \implies \exists J \subseteq \mathcal{O}_L$ non-zero ideal such that $JK = (\alpha)$. We have $[J] = [K]^{-1} = [I]$, so $J \in [I]$. Also, $N(J) = N(\alpha) / N(K) \leq c_L$.
\end{proof}
\end{coro}

\begin{thm} (6.7)\\
The group $Cl(\mathcal{O}_L)$ is finite. (we'll prove this for any $L$ next time).
\begin{proof}
We've shown every class $[I] \in Cl(\mathcal{O}_L)$ has a representative of norm $\leq c_L$. It therefore suffices to show that $\forall m \in \Z, m \geq 1$, the number of ideals $I \subseteq \mathcal{O}_L$ of norm $N(I) = m$ is finite. If $N(I) = m$, then $[\mathcal{O}_L:I] = m$, so by Lagrance, $m \in I$. Thus $I$ comes from an ideal of the finite ring $\mathcal{O}_L / m\mathcal{O}_L$.
\end{proof}
\end{thm}

Note: we see $CL(\mathcal{O}_L)$ is generated by ideal classes $[P]$, where $P \subseteq \mathcal{O}_L$ is a non-zerp prime ideal of norm $N(P) \leq c_L$. Why? Any class has the form $[I]$, where $N(I) \leq c_L$. If $I = \prod_{i=1}^r p_i^{e_i}$, then $[I] = \pod_{i=1}^r [P_i]^{e_i}$ and $N(I) = \prod_{i=1}^r N(P_i)^{e_i}$, so $N(P_i) \leq N(I) \leq c_L$ for each $i=1,...,r$.

\begin{eg}
Consider $d=-7$. $d \equiv 1\pmod 4$, so $D_L = -d$, $c_l = \frac{2}{\pi} \sqrt{7} < \frac{2}{3} \sqrt{7} < 2$.\\
$Cl(\mathcal{O}_L)$ is generated by ideals of norm $<2$. There are none except $\mathcal{O}_L$, so $Cl(\mathcal{O}_L)$ is the trivial group. Hence $\mathcal{O}_L = \Z[\frac{1+\sqrt{-7}}{2}]$ is a UFD.\\
$d=-5$: $D_L = -4d$, $c_L = \frac{2}{\pi} \sqrt{70} = \frac{4}{\pi} \sqrt{5} < \frac{4}{3}\sqrt{5} < 3$. Hence $Cl(\mathcal{O}_L)$ is generated by prime ideals $P \subseteq \mathcal{O}_L$ of norm $N(P) = 2$. We know by Dedekind's criterion that $2\mathcal{O}_L = P^2$. Hence $Cl(\mathcal{O}_L)$ is generated by $[P]$, and $[P]^2 = [2\mathcal{O}_L]$ is the trivial class.\\
Hence there are two possibilities: if $P$ is principal, then $Cl(\mathcal{O}_L)$ is trivial; if $P$ is not principal, then $Cl(\mathcal{O}_L) \cong \Z/2\Z$. We know $\mathcal{O}_L$ is not a UFD, so we must have $Cl(\mathcal{O}_L) \cong \Z/2\Z$.
\end{eg}

Last time we see that if $L$ is an imaginary quadratic field, then $Cl(\mathcal{O}_L)$ is finite, generated by $[P]$ where $P$ is a prime ideal of norm $N(P) \leq C_L$, where $C_L = \frac{2}{\pi} \sqrt{|D_L|}$.

This time we will show the case of a general number field $L$.

\begin{thm} (6.8, Minkowski's theorem)\\
Let $\Lambda \subseteq \R^n$ be a lattice, and let $E \subseteq \R^n$ be a measurable subset which is conve, and centrally symmetric ($E=-E = \{x \in \R^n | -x \in E\}$). Then:\\
(i) If $vol(E) > 2^n A(\Lambda)$, then $\exists \lambda \in \Lambda \setminus \{0\}$ such that $\lambda \in E$;\\
(i) If $vol(E) \geq 2^n A(\Lambda)$ and $E$ is compact, then $\exists \lambda \in \Lambda \setminus \{0\}$ such that $\lambda \in E$.\\
(we used this last time in the special case $n=2$, $E$=closed disk).
\begin{proof}
Let $\Lambda = \oplus_{i=1}^n \Z v_i$, $P = \{\sum_{i=1}^n t_i v_i | t_i \in [0,1)\}$. Then $vol(P) = A(\Lambda)$, and $\R^n = \sqcup_{\lambda \in \Lambda} (P+\lambda)$.\\
(i) $vol(P) < \frac{1}{2^n} vol(E) = vol(\frac{1}{2} E) = \sum_{\lambda \in \Lambda} vol([\frac{1}{2} E] \cap [\lambda+P]) = \sum_{\lambda \in \Lambda} vol ([\frac{1}{2} E - \lambda] \cap P)$.\\
We claim that there exists $\lambda\neq\mu \in \Lambda$ such that $(\frac{1}{2} E - \lambda) \cap (\frac{1}{2} E - \mu)$ is non-empty. Why? If not, sets $\frac{1}{2} E - \lambda$ are pairwise disjoint, so $vol(P) < \sum_{\lambda \in \Lambda} vol([\frac{1}{2} E-\lambda]\cap P) \leq vol(P)$, contradiction.\\
Hence $\exists z,w \in E$ such that $\frac{z}{2} - \lambda = \frac{w}{2} - \mu$, where $\lambda \neq \mu \in \Lambda$, so $\lambda-\mu = \frac{z}{2} - \frac{w}{2} = \frac{z}{2} + \frac{(-w)}{2}$. Since $E$ is centrally symmetric, $-w \in E$, and $E$ is convex implies that $\frac{z}{2} + \frac{(-w)}{2} \in E$, so $\lambda -\ mu \in (\Lambda \setminus \{0\}) \cap E$.\\
(ii) $E$ compact implies that $E$ is closed and bounded. $vol(E) \geq 2^n A(\Lambda)$ so $\forall m \geq 1$, $vol((1+\frac{1}{m}) E) > 2^n A(\Lambda)$. By (i), $\forall m \in \N \exists s\lambda_m \in (\Lambda \setminus \{0\})\cap((1+\frac{1}{m})E)$, and $(1+\frac{1}{m})E \subseteq 2E$, and $2E \cap \Lambda$ is finite as $2E$ is bounde. By pigeonhole principle we can assume $\exists \lambda \in \Lambda \setminus \{0\}$ such that $\lambda_m = \lambda \forall m \geq 1$. $E$ closed and $\lambda \in (1+\frac{1}{m})E \forall m \geq 1$ $\implies \lambda \in E$. Now let $L$ be a number field. Let $n = [L:\Q]$, let $\tau_1,...,\tau_r:L \to \R$ be the real embeddings of $L$, and let $\sigma_1,\bar{\sigma}_1,...,\sigma_s,\bar{\sigma}_s:L \to \C$ be the remaining distinct complex embeddings of $L$. Then $r+2s = n$.

Define a map $S:l \to \R^r \times \C^s$ by $\alpha \to (\tau_1(\alpha),...,\tau_r(\alpha),\sigma_1(\alpha),...,\sigma_s(\alpha))$. This is a homomorphism of additive groups.
\end{proof}
\end{thm}

\begin{lemma}
If $I \subseteq \mathcal{O}_L$ is a non-zero ideal, then $S(I)$ is a lattice.
\begin{proof}
Let $\alpha_1,...,\alpha_n$ be an integral basis of $I$. Then $S(I) = \oplus_{i=1}^n \Z s(\alpha_i)$ and $\R^r \times \C^3$ has dimension $n$ as $\R$-vector space. So we must show that $S(\alpha_1),...,S(\alpha_n)$ are independent or equivalently that
\begin{equation*}
\begin{aligned}
\det \begin{pmatrix}
\tau_1 (\alpha)1) ... \tau_1(\alpha_n)\\
...\\
\tau_r(\alpha_1) ... \tau_r(\alpha_n)\\
Re \sigma_1 (\alpha_1)...Re \sigma_1(\alpha_n)\\
Im \sigma_1 (\alpha_1)...Im \sigma_1(\alpha_n)\\
...\\
Im \sigma_n (\alpha_1)...Im \sigma_s(\alpha_n)
\end{pmatrix} \neq 0
\end{aligned}
\end{equation*}
Note: for $z \in \C$,
\begin{equation*}
\begin{aligned}
\begin{pmatrix}
z\\
z
\end{pmatrix} =
\begin{pmatrix}
1 & i\\
1 & -i
\end{pmatrix}
\begin{pmatrix}
Re z\\
Im z
\end{pmatrix}
\end{aligned}
\end{equation*}
So this determinant equals
\begin{equation*}
\begin{aligned}
(\frac{1}{-2i})^s  \det \begin{pmatrix}
\tau_1 (\alpha)1) ... \tau_1(\alpha_n)\\
...\\
\tau_r(\alpha_1) ... \tau_r(\alpha_n)\\
\sigma_1 (\alpha_1)...\sigma_1(\alpha_n)\\
...\\
sigma_n (\alpha_1)...\sigma_s(\alpha_n)
\end{pmatrix} \neq 0
\end{aligned}
\end{equation*}
as $disc(I) \neq 0$.
\end{proof}
\end{lemma}

\begin{lemma} (6.10)\\
If $I \subseteq \mathcal{O}_L$ is a non-zero ideal, then
\begin{equation*}
\begin{aligned}
A(S(I)) = \frac{1}{2^s} \sqrt{|disc(I)|} = \frac{N(I)}{2^s} \sqrt{|D_L|}
\end{aligned}
\end{equation*}
\end{lemma}

\begin{prop} (6.11)\\
If $I\subseteq \mathcal{O}_L$ is a non-zero ideal, then there exists $\alpha \in I \setminus \{0\}$ such that $N(\alpha) \leq C_L N(I)$, where $C_L = (\frac{4}{\pi})^s \frac{n!}{n^n} \sqrt{|D_L|}$.\\
Here $C_L$ is called the Minkowski constant of $L$.
\begin{proof}
We apply Minkowski's theorem to the lattice $S(I)$, and region $B_{r,s}(t) = \{(\mathbf{x},\mathbf{z}) \in \R^r \times \C^s | \sum_{i=1}^r |X_i| + 2\sum_{i=1}^s |z_i| \leq t\}$.\\
Note: $B_{r,s} (t)$ is convex, centrally symmetric and compact.\\
If $vol(B_{r,s}(t)) \geq 2^n A(S(I))$, then there exists $\alpha \in I \setminus \{0\}$ such that $S(\alpha) \in B_{r,s}(t)$.\\
We use a tuck with the AM-GM inequality to bound $N(\alpha)$:
\begin{equation*}
\begin{aligned}
N(\alpha)^{1/n} = (\prod_{i=1}^r |\tau_i(\alpha)) \prod_{i=1}^s |\sigma_i(\alpha)|^2)^{1/n} \leq \frac{(\sum_{i=1}^r |\tau_1(\alpha)| + 2\sum_{i=1}^s |\sigma_i(\alpha)|)}{n}
\end{aligned}
\end{equation*}
Hence $N(\alpha) \leq t^n/n^n$. To get optimal bound, choose $t$ so that $vol(B_{r,s}(t)) = 2^n A(S(I))$.\\
Exercise: $vol(B_{r,s}(t)) = 2^r (\frac{\pi}{2})^s t^n/n!$ (Induction on $r$ and $s$).\\
We have
\begin{equation*}
\begin{aligned}
2^r (\pi/2)^s t^n / n! &= 2^n A(S(I)) = 2^{r+s} N(I) \sqrt{|D_L|}\\
\implies t^n &= (4/\pi)^s n! N(I) \sqrt{|D_L|}\\
\implies N(\alpha) &\leq t^n/n^n = C_L N(I)
\end{aligned}
\end{equation*}
\end{proof}
\end{prop}

\begin{coro} (6.12)\\
For any class $[I] \in Cl(\mathcal{O}_L)$, there exists $J \in [I]$ such that $N(J) \leq C_L$.
\end{coro}

\begin{coro} (6.13)\\
The group $Cl(\mathcal{O}_L)$ is finite, generated by $[P]$ where $P$ is a prime ideal of norm $N(P) \leq C_L$.
\end{coro}

These corollaries are deduced from the proposition exactly as in the case $L=\Q(\sqrt{d})$, $d<0$.

\begin{rem}
In practice this bound is very effective. For example consider $f(x) = x^5 -x+1$, this is irreducible mod 5, so over $\Q$. Let $L=\Q(\alpha)$ where $\alpha$ is a root of $f(x)$. In this case $r=1,s=2$, the discriminant $disc f = 2869 = 19 \cdot 151$ is square-free, so $\mathcal{O}_L = \Z[\alpha]$, and $D_L = disc f$, so $c_L =(4/\pi)^2 (5^!/5^5) \sqrt{2869} < 4$. Hence $Cl)\mathcal{O}_L$ is generated by $P$ of norm $N(P) = 2$ or $3$. By Dedekind's criterion, such primes exist iff $f(x)$ has a root in $F_2$ or $F_3$. But there are no such roots. Hence $Cl(\mathcal{O}_L)$ is trivial, hence $\Z[\alpha]$ is a UFD.
\end{rem}

Last time we showed $CL(\mathcal{O}_L)$ is generated by $[P]$ where $[P]$ is a prime ideal of norm $N(P) \leq C_L = (4/\pi)^3 n!/n^n \sqrt{|D_L|}$. For example, if $L=\Q(\sqrt{10})$, $C_L = \frac{1}{2} \sqrt{4 \cdot 10} = \sqrt{10} < 4$. $Cl(\mathcal{O}_L$ is generated by $[P]$ where $N(P) = 2$ or $3$. \\
Dedekind's criterion: $2\mathcal{O}_L = P_2^2$, where $P_2 = (2,\sqrt{10})$. $x^2 -10 \equiv x^2-1 \pmod 3$ so $3\mathcal{O}_L = P_3 P_3'$, where $P_3 = (3,1+\sqrt{10})$. To find relatoins in $Cl(\mathcal{O}_L)$, we can calculate norms, e.g. $N(2+\sqrt{10}) = |4-10| = 6$, so $(2+\sqrt{10}) = P_2 P_3$ or $P_2 P_3'$. In either case we see that $[P_2]$ generates $Cl(\mathcal{O}_L$. So either $Cl(\mathcal{O}_L)$ is trivail, or $Cl(\mathcal{O}_L \cong \Z/2\Z$ with the second case occuring iff So $P_2$ is not principal. $P_2$ is principal $\iff \exists a+b\sqrt{10} \in \mathcal{O}_L$ such that $(a+b\sqrt{10}) = P_2$ $\iff \exists a,b \in \Z$ s.t. $a^2 - 10b^2 = \pm 2$.\\
If $a^2-10b^2 = \pm 2$, then either $2$ or $-2$ is a quadratic residue $\pmod 5$. So in fact $P_2$ is not principal. So $Cl(\mathcal{O}_L) \cong \Z/2\Z$.

Now take $L = \Q(\sqrt{-17})$. $C_l = \frac{4}{\pi}\cdot \frac{1}{2} \sqrt{4\cdot 17} = 4/\pi \sqrt{17} < \frac{4}{3} \sqrt{17} < 6$. So $Cl(\mathcal{O}_L)$ is generated by primes of norm $2,3$ or $5$. Dedekind's criterion: $x^2 + 17 \equiv x^2 + 2 \pmod 5$, so $5 \mathcal{O}_L$ is prime of norm 25. $x^2+17 \equiv x^2-1 \pmod 3$, so $3 \mathcal{O}_L = Q_3 Q'_3$ where $Q_3 = (3,1+\sqrt{-17})$, $Q_3' = (3,1-\sqrt{-17})$. $x^2+17 = (x+1)^2 \pmod 2$, so $2 \mathcal{O}_L = Q_2^2$ where $Q_2 = (2,1+\sqrt{-17})$.\\
Now $N(1+\sqrt{-17}) = 18 = 2 \times 3^2$. Note $1+\sqrt{-17} \in Q_3 \implies Q_3 | (1+\sqrt{-17})$. So we must have either $(1+\sqrt{-17}) = Q_2Q_3Q'_3$, or $(1+\sqrt{-17}) = Q_2 Q_3^2$. To decide between these, we compute
\begin{equation*}
\begin{aligned}
Q_3^2 &= (0,3+3\sqrt{-17},(1+\sqrt{-17})^2)\\
&= (9,3+3\sqrt{-17},-16+2\sqrt{-17})\\
&= (9,3+3\sqrt{-17},2+2\sqrt{-17})\\
&= (9,1+\sqrt{-17})
\end{aligned}
\end{equation*}

We see $1+\sqrt{-17} \in Q_3^2$ so $Q_3^2 | (1+\sqrt{-17})$, hence $(1+\sqrt{-17}) = Q_2Q_3^2$. We see $[Q_3]$ generates $Cl(\mathcal{O}_L)$ and if $Q_2$ is not principal then $Cl(\mathcal{O}_L) \cong \Z/4\Z$. But $Q_2$ is principal iff we can solve $a^2+17b^2 = 2$ with $a,b \in \Z$. This is impossible, so $Cl(\mathcal{O}_L) \cong \Z/4\Z$.

\begin{rem}
Ther are many open questions about ideal class groups even for quadratic fields.\\
Things we know: Number of $Cl(\mathcal{O}_{\Q(\sqrt{d})} \to \infty$ as $d \to -\infty$ through squaree-free integers. There are exactly 9 imaginary quadratic fields with trivial ideal class group (hard).\\
Things we don't know: are there infinitely many real quadratic fields of trivial ideal class group?
\end{rem}

Cohen-Lenstra heuristics: let $p$ be an odd prime, and let $A$ be a finite abelian group of $p$-power order. Then for $d<0$ square-free, $\P(Cl(\mathcal{O}_{\Q(\sqrt{d})}) \cong A) = \frac{\prod_{i=1}^\infty (1-1/p^i)}{\text{Number of }Aut(A)}$.

For $M$ a finite abelian group, $M_p$ is the (unique) $p$-sylow subgroup.

By definition, The above probablity is the ratio between the number of $d<0$ square-free, $Cl(\mathcal{O}_{\Q(\sqrt{d})})_p \cong A$, $|d| < X$ and the number of $d<0$ square-free, $|d|<x$.

\newpage

\section{Dirichlet's unit theorem}
Let $L$ be a number field of degree $n = [L:\Q]$, $\tau_1,...,\tau_r:L \to \R$ are real embeddings, $\sigma_1,...,\sigma_s,\bar{\sigma}_1,...,\bar{\sigma}_s: L \to \C$ are distinct complex embeddings.

\begin{thm} (7.1)\\
There is an isomorphism $\mathcal{O}_L^* \cong \mu_L \times \Z^{r+s-1}$, where $\mu_L \subseteq \mathcal{O}_L^*$ is the finite cyclic group of roots of unity in $\mathcal{O}_L^*$.\\
In fact the proof shows omre: define a map $l:\mathcal{O}_L^* \to \R^{r+s}$: $l(\alpha) = (\log |\tau_1(\alpha)|,...,\log|\tau_r(\alpha)|,2\log |\sigma_1(\alpha)|,...,2\log|\sigma_s(\alpha)|)$, then this is a homomorphism of abelian groups, and $l(\mathcal{O}_L^*)$ is contained in the hyperplane $H =\{\mathbf{x} \in \R^{r+s} | \sum_{i=1}^{r+s} x_i = 0\}\subseteq \R^{r+s}$. This expresses the condition $\alpha \in \mathcal{O}_L^* \implies \log N(\alpha) = \sum_{i=1}^r \log |\tau_i(\alpha)|+2\sum_{i=1}^s |\sigma_i(\alpha)|$.

The proof of the theorem will show $l(\mathcal{O}_L^*)$ is a lattice in $H$.

Example: $\mathcal{O}_L^*$ is finite $\iff r+s = 1$, i.e. $r=1,s=0$ ($L=\Q$), or $r=0,s=1$ ($L=\Q(\sqrt{d})$,$d<0$ square-free). The first case where $\mathcal{O}_L^*$ is infinite is $L=\Q\sqrt{d}), d>0$, square-free. Then $+s-1=1$, so $l(\mathcal{O}_L^*)$ is infinite cyclic. Let's fix $\sigma:\Q(sqrt{d}) \to \R$ to be the real embedding with $\sigma(\sqrt{d}) \geq 0$. $\sigma(\mu_L) \subseteq \R^*$, so $\mu_L = \{\pm 1\}$ in this case. In this case, we can consider the map $l':\mathcal{O}_L^* \to \R$ by $\alpha \to \log |\sigma(\alpha)|$. We know that $l'(\mathcal{O}_L^*) \subseteq \R$ is a lattice, in particular there is a uniquely characterised unit $\alpha \in \mathcal{O}_L^*$ satisfying $\sigma(\alpha)>0$, $\log |\sigma(\alpha)| > 0$ and as small as possible. In other words, $\alpha \in \mathcal{O}_L^*$ is the unit for which $\sigma(\alpha)>1$ and $\sigma(\alpha)$ is minimal with respect to this property. We call $\alpha$ the fundamental unit of $L=\Q(\sqrt{d})$. Then we have $\mathcal{O}_L^* = \{\pm \alpha^n | n \in \Z\}$.
\end{thm}

Example sheet 3 is now online!

Last time we have: if $L$ is a number field, then $\mathcal{O}_L^* \cong \mu_L \times \Z^{r+s-1}$, where $\mu_L$ are roots of unity.

Now suppose $L = \Q(\sqrt{d})$ where $d \in \Z$ is a square free integer, $d>1$. We identify $L$ with a subfield of $\R$, where $\sqrt{d}$ is the positive square root.

We saw that the Dirichlet's unit theorem implies $\exists u \in \mathcal{O}_L^*$ such that $u = \min \{v \in \mathcal{O}_L^* | v>1\}$. $u$ is called the fundamental unit, and $\mathcal{O}_L^* = \{\pm u^n | n \in \Z\}$.

\begin{lemma} (7.2)\\
(1) If $d \equiv 2,3 \pmod 4$ and $v \in \mathcal{O}_L^*$ satisfies $v > 1$, then $v = a+b\sqrt{d}$ where $a \geq b \geq 1$;\\
(2) If $d \equiv 1 \pmod 4$, and $v \in \mathcal{O}_L^*$ satisfies $v>1$, then $v = \frac{1}{2} (a+b\sqrt{d})$ wher $a \geq b \geq 1$.
\begin{proof}
(1) Let $v' = a-b\sqrt{d}$. Then $vv' = a^2 - db^2 = N_{L/\Q}(v) = \pm 1$. So $v>1 \implies |v'| < 1$. Hence $v+v' = 2a > 0$, $v-v' = sb\sqrt{d} > 0$. As $a,b$ are integers, we must have $a \geq 1, b \geq 1$.\\
Also, $(a/b)^2 = d \pm 1/b^2 \geq 1$ as $d \geq 2$.\\
(2) Let $v' = \frac{1}{2} (a-b\sqrt{d}$). Then $vv' = \pm 1$ and $a^2-db^2 = \pm 4$. Then $v+v' = a > 0$, and $v-v' = b\sqrt{d} > 0$. Hence $a \geq 1, b \geq 1$. Also, $(a/b)^2 = d \pm 4/b^2$ as $d \geq 5$ as $d \equiv 1 \pmod 4$.
\end{proof}
\end{lemma}

We can use this to find the fundamental unit $u \in \mathcal{O}_L^*$. First suppose $d \equiv 2,3 \pmod 4$ and let $u = a+b\sqrt{d}$. Let $u^k = a_k + b_k \sqrt{d}$. Then $u^{k+1} = (a_1 + b_1 \sqrt{d})(a_k + b_k \sqrt{d}) = (a_1a_k + db_1b_k) + (b_1a_k + a_1b_k) \sqrt{d}$. Hence $b_{k+1} = b_1a_k + a_1b_k > b_k$.\\
Hence the sequence $b_1,b_2,b_3$ is strictly increasing.

We can therefore characterise $u$ as follows: let $b \in \N$ be the least positive integer such that $db^2+1$ or $db^2-1$ is of the form $a^2$ for some $a \in \N$. Then $u = a+b\sqrt{d}$.\\
Now suppose $d \equiv 1 \pmod 4$, and let $u = \frac{1}{2} (a+b\sqrt{d})$, $a,b \in \Z$. Let $u^k = \frac{1}{2} (a_k + b_k \sqrt{d})$. Then $b_{k+1} = \frac{1}{2} (a_1b_k + b_1 a_k$). Using lemma 7.2, we see $b_{k+1} \geq b_k$. If (??)\\
This is wrong. Let's correct this next time. Sorry!

\begin{eg}
$d=2$.$L = \Q(\sqrt{2})$. $b=1$ works: $2-1=1^2$. So $1+\sqrt{2}$ is a fundamental unit.\\
$d=7$. Try $b=1$: $7 \pm 1$ is not a square; $b=2,$ doesn't work either; $b=3$: $9 \cdot 7 \pm 1 = 8^2$. So $8+3\sqrt{7}$ is a fundamental unit.
\end{eg}

Note: This procedure is not always efficient. For example, the fundamental unit in $\Q(\sqrt{22})$ is $197+42\sqrt{22}$.

There is a more efficient algorithm which uses continued fractions, but it is not discussed in this course (see number theory).

We now prove the unit theorem (this is non-examinable).

We recall the setup: $L$ is a number field, $\tau_1,...,\tau_r: L \to \R$, $\sigma_1,\bar{\sigma}_1,...,\sigma_s,\bar{\sigma}_s:L \to \C$ are real and complex embeddings of $L$ respectively.\\
Last time we defined a map: $l: \mathcal{O}_L^* \to \R^{r+s}$ by $\alpha \to (\log (\tau_1(\alpha)),...,\log(\tau_r (\alpha)),2\log (\sigma_1(\alpha)),...,2\log(\sigma_s(\alpha)))$.\\
The image is contained inside the subspace $H = \{\mathbf{x} \in \R^{r+s}| \sum_{i=1}^{r+s} x_i = 0\}$.

\begin{lemma} (7.3)\\
Let $\alpha \in \mathcal{O}_L \setminus \{0\}$ be such that the above image vector is $(a_1,...,a_{r+s}) \in \R^{r+s}$. Fix an integer $1 \leq k \leq r+s$. Then ther exists $\beta \in \mathcal{O}_L \setminus\{0\}$ such that if $l(\beta) = (b_1,...,b_{r+s}) \in \R^{r+s}$, then $b_i < a_i$ if $i \neq k$. Moreover, $N(\beta) \leq (\frac{2}{\pi})^s \sqrt{|D_L|}$.
\begin{proof}
Let $c_1,...,c_{r+s} \in \R_{>0}$, and let
\begin{equation*}
\begin{aligned}
E=\{(\mathbf{x},\mathbf{z})\in \R^r \times \C^s | |x_1| \leq c_1,...,|x_r| \leq c_r,|z_1|^2 \leq c_{r+1},...,|z_r|^2 \leq c_{r+s}\}
\end{aligned}
\end{equation*}
Then if $vol(E) \geq 2^{r+2s} A(S(\mathcal{O}_L)) = 2^{r+s} \sqrt{|D_L|}$, then $(S:\mathcal{O}_L \to \R^r \times \C^s)$.

There exists $\beta \in \mathcal{O}_L \setminus \{0\}$ such that $S(\beta) \in E$ (by Minkovski's theorem). In particular, $N(\beta) = \prod_{i=1}^r |\tau_1(\beta)| \prod_{i=1}^s |\sigma_i(\beta)|^2 \leq c_1...c_{r+s}$ (by defiintion of $E$).

We choose $c_i$ so that $0 < c_i < e^{a_i}$ if $i \neq k$, and $vol(E) = \pi^s 2^r c_1...c_{r+s} = 2^{r+s} \sqrt{|D_L|}$.

The first property gives $b_i < a_i$ if $i \neq k$, and the second property gives $N(\beta) \geq c_1...c_{r+s} = (\frac{2}{\pi})^s \sqrt{|D_L|}$.
\end{proof}
\end{lemma}

\begin{coro} (7.4)\\
Fix an integer $1 \leq k \leq r+s$. Then there exists $\varepsilon \in \mathcal{O}_L^*$ such that if $l(\varepsilon) = (a_1,...,a_{r+s})$ then $a_i < 0$ if $i \neq k$, and $a_k > 0$.
\begin{proof}
By the lemma, we can find elements $\alpha_1,\alpha_2,...$ of $\mathcal{O}_L\setminus \{0\}$ such that $N(\alpha_1) \leq (\frac{2}{\pi})^s \sqrt{|D_L|}$ $\forall i \in \N$, and if $l(\alpha_i) = (b_{i_1},...,b_{i,r+s})$, then $b_{ij} < b_{i-1,j}$ if $j \neq k$ $\forall i = 2,3,...$. The ideals $(\alpha_i)$ have bounded norm, so are finite in number, so there exist elements $\alpha_N, \alpha_M$ with $(\alpha_N) = (\alpha_M)$. Then the element $\varepsilon = \alpha_N/\alpha_M \in \mathcal{O}_L^*$ has the desired property.
\end{proof}
\end{coro}

We continue with the non-examinable proof of Dirichlet's unit theorem.

We proved propoition: let $\alpha \in \mathcal{O}_L \setminus \{0\}$ be such that $l(\alpha)$ fix $1 \leq k \leq r+s$. Then $\exists \beta \in \mathcal{O}_L \setminus \{0\}$ such that $N(\beta) \leq (\frac{2}{\pi})^s \sqrt{|D_L|}$, and if $l(\beta) = (b_1,...,b_{r+s})$ then $b_i < a_i$ if $i \neq k$.

We deduced Corllary 7.4: fix $1 \leq k \leq r+s$. Then there exists $\varepsilon \in \mathcal{O}_L^*$ such that if $l(\varepsilon) = (a_1,...,a_{r+s})$, then $a_i < 0$ if $i \neq k$.
\begin{proof}
Choose $\alpha \in \mathcal{O}_L \setminus \{0\}$. By the proposition, we can find elements $\alpha_1,...$ such that $N(\alpha_i) \leq (2/\pi)^s \sqrt{|D_L|}$, and if $l(i) = (b_{i1,...,ir+s})$ then $b_{ij} > b_{i+1j}$ if $j \neq k$ for all $i \geq 1$.

We now look at the ideals $(\alpha_1),(\alpha_2),...$. These have norm at most $(2/\pi)\sqrt{|D_L|}$. We know there are only finitely many ideals of $\mathcal{O}_L$ of norm at monst that, so there must exist $N<M$ such that $(\alpha_N) = (\alpha_M)$. Hence $\exists u \in \mathcal{O}_L^*$ such that $\alpha_M = u \alpha_N$. Also, $u = \alpha_M / \alpha_N \implies l(u) = (b_{m1}-b_{N1},...,b_{mr+s}-b_{Nr+s})$. But $N<M$, so $b_{Nj} > b_{Mj}$ if $j \neq k$. So $B_{Mj} - b_{Nj} < 0$ if $j \neq k$.
\end{proof}

\begin{lemma} (7.5)\\
Let $N \geq 1$, and let $A \in M_{N \times N} (\R)$ be such that:\\
$\bullet$ $\sum_{i=1}^N A_{ij} = 0$ for all $j=1,...,N$;\\
$\bullet$ $A_{ij} > 0$ if $i = j$, and $<0$ if $i \neq j$.\\
Then $A$ has rank $N-1$.
\begin{proof}
The rank is at most $N-1$. We show the first $N-1$ rows of $A$ are LI.\\
Suppose there exist $t_i \in \R,i = 1,...,N-1$ not all zero s.t. $\sum_{i=1}^{N-1} t_i A_{ij} = 0$ for each $j = 1,...,N$. WLOG after rescaling ther exists $k$ that $t_k=1$ and $t_i \leq 1$ if $i \neq k$. Then $0 = \sum_{i=1}^{N-1} t_i A_{ik} \geq \sum_{i=1}^{N-1} A_{ik} > \sum_{i=1}^N A_{ik} = 0$, contradiction.
\end{proof}
\end{lemma}

\begin{lemma} (7.6)\\
Fix $B>0$. Let $X_B = \{\alpha \in \mathcal{O}_L | \forall \sigma: L \to \C, |\sigma(\alpha)| \leq B\}$. THen $X_B$ is finite.
\begin{proof}
Recall the map $S:\mathcal{O}_L \to \R^r \times \C^s$. $S(\mathcal{O}_L)$ is a lattice in $\R^r \times \C^s$. $S(X_B)$ is the intersection of the lattice $S(\mathcal{O}_L)$ with a compact subset of $\R^r \times \C^s$. Therefore it must be finite.
\end{proof}
\end{lemma}

\begin{prop} (7.7)\\
$l(\mathcal{O}_L^*)$ is a lattice in $H \leq \R^{r+s}$.
\begin{proof}
We must show there exist units $v_1,...,v_{r+s-1} \in \mathcal{O}_L^*$ such that $l(v_1),...,l(v_{r+s-1})$ span $H$ as an $\R$-vector space and generate $l(\mathcal{O}_L^*)$ as an abelian group.\\
By corollary 7.4, we can find $\varepsilon_1,...,\varepsilon_{r+s} \in \mathcal{O}_L^*$ such htat if $l(\varepsilon_j) = (A_{ij},...,A_{r+sj})$, then $A_{ij} < 0$ if $i \neq j$ and $A_{ij} > 0$ if $i = j$. By lemma 7.5, the matrix $A$ has rank $r+s-1$, so we can find $v_1,...,v_{r+s-1} \in \mathcal{O}_L^*$ such that $l(v_1),...,l(v_{r+s-1})$ span $\mathcal{O}_L^*$ as an $\R$-vector space.\\
Let $\Lambda = \oplus_{i=1}^{r+s-1} \Z l(v_i) \leq H$. This is a lattice in $H$. Then $\Lambda \leq l(\mathcal{O}_L^*)$ and if $u \in \mathcal{O}_L^*$, then $\exists \lambda \in \Lambda$ such that $l(u) - \lambda \in \{\sum_{i=1}^{r+s-1} t_i l(v_i) | t_1,...,t_{r+s-1} \in [0,1]\}=P$. But the set of units $l(P)$ is finite by Lemma 7.6. Hence the quotien $l(\mathcal{O}_L^*) / \Lambda$ is finite. By Lagrange's theorem, $\exists N \in \Z, N > 1$ such that $N l(\mathcal{O}_L^*) \leq \Lambda$. Hence $\Lambda \leq l(\mathcal{O}_L^*) \leq \frac{1}{N}\Lambda$. By the sandwich lemma, $l(\mathcal{O}_L^*)$ is a free abelian group of rank $r+s-1$. In particular, it is a lattice in $H$.
\end{proof}
\end{prop}

Let's now finish the proof of the unit theorem, i.e. show there's an isomorphism $\mathcal{O}_L^* \cong \mu_L \times \Z^{r+s-1}$, where $\mu_L$ is the (finite) group of roots of unity in $\mathcal{O}_L$.

\begin{proof}
We have $\mu_L = \ker l$. If $\xi \in \mu_L$, then $\xi^N = 1$ for some $N\geq 1$, hence $l(\xi^N) = 0 = N l(\xi) \implies l(\xi) = 0$ as $l(\xi \in \R^{r+s}$. If $\alpha \in \mathcal{O}_L^*$ and $l(\alpha) = 0$ then $\forall \sigma: L \to \C$, $|\sigma(\alpha)| = 1$. By lemma 7.6, $\ker l$ is finite. By Lagrange's theorem, it consists of roots of unity.

Choose $v_1,...,v_{r=s-1} \in \mathcal{O}_L^*$ such that $l(v_1),...,l(v_{r+s-1})$ is a $\Z$-basis of $l(\mathcal{O}_L^*)$. Define a map $f:\mu_L \times \Z^{r+s-1} \to \mathcal{O}_L^*$ by $(\xi,n_1,...,n_{r+s-1}) \to \xi v_1^{n_1} ... v_{r+s-1}^{n_{r+s-1}}$.
\end{proof}

Exercise: this is an isomorphism.

Return to the examinable parts:

We now show how to find the fundamental unit in $\Q(\sqrt{d})$, where $\sqrt{d} \in \R_{>0}$ and $d \in \Z$ is a positive square-free integer.

$d>1,d \equiv 1 \pmod 4$:

Recall: the fundamental unit $u \in \mathcal{O}_L^*$ is the least unit $u>1$. We saw last time that if $v=\frac{1}{2}(a+b\sqrt{d}) \in \mathcal{O}_L^*$ is any unit with $v>1$, then $a \geq b \geq 1$.

Let $u^k = \frac{1}{2} (a_k+b_k\sqrt{d})$. Then $b_{k+1} = \frac{1}{2} (a_1 b_k + b_1a_k) \geq \frac{1}{2} (a_1+b_1)b_k \geq b_k$. We see $b_{k+1} \geq b_k$, with equality iff $a_k = b_k$ and $a_1=b_1=1$. Note: if $a_1=b_1 =1$, then $N(u) = |\frac{1-d}{4}| = 1 \implies d=5$. Assume first that $d>5$. Then the sequence $b_1<b_2<b_3<...$ is strictly increasing. The fundamental unit $u$ can therefore be found as following: let $b \in \N$ be the least positive integer such that $db^2 + 4 = a^2$ or $db^2 - 4 = a^2$, where $a \in \N$. Then $\frac{1}{2} (a+b\sqrt{d})$ is the fundamental unit.

Now suppose $d=5$. Then at least $b_1 \leq b_2 \leq ...$ is non-decreasing, and each value $b_i$ can appear at most twice: this is because occurrences correspond to solutions to $b_i^2 d \pm 4 = a_i^2$. We can therefore characterize the fundamental unit $u$ as follows: let $b\ in \N$ be the least positive integer for which $db^2+4 = a^2$ or $db^2-4 = a^2$ for $a,a' \in \N$ (units $\frac{1}{2}(a+b\sqrt{d}) and \frac{1}{2} (a'+b\sqrt{d})$). Recall that the fundamental unit is the least unit with $u>1$. Of these two possibilities, choose the unit with the smaller value of $a$ or $a'$. In this case, $b=1$ gives $d+4=3^2,d-4=1$. So $\frac{1}{2} (1+\sqrt{5})$ is the fundamental unit in this case.

\iffalse
\begin{equation*}
\begin{aligned}

\end{aligned}
\end{equation*}
\fi

\end{document}


\subsection{Query 1}
\label{sec:solution-q1}

The goal of the first query is to determine the updates of the top-3 (i.e., $k=3$) most influential posts in the social network, that is, those that maintain over time a high rate of interaction via comments.
The scoring discipline encapsulate this concept, and is computed as follows.
When a post/comment is created, its initial score is set to $10$ and is then decreased by 1 once a day. The post score is the sum of its value plus the score of every direct and indirect comment rooted in it. Once the score reaches zero, the post expires and does not compete for any ranking.
%
We now describe the operators involved in the first query, whose topology is shown in Figure~\ref{fig:q1-architecture}.

\begin{figure*}[ht]
	\centering
	\includegraphics[width=.75\textwidth]{fig/query-2-architecture}
	\caption{The topology for Query 2.}
	\label{fig:q2-architecture}
\end{figure*}

The \textit{Event Dispatcher} is a centralized operator that reads in parallel posts and comments from several data sources. It converts each entry in a tuple containing only the fields that are strictly necessary to the following operators. Taking advantage of the ascending events timestamp, it can efficiently carry out the interlacing of events.
%
Afterwards, the \textit{Comment Mapper} maps each comment to the related post.
%
Observe that this operator also assigns indirect comments (i.e., comments of a post's comment) to the related post by maintaining in-memory a mapping table that links each post with its comments.
%
Storing this mapping table is challenging because it can potentially grow unlimited, thus saturating the memory.
%
To handle this situation, we exploit the feedback mechanism, which allows us to store only the mapping of not yet expired posts.
%
The \textit{Post Score Updater} is a parallel operator that receives the streams of posts and comments and updates the post score and number of commenters of the post.
To preserve the application integrity with parallelism, the incoming streams are partitioned by the post identifier.
%
To synchronize the parallel instances of this operator, so to properly update the scores, the upstream operator \textit{Comment Mapper} broadcasts a \textit{time sync} message when the time associated to posts and comments moves forward.
%
The \textit{Aggregator} is a parallel rolling counting operator that merges the single score updates to produce the total score for each post.
%
Moreover, this operator is in charge of emitting the \textit{Feedback Stream}.
%
The \textit{Post Rank} defines the partial top-3 ranking of posts handled by its upstreams operators; the ranking is sorted by the post score.
%
Then, the \textit{Post Rank Merger} merges all the partial rankings into a global one and identifies the top-3 posts that trigger the most activity in the social network.
%
The latest two operators optimize the sorting operations by
%
(1) reducing the elements to be sorted thanks to the parallelization;
%
(2) pruning the expired posts; and
%
(3) avoiding to sort the elements that cannot actually generate a ranking update (i.e., elements whose score is less than the lowest in the rankings).
%
%
Finally, the \textit{Post Rank Filter} produces an updated result every time the top-3 most active posts change.


\documentclass{beamer}
\usepackage{csquotes}
\usepackage{tikz}
\usetikzlibrary{arrows,positioning,shapes.geometric, calc}
\usepackage{amsmath}
\usepackage{listings, xcolor}
\usepackage{lmodern}
\usepackage{adjustbox}
\usepackage{booktabs}
\usepackage{colortbl}
\usepackage{caption}
\usepackage{icomma}
\usepackage{bigstrut}
\usepackage{geometry}
\usepackage{subfigure}

\DeclareMathOperator*{\argmin}{argmin}

\usetheme{metropolis}           % Use metropolis theme
\title{Penalized Regression for predicting trait from genotypes}
\date{\today}
\author{Robert M. Porsch}
\institute{Center of Genomic Science}
\begin{document}
\maketitle

\begin{frame}[t]{Introduction}
  Our goal is to use raw genotypes to predict traits and compare it to PGS\@.
  \begin{enumerate}[(i)]
    \item Application of penalized regressions (linear models)
      \begin{itemize}
        \item $L_1$, $L_2$, and $L_0$ norm
      \end{itemize}
    \item Application of non-linear frameworks
      \begin{itemize}
        \item Neural Networks, SVM
      \end{itemize}
  \end{enumerate}
  Identify and estimate epistatic effects.
\end{frame}

\section{Simple Simulation}
\label{sec:simulation}

\begin{frame}[t]{Some simple simulation}
  I simulated some toy data with $n = 10,000$ with $p = 10,000$.
  Effect sizes of 1\% of variables were drawn from a $\mathcal{N}(0, 0.01)$, the rest are set to 0.

  \begin{figure}[htpb]
    \centering
    \includegraphics[width=0.99\linewidth]{./performance.png}
  \end{figure}
  \begin{columns}
    \begin{column}{0.35\textwidth}
      \includegraphics[width=0.99\linewidth]{./joyplot_l0.png}
    \end{column}
    \begin{column}{0.35\textwidth}
      \includegraphics[width=0.99\linewidth]{./joyplot_l1.png}
    \end{column}
    \begin{column}{0.35\textwidth}
      \includegraphics[width=0.99\linewidth]{./percentage_0_l0l1.png}
    \end{column}
  \end{columns}
\end{frame}

\begin{frame}[t]{}
  \begin{figure}[htpb]
    \centering
    \includegraphics[width=0.999\linewidth]{./performance_epochs_l1l0.png} \\
  \end{figure}
  \begin{figure}[htpb]
    \centering
    \includegraphics[width=0.999\linewidth]{./loss_l1l0.png} \\
  \end{figure}
\end{frame}

\section{Combing $L_0$ and $L0_1$}
\label{sec:combing_l_0_and_l0_1_}

\begin{frame}[t]{Connection between the $L_2$-norm and Gaussian prior}
  \small
  Let $\mathcal{D}$ the dataset consisting of $N$ input-output pairs $\{(x_1, y_1), \ldots, (x_N, y_N)\}$.

  Lets also assume that the output is linear related via $\theta$ and that the data is corrupted by some noise $\epsilon\sim N(0, \sigma^2)$.
  Then the Gaussian likelihood is:
  \begin{equation}
    \prod^N_{i=1} \mathcal{N}(y_i|\theta x_i, \sigma^2)
  \end{equation}
  We can then impose a Gaussian prior $\mathcal{N}(\theta|0, \lambda^{-1})$, in which $\lambda$ is a positive scalar.
  Then we have for the likelihood:
  \begin{equation}
    \prod^N_{i=1} \mathcal{N}(y_i|\theta x_i, \sigma^2)\mathcal{N}(\theta|0, \lambda^{-1})
  \end{equation}
  Taking the logarithm of the above and removing unecessary constants we get:
  \begin{equation}
    \sum^N_{i=1} - \frac{1}{\sigma^2} (y_i - \theta x_i)^2 - \lambda \theta^2 + const.
  \end{equation}
\end{frame}

\begin{frame}[t]{Combining $L_0$ and $L_2$}
  \small
  Often it is desirable to shrink paramenters, hence on should combine $L_0$ and other norms.
  Using the $L_2$-norm and under a bernulli gating mechanism:
  \begin{equation}
    \mathbb{E}_{q(z|\pi)} [||\theta||^2_2] = \sum^P_{j=1} \mathbb{E}_{q(z_j|\pi_j)} [z_j^2\tilde{\theta}^2_j] = \sum^P_{j=1} \pi_j \tilde{\theta}^2_j
  \end{equation}
  in which $\pi_j$ is the probability of the gate being open.

  Since the $L_2$ norm is proportional to the negative log density of $\mathcal{N}(0, \sigma^2)$ we can assume that the $\sigma$ for each $\theta$ is controlled by $z$.
  That is if $z=0 \rightarrow \sigma=1$, while if $z > 0 \rightarrow \sigma = z$
  Hence the $L_2$ norm is then (where $\hat{\theta} = \frac{\theta}{\sigma}$)
  \begin{equation}
    \begin{split}
      \mathbb{E}_{q(z|\pi)} [||\hat{\theta}||^2_2] =  & \sum^P_{j=1} (1 - Q_{\hat{s}_j}(0 | \phi_j)) \mathbb{E}_{q(z_j|\pi_j, \hat{s}_j > 0)} [\frac{\tilde{\theta}^2_j z^2_j}{z^2_j}] \\
                                                & \sum^P_{j=1} (1 - Q_{\hat{s}_j}(0 | \phi_j)) \tilde{\theta}^2_j
      \end{split}
  \end{equation}
\end{frame}

\begin{frame}[t]{Results of $L_{0,2}$}

\end{frame}

\begin{frame}[t]{Simulation Framework}
  \begin{center}
  \textbf{Completely blind. You need to ask Tim.}
  \end{center}
  \\
  But here is what I know:
  \begin{itemize}
    \item Linear and non-linear effects present
    \item Currently only using chromosome 10
    \item $R^2$ between $0.04$ and $0.01$
    \item Performance of Lassosum is a correlation of $0.05$
  \end{itemize}
\end{frame}

\begin{frame}[t]{Task processing}
  \begin{figure}[htpb]
    \centering
    \includegraphics[width=0.8\linewidth]{Dask_processing.png}
  \end{figure}
\end{frame}

\begin{frame}[t]{Some Results}
  \begin{figure}[htpb]
    \centering
    \includegraphics[width=0.99\linewidth]{performance.png}
    \caption{Performance}
  \end{figure}
\end{frame}

\begin{frame}[t]{Parameter Space}
  \begin{figure}[htpb]
    \centering
    \includegraphics[width=0.99\linewidth]{null_proportion.png}
    \caption{Proportion of parameters equal or close to 0}
  \end{figure}
  \begin{itemize}
    \item $L_0$ comes very quickly to sparse solution (parameters are not accurate, small effects)
    \item $L_1$ and $L_2$ seem to take a bit longer
  \end{itemize}
\end{frame}

\section{Suplementary for $L_0$-norm}
\label{sec:suplementary}

\begin{frame}[t]{Penalized Regression}
  Let $\mathcal{D}$ the dataset consisting of $N$ input-output pairs $\{(x_1, y_1), \ldots, (x_N, y_N)\}$ and consider the following regularized minimization procedure

  \begin{equation}
    \mathcal{R}(\theta) = \frac{1}{N} ( \sum^N_{i=1} \mathcal{L}(h(x_i; \theta), y_i)) + \lambda\mathcal{P}(\theta)
  \end{equation}

  With $\theta^* = \underset{\theta}{\argmin}\{\mathcal{R}(\theta)\}$.

  $\mathcal{P}(\theta)$ is a penalization function for the parameters $\theta$.
\end{frame}

\begin{frame}[t]{The General Recipe}
  The first step is to reformulate the $L_0$ norm under the parameters $\theta$.
  Hence let,
  \begin{equation}
    \begin{matrix}
      \theta_j = \tilde{\theta_j}z_j, & z_j \in \{0, 1\}, & \tilde{\theta}_j \neq 0
    \end{matrix}
  \end{equation}
  Therefore $z_j$ can be considered as binary gates (parameter has an effect).

  Then we can reformulate the minimization from Eq. 1 by letting $q(z_j|\pi_j) = Bern(\pi_j)$
  \begin{equation}
    \mathcal{R}(\tilde{\theta}, \pi) = \mathbb{E}_{q(z|\pi)} [\frac{1}{N} ( \sum^N_{i=1} \mathcal{L}(h(x_i; \tilde{\theta} \otimes z), y_i)] + \lambda \sum^{p}_{j=1} \pi_j
  \end{equation}
  with  $\tilde{\theta}^*, \pi^* = \underset{{\tilde{\theta}, \pi}}{\argmin} \{\mathcal{R}(\tilde{\theta}, \pi)\}$

  However, the discrete nature of $z$ makes it still difficult to minimize $\pi$.
\end{frame}

\begin{frame}[t]{Hard-sigmoid function}
  \small
  Let s be a continuous random variable with a distribution $q(s)$ with parameter $\phi$.
  Then we can give the gates $z$ a hard sigmoid function with:
  \begin{equation}
    \begin{align*}
      s \sim& q(s|\phi) \\
      z =& \min(1, \max(0, s))
    \end{align*}
  \end{equation}
  Then the probability of the gates being non-zero is
  \begin{equation}
    q(z \neq 0 | \delata) =  1 - Q(s\leq 0 | \phi)
  \end{equation}
  in which $Q(\cdot)$ is the cumulative distribution function of s
  \begin{equation}
    \begin{split}
      \mathcal{R}(\tilde{\theta}, \phi) = \mathbb{E}_{q(s|\phi)} [\frac{1}{N} ( \sum^N_{i=1} \mathcal{L}(h(x_i; \tilde{\theta} \otimes g(s)), y_i)] + \\
      \lambda \sum^{|\theta|}_{j=1} (1 - Q(s_j \leq 0 | \phi_j))
    \end{split}
  \end{equation}
  with $\theta^*, \phi^* = \underset{{\tilde{\theta}, \phi}}{\argmin} \{\mathcal{R}(\tilde{\theta}, \phi)\}$ and $g(\cdot) = \min(1, \max(0, \cdot))$
\end{frame}

\begin{frame}[t]{The Hard Concrete Distribution}
  The literature suggests to use a hard concrete distribution as a smoothing function $q(s)$.
  The parameters of the distribution are $\phi = (\log \alpha, \beta)$ and can be stretched to $(\gamma, \zeta)$ intervals.
  \begin{equation}
    \begin{align*}
      u \sim \mathcal{U}(0,1) \\
      s = Sigmoid((\log u - \log(1-u) + \log\alpha)/\beta) \\
      \bar{s} = s(\zeta - \gamma) + \gamma
    \end{align*}
  \end{equation}
  \begin{figure}[htpb]
    \centering
    \includegraphics[width=0.5\linewidth]{hard_concrete.png}
    \caption{Sample from the Hard Concrete Distribution}\label{fig:hard_concrete}
  \end{figure}
\end{frame}

\begin{frame}[t]{Connection between the $L_2$-norm and Gaussian prior}
  \small
  Let $\mathcal{D}$ the dataset consisting of $N$ input-output pairs $\{(x_1, y_1), \ldots, (x_N, y_N)\}$.

  Lets also assume that the output is linear related via $\theta$ and that the data is corrupted by some noise $\epsilon\sim N(0, \sigma^2)$.
  Then the Gaussian likelihood is:
  \begin{equation}
    \prod^N_{i=1} \mathcal{N}(y_i|\theta x_i, \sigma^2)
  \end{equation}
  We can then impose a Gaussian prior $\mathcal{N}(\theta|0, \lambda^{-1})$, in which $\lambda$ is a positive scalar.
  Then we have for the likelihood:
  \begin{equation}
    \prod^N_{i=1} \mathcal{N}(y_i|\theta x_i, \sigma^2)\mathcal{N}(\theta|0, \lambda^{-1})
  \end{equation}
  Taking the logarithm of the above and removing unecessary constants we get:
  \begin{equation}
    \sum^N_{i=1} - \frac{1}{\sigma^2} (y_i - \theta x_i)^2 - \lambda \theta^2 + const.
  \end{equation}
\end{frame}

\begin{frame}[t]{Combining $L_0$ and $L_2$}
  \small
  Often it is desirable to shrink paramenters, hence on should combine $L_0$ and other norms.

  Using the $L_2$-norm and under a bernulli gating mechanism:
  \begin{equation}
    \mathbb{E}_{q(z|\pi)} [||\theta||^2_2] = \sum^P_{j=1} \mathbb{E}_{q(z_j|\pi_j)} [z_j^2\tilde{\theta}^2_j] = \sum^P_{j=1} \pi_j \tilde{\theta}^2_j
  \end{equation}
  in which $\pi_j$ is the probability of the gate being open.

  Since the $L_2$ norm is proportional to the negative log density of $\mathcal{N}(0, \sigma^2)$ we can assume that the $\sigma$ for each $\theta$ is controlled by $z$.
  That is if $z=0 \rightarrow \sigma=1$, while if $z > 0 \rightarrow \sigma = z$
  Hence the $L_2$ norm is then (where $\hat{\theta} = \frac{\theta}{\sigma}$)
  \begin{equation}
    \begin{split}
      \mathbb{E}_{q(z|\pi)} [||\hat{\theta}||^2_2] =  & \sum^P_{j=1} (1 - Q_{\hat{s}_j}(0 | \phi_j)) \mathbb{E}_{q(z_j|\pi_j, \hat{s}_j > 0)} [\frac{\tilde{\theta}^2_j z^2_j}{z^2_j}] \\
                                                & \sum^P_{j=1} (1 - Q_{\hat{s}_j}(0 | \phi_j)) \tilde{\theta}^2_j
      \end{split}
  \end{equation}

\end{frame}

\begin{frame}[t]{Challenges}
  \textbf{Challenge:} \\
  The plumbing (data engineering)
  \\
  \begin{itemize}
    \item Process of the UKB in parallel efficiently
    \item Large Memory requirements for the UKB (scaling)
    \item Currently using Dask on our cluster
  \end{itemize}
  The optimization/implementation:
  \begin{itemize}
    \item Works fine on test data (1k Genome Project)
    \item Issues with larger data
    \item size of mini-batch size
    \item choosing appropriate learning rates
  \end{itemize}
\end{frame}

\end{document}


\subsection{Agent Function Design}\label{subsec:AgentFNArch}
%\note{This section shows how the agent function is implemented. Diagram shows how DerivedOccupancyGridAgent actually operates.}
Putting together the components outlined so far in this chapter, we end up with an agent function that is shown in Figure \ref{fig:BasicAgentArchitecture}.

\note{Fix this figure}
\begin{figure}[H]
    \centering
    \includegraphics[width = 0.88\linewidth]{Chapters/MultiAgentTargetDetection/Figs/AgentFnArchitecture/BasicAgentFunctionNoCommunication.PNG}
    \caption{The structure of the agent function.}
    \label{fig:BasicAgentArchitecture}
\end{figure}

The diagram outlines how the agent interacts with its environment and abstracts away technical details such as the search status. This is a concrete version of the model-based reflex agent outlined in \cite[p.~51]{AIAMA}. The estimated state update rules are applied as outlined in the previous sections in this chapter. Algorithm \ref{alg:SingleTargetLocalisation} shows the algorithm which describes the agent function.
\note{might be able to bulk this up a bit, come back to it}

\begin{algorithm}[H]
\caption{Single Target Localisation Algorithm}
\label{alg:SingleTargetLocalisation}

\begin{algorithmic}[1]
\renewcommand{\algorithmicrequire}{\textbf{Input:}}
\renewcommand{\algorithmicensure}{\textbf{Output:}}
%Input
\REQUIRE $ \newline initial\_estimated\_state = P(x_{0} | e_{0}, u_0): \quad \text{ The initial distribution of the estimated state}
\newline select\_action: \quad \text{Returns an action given an estimated state}
\newline estimated\_state\_updater: \quad \text{Returns the updated estimated state given the current} \newline \text{ estimated state, a percept and an action}
\newline SPRT: \quad \text{An object calibrated with a specified Type \Romannum{1}, Type \Romannum{2} error which has a } \newline \text{method to implement the SPRT and an attribute which returns the result of the}
\newline
\text{application of the SPRT.}
$
%Output
\ENSURE  $\newline \text{The grid cell containing the source, } x_i \text{ for } i \in \{1, ..., N\}, \text{ or } x_{N+1} \text{, indicating} \newline \text{ the source is not present}$

\hfill\pagebreak

\STATE $estimated\_state$ $\leftarrow$ $initial\_estimated\_state$
\WHILE{not $SPRT.should\_terminate\_search(estimated\_state)$}
\STATE $action \leftarrow select\_action(estimated\_state)$
\STATE $perform\_action(action)$
\STATE $percept \leftarrow get\_percept()$
\STATE $estimated\_state \leftarrow estimated\_state\_updater(estimated\_state, percept, action)$
\ENDWHILE
\RETURN $\argmax_{x_i}{p(x_i | e_{1:t}, u_{1:t}) = \argmax estimated\_state}$
\end{algorithmic}
\end{algorithm}