# Ideal Class Group is finite

I'm having trouble understanding the proof given in class that the ideal class group of finite algebraic extension is finite.

Here it is (with my questions):

First, a lemma: Any class of ideals in $Cl(K)$ contains a representative $L$ for which $\mathcal{O}_K \subset L$ and $[L:\mathcal{O}_K]<(\frac{4}{\pi})^{r_2}\cdot\frac{n!}{n^n}\cdot\sqrt{|Disc(\mathcal{O})|}$.
Proof:

I'm already confused because the proof does not start by taking a certain class in $Cl(K)$ so I don't understand its structure.

Take a lattice $M$ in $K$ such that $\mathcal{O}_K=\mathcal{O}(M)$ where $\mathcal{O}(M)=\{x\in K \mid xM \subset M\}$.

Why does such $M$ exist?

W.L.O.G we have $M \subset \mathcal{O}_K$ (I'm fine with this step because we can multiply $M$ by some scalar to make it so). By Minkowski's theorem, we have $|Nm_{K/\mathbb{Q}}|\leq (\frac{4}{\pi})^{r_2}\cdot\frac{n!}{n^n}\cdot\sqrt{|Disc(\mathcal{O}_K)|}\cdot[\mathcal{O}_K:M]$.

Where did the $[\mathcal{O}_K:M]$ factor come from? I don't see it in Minkowski's theorem.

So $\alpha\mathcal{O}_K \subset M$ because $M$ is both a lattice and a ring.

I don't understand this implication.

Take $L=\alpha^{-1}M$.

Since we didn't start with a specific class, I don't understand what this $L$ has to do with anything.

We now have $[\alpha^{-1} M:\mathcal{O}]\leq[\alpha^{-1}M:M]=|Nm_{K/\mathbb{Q}}(\alpha)|\leq (\frac{4}{\pi})^{r_2}\cdot\frac{n!}{n^n}\cdot\sqrt{|Disc(\mathcal{O})|}$.

That ends the proof of the lemma. The proof of the theorem itself is then more clear.

• Are you reading a particular book? – Álvaro Lozano-Robledo Mar 3 '13 at 20:02
• @ÁlvaroLozano-Robledo: No. Unfortunately, the course does not follow a book. – Gils Mar 3 '13 at 20:05
• @ÁlvaroLozano-Robledo: Is there anything I should add to make helping me easier? – Gils Mar 3 '13 at 20:17
• Your question is very detailed, but this is a very common lemma when proving the finiteness of the class group. Read Chapter 5 in Marcus' "Number Fields", in particular, your question is Theorem 37 (and see its corollaries 1 and 2), in page 135. – Álvaro Lozano-Robledo Mar 3 '13 at 20:30
• Quick guesses: 1) $M$ will represent an arbitrary class, 2) any fractional ideal has this property (I thinks that's the way it plays out, but haven't fully checked), 3) $M\le{\cal O}_K$, so $$|Disc(M)|=|Disc({\cal O}_K|\cdot [{\cal O}_K:M]^2,$$ 4) The assumption was that the product of any element of $M$ and any algebraic integer is still in $M$, 5) $L$ and $M$ are in the same class in the class group. – Jyrki Lahtonen Mar 3 '13 at 20:46

Let $M'$ be a fractionnal ideal. Write $M'=aM$ with $M\subseteq O_K$. By Minkowski, there exists $\alpha\in M$ non-zero such that $$\mathrm{Nm}_{K/\mathbb Q}(\alpha) \le C [O_K: M]$$ where $C=(4/\pi)^{r_2}(n!/n^n)\sqrt{|Disc(O_K)|}$. Let $L=\alpha^{-1}M$. Then $$[L:O_K]=[\alpha^{-1}M: O_K]=[M: \alpha O_K]=\mathrm{Nm}_{K/\mathbb Q}(\alpha)/[O_K: M]\le C.$$ By construction, $L=(a\alpha)^{-1}M'$ is a representative of the class of $M'$.