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 $[L:\mathcal{O}_K]<(\frac{4}{\pi})^{r_2}\cdot\frac{n!}{n^n}\cdot\sqrt{|Disc(\mathcal{O})|}$.

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.

  • 1
    $\begingroup$ Are you reading a particular book? $\endgroup$ – Álvaro Lozano-Robledo Mar 3 '13 at 20:02
  • $\begingroup$ @ÁlvaroLozano-Robledo: No. Unfortunately, the course does not follow a book. $\endgroup$ – Gils Mar 3 '13 at 20:05
  • $\begingroup$ @ÁlvaroLozano-Robledo: Is there anything I should add to make helping me easier? $\endgroup$ – Gils Mar 3 '13 at 20:17
  • $\begingroup$ 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. $\endgroup$ – Álvaro Lozano-Robledo Mar 3 '13 at 20:30
  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Mar 3 '13 at 20:46

Usually, the proof goes in the following way.

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 $M'$.

  • $\begingroup$ @user18119 To prove only the finiteness of the class group of a number field K, you don't need Minkowski's bound. The mere existence of the constant C can be shown elementarily, see e.g. Marcus' "Number Fields", chap.5, thm.35 $\endgroup$ – nguyen quang do Apr 22 '16 at 5:57

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