# Ideal class group of $\mathbb{Q} ( \sqrt{-23} )$ is $\cong \mathbb{Z}/3$ by Minkowski bound.

I am trying to show the above result. First, I compute that the Minkowski bound must be strictly lower than 4. Therefore, for every ideal class, there is a representative, so that its norm is at most 3. Now every ideal $a$ can be decomposed into prime ideals $p_1, \ldots, p_k$. Noting that every rational prime $p$ is contained in some $p_i$, so $N(p_i)$ is a power of $p$. Now $N(a) = N(p_1) \ldots N(p_k) \leq 3$. Therefore $p \leq 3$. Next I would try to decompose (2) and (3) and look at their factors to get the result. My problem actually is to decompose them into prime ideals. How do I do that?

• Now I found that $(2) = (2 , 1 + \frac{\sqrt{-23} +1}{2} )(2 , \frac{\sqrt{-23} +1}{2} )$ and $(3) = (3 , 1 + \frac{\sqrt{-23} +1}{2} )(3 , \frac{\sqrt{-23} +1}{2} )$, but I am struggling to get relations. – MPB94 Nov 20 '17 at 10:05
• Maybe these equalities help $N\left(\dfrac{1+\sqrt{-23}}{2}\right) = 6$, $N\left(\dfrac{3+\sqrt{-23}}{2}\right) = 8$. – eduard Nov 20 '17 at 10:23
• The Minkowski bound is $\frac{2}{\pi}\sqrt{23}\sim 3.0531211726846143917749841849168361874$. – Dietrich Burde Nov 20 '17 at 10:23
• @eduard , how do the norms help? I wanted now to looked at the squares of the ideals and see what happens. – MPB94 Nov 20 '17 at 10:42
• Those are norms of principal ideals. The norm $6$ ideal factor as (two) prime ideals. Those prime ideals will be mutually inverse in the class group. – eduard Nov 20 '17 at 10:50

More precisely: in your case the ring of integers is $\mathcal O_K:=\mathcal O_{\mathbb Q(\sqrt{-23})}=\mathbb Z[\alpha]$ with $\alpha=\dfrac{1+\sqrt{-23}}{2}$ and irreducible polynomial $P(X)=X^2 - X + 6$. With this notation $\mathbb Z[X]/(P(X))\overset{\simeq}{\rightarrow}\mathbb Z[\alpha]$.
Let $p$ be prime. We want to decompose $p$ in $\mathcal O_K$ and we look for maximal ideals containing $p$. Those are the maximal ideals of $\mathcal O_K/(p)\simeq \mathbb F_p[X]/(\bar{P}(X))$.