# 2-class group of a quadratic imaginary extension

Let $p\equiv 5 $ be a prime number, and consider $K=\mathbb{Q}(\sqrt{-p})$.

I would like to check that the $2$-Sylow subgroup of the class group $C_K$ has order $2$ (I'm pretty sure it's true).

Apparently, this can be done using genus theory, but I don't know anything about it or class field theory , really.

I know that $M=\mathbb{Q}(i,\sqrt{p})$ is the genus field of $K$. Knowing this, using the few results I found in the literature, I can show that, if $K'/K$ is an unramified abelian extension of degree $4$, then $K'/\mathbb{Q}$ is Galois, with Galois group the quaternionic group $\mathbb{H}_8$, and I'm stuck...

If someone knows how to prove this, I would be happy to read a proof.

Greg

• Greg,---In the edit to my answer I should have said that a generator of one of the 2 ideals is a square root of r(a+bi), while that of the other is a square root of r(a-bi). Anyway, does my answer solve your follow-up question to your satisfaction? I think I've kept the use of class-field theory to a minimum. – paul Monsky Sep 9 '13 at 13:14

Here's a simple argument using Hilbert's theorem 90. No doubt it's the same as Will's (deleted) argument couched in quadratic form language, and I suppose it's the same as the Frohlich -Taylor argument as well.

First one shows that the 2-torsion in the class group has order 2 and is generated by the prime P of norm 2. For suppose I lies in an order 2 ideal class C. Then the fractional ideal I/I(bar) is principal on some alpha, and the norm of alpha can only be 1. Using Hilbert's theorem 90, one finds an ideal J in the class of C with J=J(bar). Since only 2 and p ramify, and the prime over p is principal, we can assume J is P or (1). Since aa+pbb isn't 2, the result follows.

It remains to show that P isn't equivalent to the square of any ideal. Suppose on the contrary that we have (beta)=P*(Q^2). Taking norms we get aa+pbb=2(non-zero square), We may assume a and b are in Z and are odd. Then the left-hand side is 6 mod 8, a contradiction.

EDIT: Here's an idea for handling your follow-up question in a reasonably elementary way, giving an explicit generator of your ideal lying over p, without heavy class-field-theory artillery. Let K be Q(root(-p)).

Lemma 1__There is no unramified degree 2 extension of K other than K(i).

For let K(root(alpha) be such an extension. Then (alpha) is the square of an ideal. So from what we know, this ideal is principal or P*(principal), and alpha is in the group generated by -1,2 and the squares. The rest is easy.

Lemma 2__There is no degree 4 cyclic unramified extension of K.

For Let D(L) and D(K) be the groups of fractional ideals of L and K. The first inequality of class field theory (whose proof, using the Herbrand quotient is elementary; no ideles required) tells us that the quotient of D(K) by the subgroup generated by the principal fractional ideals and the norms of elements of D(L) has order at least 4. It follows that C/C^4 has order at least 4, where C is the ideal class group of K. But we know that C^4=C^2,and that C/C^2 has order 2.

Corollary__There is no unramified degree 2 extension L of K(i).

For in this case, L/K would be of degree 4 and solvable, and the smallest Galois extension of K containing L would contain a degree 4 unramified abelian extension of K, contradicting Lemmas 1 and 2.

Now write p as a^2+b^2 with b even, and let r be the fundamental unit in Z[(1+root(p))/2]. Then it should be possible to show that the extension of K(i) obtained by adjoining a square root of r(a+bi) is unramified. So by the corollary r(a+bi) has a square root in K(i), and this square root is what you're looking for.

• Of course Hilbert's theorem 90 in this situation is a triviality, and in some sense goes back to Gauss. – paul Monsky Sep 3 '13 at 21:09
• Thanks! Very nice. Now, I'd like to ask another thing, which is related. I'd like to prove that the class group of M=Q(i,p√) has no element of order 2, in an elementary way. I know how to prove that M has no unramified 2-extension; properties of Hilbert class field then implies the result. However, I wonder if there is an elementary way to see this, in the spirit of the previous answer. My ultimate goal is to prove that the ideals lying above p are principal in an elementary way. Since the square of these ideals are principal, I would be done. – GreginGre Sep 4 '13 at 11:13
• Is this true when p=37? I don't think that either root(37) or(6+root(37))(root(37)) is a norm from the ring of integers in Q(i,root(37)). – paul Monsky Sep 4 '13 at 13:09
• The element $\alpha=(-3+4i)+(1-i)\dfrac{1+\sqrt{37}}{2}$ is an algebraic integer with absolute norm $37$. In fact, its norm over $\mathbb{Q}(i)$ is $-6+i$. – GreginGre Sep 4 '13 at 14:12
• Proof of the fact that $M$ has no unramified $2$-extension: let $G$ be the Galois group of the the maximal unramified $2$-extension of $K$. The previous result shows that $G^{ab}$ has order $2$. A classical result of group theory says that $G$ has order $2$ in this case. So the maximal unramified $2$-extension of $K$ has degree $2$ over $K$. So, this is $M$, since $M/K$ is unramified. In particular, $M$ has no unramified $2$-extensions. So $C_M$ is odd. But the square of an ideal $\mathcal{P}$ above $p$ is principal (generated by a prime divisor of $p$ in $\mathbb{Z}[i]$) and we are done. – GreginGre Sep 4 '13 at 14:22

The result is contained in Theorem 41 of Fröhlich-Taylor: Algebraic number theory. See also Theorem 39 in that book.

• Thanks! I had a look to Fröhlich-Taylor . The proof seems quite involded. I thought this would be easier to show, using elementary arguments. At least, the result is true :-) – GreginGre Sep 3 '13 at 19:51