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On page 156 of Milne's Class field theory notes available online here, he claims that the Hilbert class field of $K = \mathbb Q(\sqrt{-6})$ is the splitting field of $x^2+3$ but I don't believe so.

The prime 5 does not ramify in $K$ but does so in $L = \mathbb Q(\sqrt{-6}, \sqrt{-3}) = \mathbb Q(\sqrt{-3}, \sqrt{2})$. To see this, I consider the order $M = \mathbb Z[\alpha]$ where $\alpha = \sqrt{-3}+\sqrt{2}$. Indeed, $M$ is not the ring of integers of $L$ but has index 8 in $\mathcal O_L$. Since $5 \not |[\mathcal O_L:M]$, the splitting of this ideal inside $\mathcal O_L$ would be the same as over $M$. But $\alpha$ has minimal polynomial $x^4+2x^2+25$. Reducing this equation modulo 5 gives a double root, from which we conclude that 5 ramifies in $L$. Since the discriminant of $K$ over $\mathbb Q$ is $-24$, so 5 doesn't ramify there.

I am not able to figure out my mistake in this calculation. I'd be glad if you could spot it. If my argument is correct and $L$ is not the Hilbert class field of $K$, then what is? Thanks.

Edit: From the answers, it seems my calculation of index of the order is wrong. It would be great to have an answer that gives an easy way to see how or why it is wrong, and why $L$ is the class field. Otherwise, I'll accept one of the answers.

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For the future: Please use a tag with two-letter prefix if (and only if) it exists. [The list is to be found right below where you need to enter them in the first place.] Thanks in advance! –  quid Feb 25 '13 at 13:37
From a practical point of view, an easy way to check if you are right or wrong is to appeal to CM theory. In Mathematica, RootApproximant[KleinInvariantJ[Sqrt[-6]]] gives $1399+988\sqrt{2}$. The j-invariant generates the HCF, so Milne is right. –  Gene S. Kopp Feb 27 '13 at 5:36

2 Answers 2

up vote 6 down vote accepted

How did you determine that the index $(\mathcal O_L : M)$ is $8$? It seems to me that it's actually $160$, which is divisible by $5$.

Incidentally, a quick way to see that Milne is correct is to note that the discriminant of $K$ is $-4\cdot 6$ and $6$ is an idoneal number: this means that the Hilbert class field of $K$ coincides with its genus field, which is easily computed to be $K(\sqrt{-3})$. See section 6 of Cox's "Primes of the form $x^2 + ny^2$".

Edit: As requested, here are some more details, as well as a low-tech way of seeing why $L=K(\sqrt{-3})$ is the Hilbert class field of $K$.

First off, an integral basis for $L$ is $\{1, (1+\sqrt{-3})/2, \sqrt2, (\sqrt 2 + \sqrt{-6})/2\}$ (taken from an exercise in Marcus's "Number Fields"), and then a simple computation gives $\text{disc } \mathcal O_L = 12^2$. On the other hand, we find that $\text{disc } M = 1920^2$ and so the equation $\text{disc } M = (\mathcal O_L : M)^2 \text{disc } \mathcal O_L$ gives $(\mathcal O_L : M) = 160$.

Now, as to why $L$ is the Hilbert class field of $K$---well, the HCF has to be a quadratic extension of $K$ (because the class number of $K$ is easily computed to be $2$), so it suffices to show that the finite primes of $K$ are unramified in $L$. For this we can use the relative discriminant of $L/K$: this is an ideal that contains, in particular, $\text{disc } \{1,\sqrt 2\} = 8$ and $\text{disc } \{1, (1+\sqrt{-3})/2\} = -3$, hence contains $1$. That is, the relative discriminant of $L/K$ is the unit ideal and so the extension is unramified.

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I might have made a mistake in computing the index. I'll take a look at Cox's book. –  Abhishek Parab Feb 24 '13 at 2:31
Nice argument, thanks! –  Abhishek Parab Feb 24 '13 at 15:09

I think Milne is right.

If a prime $p$ does not ramify in fields $K_i$, then it does not ramify in their compositum. So you know that 5 can't ramify in $L$, because it visibly doesn't ramify in any of the quadratic fields.

Moreover, PARI/GP tells me that the field generated by $x^4 + 2 x^2 + 25$ has discriminant $576$, but that the polynomial has discriminant $3686400$, so $5$ does divide the index of $M$ in $\mathcal{O}_L$.

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