On page 156 of Milne's Class field theory notes available online [here](http://www.jmilne.org/math/CourseNotes/cft.html), 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. 



Exercise in Milne's CFT notes
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.


