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.