Recently, I’ve been preparing for my advanced topics exam at Purdue and have been going over some algebraic number theory. The question in consideration was to find the Hilbert class field (i.e., the maximal abelian unramified extension) of ${\mathbb Q(\sqrt{-6})}$. The problem boiled down to determining how a prime split in an extension, which reduced to finding the ring of integers of a number field. The usual method (following Kummer) is to find an algebraic integer that generates the ring of integers of the extension over the ring of integers of the base field. But such an integer is not guaranteed to exist. I posted this on math.stackexchange and got an interesting answer in the form of a comment which I now blog.

Theorem: Let ${K}$ and ${L}$ be linearly disjoint number fields, i.e., $\displaystyle K \otimes_{\mathbb Q} L \rightarrow KL$

is an isomorphism. Then, ${D_{KL} | D_K^s D_L^r}$, where ${D_K}$ denotes the discriminant of ${K}$ over ${\mathbb Q}$, etc. and ${r = [K:\mathbb Q]}$ and ${s = [L : \mathbb Q]}$. Moreover, if ${D_K}$ and ${D_L}$ are relatively prime over ${\mathbb Z}$, then ${D_{KL} = D_K^s D_L^r}$ and ${\mathcal O_K \otimes_{\mathbb Z} \mathcal O_L \hookrightarrow \mathcal O_{KL}}$ is an isomorphism. So, the basis of ${\mathcal O_{KL}}$ can be obtained from the bases of ${\mathcal O_K}$ and ${\mathcal O_L}$ in a natural way.

Proof:
Let ${M}$ be the image of $\displaystyle \mathcal O_K \otimes_{\mathbb Z} \mathcal O_L \hookrightarrow \mathcal O_{KL}.$

Then ${M}$ is an order of ${\mathcal O_{KL}}$ (i.e., a subring of the ring of integers). If ${x = ab \in M}$ is the image of a pure tensor ${a \otimes b \in \mathcal O_K \otimes \mathcal O_L}$, then the trace form on ${KL}$ acts on ${x}$ by ${T_{KL}(ab) = T_K(a).T_L(b)}$. (For a proof of this, go to the Galois closure and use the canonical isomorphism ${\text{Gal}(KL/\mathbb Q) \cong \text{Gal}(K/\mathbb Q) \times \text{Gal}(L/\mathbb Q)}$ given by linear disjointness). Thus, ${T_{KL} : M \rightarrow M}$ acts by ${T_K \otimes T_L}$. The matrix of ${T_{KL}}$ is then the Kronecker product of the matrices of ${T_K}$ and ${T_L}$. It follows that ${D_M = D_K^s D_L^r}$. By transitivity of discriminants (see below), ${D_{KL} | D_K^s D_L^r}$. By considering the towers ${\mathbb Q \subseteq K \subseteq KL}$ and ${\mathbb Q \subseteq L \subseteq KL}$, we also know that ${D_K^s D_L^r}$ is divisible by ${D_K^s}$ and ${D_L^r}$, hence the assertion that ${D_{KL} = D_K^s D_L^r}$. It follows in this case (again by transitivity of discriminant) that ${M = \mathcal O_{KL}}$.

Transitivity of discriminants:
Let ${K\subseteq K' \subseteq K''}$ be a finite extension of number fields and ${A, A', A''}$ be their rings of integers. Then the discriminants satisfy $\displaystyle D_{A''/A} = D_{A'/A}^{[K'':K']}.\mathrm{N}_{ K'/ K}(D_{ A''/ A'}).$