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 . 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 and be linearly disjoint number fields, i.e.,

is an isomorphism. Then, , where denotes the discriminant of over , etc. and and . Moreover, if and are relatively prime over , then and is an isomorphism. So, the basis of can be obtained from the bases of and in a natural way.

** Proof:**

Let be the image of

Then is an order of (i.e., a subring of the ring of integers). If is the image of a pure tensor , then the trace form on acts on by . (For a proof of this, go to the Galois closure and use the canonical isomorphism given by linear disjointness). Thus, acts by . The matrix of is then the Kronecker product of the matrices of and . It follows that . By transitivity of discriminants (see below), . By considering the towers and , we also know that is divisible by and , hence the assertion that . It follows in this case (again by transitivity of discriminant) that .

** Transitivity of discriminants:**

Let be a finite extension of number fields and be their rings of integers. Then the discriminants satisfy

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