Last time we took a look at Dedekind domains with fraction fields and found that if was any finite field extension of that the integral closure of in is Dedekind. The proof in this case is somewhat involved, but becomes slightly less so and shows that is also a finitely generated -module under the assumption that is also separable. In the following post we look at some of what makes life so much easier in the separable case.

Recall first that a separable extension of fields is called separable if it is generated by separable polynomials, i.e. ones which do not have repeated roots. This quality is easily understood via the following quantity.

**Definition: **The *discriminant* of a polynomial of degree over a domain is the product of the roots(in the algebraic closure of the domain’s fraction field) in particular if is not separable, i.e. for the discriminant is zero. Moreover this quantity lies in the base ring of the polynomial as the discriminant can be realized as the resultant of with its polynomial first derivative .

One very nice property of separable field extensions is the following famous theorem.

**Theorem(of the Primitive Element):** If is a finite separable extension, then there exists some such that .

This alone gives us a foot in the door towards solving what might be termed the foundational problem in algebraic number theory: factoring primes in integral extensions. Say that is a Dedekind Domain with field of fractions , and has minimal polynomial . Consider . We can easily determine the factorization of prime of in , it is the same as the factorization of in .

To see this, consider that . This can be easily deduced by starting with and considering with the composite of quotient maps by the ideals generated by and in that order, and then the reverse order.

Thus understanding the ideals of amounts to understanding the ideals of containing , and since is a field, these correspond to polynomials of the form where . Moreover by definition and the primes of correspond to , specifically they will be of the form for some which reduces to . We note that since has degree , so that would have to be the prime factorization of in if were a Dedekind domain. We see this since implies that but then they must be equal by considering the dimension of their quotient by over .

In many cases we can find an such that the integral closure of in a separable extension is . For a given or even for a general ring extension we can determine how far this is from the truth using the following quantity.

**Definition:**If are two rings then the *conductor* of into is the largest ideal of which is also an ideal of , i.e.

If is a Dedekind domain, factors uniquely into a finite product of prime ideals, so the following theorem is generically how one factors primes in a finite separable extension.

**Theorem(Kummer’s Factorization Theorem):** Let be a Dedekind domain with fraction field and a finite separable extension field. Let be the integral closure of in , an integral primitive element of with minimal polynomial and a prime of such that is coprime to .

Then if factors as with monic and irreducible then

with .

The proof follows from the above discussion if we can show that . To get this isomorphism, we certainly have the composite map . Since is coprime to , it is comaximal so . Thus is surjective and has kernel .

**Remark:**One might ask here why it was important that be Dedekind in this situation. It doesn’t seem particularly important in the exposition that be integrally closed, just that nonzero primes are maximal and that is finitely generated. Well given the proof we cited that , it’s very important, but what if we used a different proof? Neukirch section 8 uses a proof which leans heavily on the fact that is a finite separable extension and is noetherian, and could work just as well if were not integrally closed. However, note that the real utility of the theorem is that since is Dedekind, any nonzero ideal can be written as a product of nonzero primes so we can just use the above theorem if is coprime to .

We now see how this theorem tells us about when a point has preimages of multiplicity greater than one in .

**Corollary: **Suppose is prime, a finite separable extension, the integral closure of in and be the unique factorization. Then if is an integral primitive element for with minimal polynomial of discriminant then if is coprime to and then for all , .

In particular the above shows that there are only finitely many primes of which *ramify* in .

Tune in next time for Dedekind Domains in Galois extensions.

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Great post.

For the proof for the Kummer’s Factorization Theorem:

How to prove the following statement?

kernel R[\alpha] \cap \mathfrak{p}S = \mathfrak{p}R[\alpha ].