Dedekind Domains in finite separable extensions

Last time we took a look at Dedekind domains $R$ with fraction fields $K$ and found that if $L$ was any finite field extension of $K$ that the integral closure $S$ of $R$ in $L$ is Dedekind. The proof in this case is somewhat involved, but becomes slightly less so and shows that $S$ is also a finitely generated $R$ -module under the assumption that $L/K$ 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 $f$ of degree $n$ over a domain is the product of the roots(in the algebraic closure of the domain’s fraction field) $\Delta(f) = \prod_{i<j} (\alpha_i -\alpha_j)^2$ in particular if $f$ is not separable, i.e. $\alpha_i = \alpha_j$ for $i \ne j$ 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 $f$ with its polynomial first derivative $f'$ .

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

Theorem(of the Primitive Element): If $L/K$ is a finite separable extension, then there exists some $\alpha\in L$ such that $L = K(\alpha)$ .

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 $R$ is a Dedekind Domain with field of fractions $K$ , $L = K(\alpha)$ and $\alpha$ has minimal polynomial $p(X)$ . Consider $R[\alpha]= R[X]/(p(X))$ . We can easily determine the factorization of prime $\mathfrak{p}$ of $R$ in $R[\alpha]$ , it is the same as the factorization of $p(X)$ in $R/\mathfrak{p}[X]$ .

To see this, consider that $R[X]/(\mathfrak{p},p(X)) \cong (R/\mathfrak{p})[X]/(\overline{p(X)}) \cong R[\alpha]/\mathfrak{p}R[\alpha]$ . This can be easily deduced by starting with $R[X]$ and considering with the composite of quotient maps by the ideals generated by $\mathfrak{p}$ and $p(X)$ in that order, and then the reverse order.

Thus understanding the ideals of $R[\alpha]/\mathfrak{p}R[\alpha]$ amounts to understanding the ideals of $R/\mathfrak{p} [X]$ containing $\overline{p(X)} = \prod_{i=1}^g \overline{p_i(X)}^{e_i}$ , and since $R/\mathfrak{p}$ is a field, these correspond to polynomials of the form $\prod_{i=1}^g \overline{p_i(X)}^{r_i}$ where $r_i \le e_i$ . Moreover by definition $\bigcap (\overline{p_i(X)}^{e_i}) = (0)$ and the primes of $R[\alpha]/(\mathfrak{p})$ correspond to $\overline{p_i(X)}$ , specifically they will be of the form $\mathfrak{P}_i = p_i(\alpha)R[\alpha] + \mathfrak{p}R[\alpha]$ for some $p_i(X) \in R[X]$ which reduces to $\overline{p_i(X)}$ . We note that since $p(X)$ has degree $[L:K]$ , so that $\mathfrak{p}R[\alpha] = \prod_{i=1}^g \mathfrak{P}_i^{e_i}$ would have to be the prime factorization of $\mathfrak{p}$ in $R[\alpha]$ if $R[\alpha]$ were a Dedekind domain. We see this since $\bigcap \overline{p_i(X)}^{e_i} \subset (o)$ implies that $\prod_{i=1}^g \mathfrak{P}_i^{e_i} \subset\mathfrak{p}R[\alpha]$ but then they must be equal by considering the dimension of their quotient by $R[\alpha]$ over $R/\mathfrak{p}$ .

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

Definition:If $R \subset S$ are two rings then the conductor of $S$ into $R$ is the largest ideal of $S$ which is also an ideal of $R$ , i.e.

$\mathfrak{f}_{S/R} = \{ \alpha \in S : \alpha S \subset R\}$

If $R$ is a Dedekind domain, $\mathfrak{f}$ 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 $R$ be a Dedekind domain with fraction field $K$ and $L$ a finite separable extension field. Let $S$ be the integral closure of $R$ in $L$ , $\alpha$ an integral primitive element of $L/K$ with minimal polynomial $p(X)$ and $\mathfrak{p} \subset R$ a prime of $R$ such that $\mathfrak{p}S$ is coprime to $\mathfrak{f}_{S/R[ \alpha ]}$ .

Then if $p(X) \bmod \mathfrak{p}$ factors as $\prod_{i=1}^g \overline{p_i(X)}^{e_i}$ with $p_i(X) \in R[X]$ monic and irreducible then

$\mathfrak{p}S = \prod_{i=1}^g \mathfrak{P}_i^{e_i}$

with $\mathfrak{P}_i = p_i(\alpha)S + \mathfrak{p}S$ .

The proof follows from the above discussion if we can show that $S/\mathfrak{p}S \cong R[\alpha]/\mathfrak{p}R[ \alpha ]$ . To get this isomorphism, we certainly have the composite map $q \circ i:R[\alpha] \to S \to S/\mathfrak{p} S$ . Since $\mathfrak{p}$ is coprime to $\mathfrak{f}_{S/R[ \alpha ]}$ , it is comaximal so $S = \mathfrak{p}S + \mathfrak{f}_{S/R[ \alpha ]}\subset \mathfrak{p}S + R[\alpha]$ . Thus $q\circ i$ is surjective and has kernel $R[\alpha] \cap \mathfrak{p}S = \mathfrak{p}R[\alpha ]$ .

Remark:One might ask here why it was important that $R$ be Dedekind in this situation. It doesn’t seem particularly important in the exposition that $R$ be integrally closed, just that nonzero primes are maximal and that $S/R$ is finitely generated. Well given the proof we cited that $[L:K] = \sum_i e_if_i$ , 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 $L/K$ is a finite separable extension and $R$ is noetherian, and could work just as well if $R$ were not integrally closed. However, note that the real utility of the theorem is that since $R$ is Dedekind, any nonzero ideal $I$ can be written as a product of nonzero primes so we can just use the above theorem if $I$ is coprime to $S/R[\alpha]$ .

We now see how this theorem tells us about when a point $\mathfrak{p} \in Spec(R)$ has preimages of multiplicity greater than one in $Spec(S)$ .

Corollary: Suppose $\mathfrak{p}\subset R$ is prime, $L/K$ a finite separable extension, $S$ the integral closure of $R$ in $L$ and $\mathfrak{p}S = \prod_{i=1}^g \mathfrak{P}_i^{e_i}$ be the unique factorization. Then if $\alpha$ is an integral primitive element for $L$ with minimal polynomial $p(X)$ of discriminant $d \in R$ then if $\mathfrak{p}S$ is coprime to $\mathfrak{f}_{S/R[ \alpha ]}$ and $d \not\equiv 0\bmod \mathfrak{p}$ then for all $i$ , $e_i = 1$ .