## Dedekind Domains in Finite Galois Extensions

Last time we examined Dedekind domains in finite separable field extensions. One advantage to using a separable field extension that we did not use is that we can base extend to a finite Galois extension, where as we see the action of Galois forces the splitting of primes to be very uniform.

First we define things. Let $R$ be a Dedekind Domain, $K$ the fraction field of $R$, $L$ a Galois extension of $K$ with Galois group $G$ and $S$ the integral closure of $R$ in $L$.

We know already that $G$ acts on $L$ and fixes $K$ and thus $R$ pointwise. This of course means that $G$ must act on $S$, in fact fixing any monic minimal polynomial in $R[X]$ which defines an element in $S$. We also know that since $G$ can be viewed as being made up of ring automorphisms of $S$, so if $\mathfrak{P}$ is a prime of $S$ then $g^{-1}\mathfrak{P}$ is a prime of $S$ for all $g \in G$. Additionally if $\mathfrak{P}\cap R = \mathfrak{p}$ then

$g(\mathfrak{P})\cap R = g(\mathfrak{P}) \cap g(R) = g(\mathfrak{P} \cap R) = g(\mathfrak{p}) = \mathfrak{p}.$

We say, in light of these findings that $G$ acts on the primes of $S$ lying over $\mathfrak{p}$. In fact as we show now, $G$ permutes the primes lying over $\mathfrak{p}$.

Theorem: Let $R,S,G, \mathfrak{P},\mathfrak{p}$ be as above. Then if $\mathfrak{P}'/\mathfrak{p}$ there exists some $g\in G$ such that $\mathfrak{P}' = g\mathfrak{P}$.

Suppose not, so $\mathfrak{P}'$ is not contained in $g\mathfrak{P}$ for any $g\in G$, and so there exists $x \in \mathfrak{P}'$ which is not in any $g\mathfrak{P}$.  Since $G$ is finite, $\{\mathfrak{P'}\}\cup\{ g\mathfrak{P}\}_{g\in G}$ is a finite set of primes lying over $\mathfrak{p}$.Then we can use the Chinese Remainder Theorem to make this precise and find some $x \in S$ such that $x \equiv 0 \bmod \mathfrak{P}'$ and $x \equiv 1 \bmod g\mathfrak{P}$. By the second condition, for all $g \in G, gx \equiv 1 \bmod \mathfrak{P}$, so $N(x) = \prod_{g \in G} gx \equiv 1 \bmod \mathfrak{P}$. Since the identity automorphism is in $G$, $N(x) = x(\prod_{g \ne e} gx) \in x S \subset \mathfrak{P}'$, but also note that $N(x)$ is Galois invariant,so $N(x) \in \mathfrak{P}'\cap K = \mathfrak{P}'\cap R = \mathfrak{p} \subset \mathfrak{P}.$ Thus we have a contradiction.

Corollary: If $\prod_{i=1}^g \mathfrak{P}_i ^{e_i} = \mathfrak{p}S$ is the factorization of $\mathfrak{p}$ in a Galois extension, then $e_i = e_1$, $f_i = f_1$ for all $1 \le i \le g$.

In the above expression, let $e = \max (e_i)$ and $\mathfrak{P}$ be a corresponding prime ideal. Thus $\mathfrak{P}^e \supset \mathfrak{p}S$. For any non-negative integer $r$, $\mathfrak{P}_i^r \supset \mathfrak{p}S$ implies $r \le e_i$. If $e_i >r$ then  by the chinese remainder theorem we can select an element of $\mathfrak{P}_i^{e_i} - \mathfrak{P}_i^r$ in $\mathfrak{p}S$.

By definition, each $e_i \le e$ but by the last theorem, for each $i$ there is some $g \in G$ such that $g\mathfrak{P} = \mathfrak{P}_i$. But then $\mathfrak{P}^e \supset \mathfrak{p}S$ implies that $\mathfrak{P}_i^e \supset g\mathfrak{p}S = \mathfrak{p}S$, so $e \le e_i$ and each of the $e_i$‘s are equal.

For the $f_i$‘s we recall that $f_i = [S/\mathfrak{P}: R/\mathfrak{p}]$.  Thus if we can induce an isomorphism $S/\mathfrak{P}\to S/g\mathfrak{P}$ fixing $R/\mathfrak{p}$ we have the result.

To do this, consider the surjective composite map $S \stackrel{g}{\to} S \to S/ g\mathfrak{P}$, which has kernel $\mathfrak{P}$. Moreover if we precompose with the inclusion $R \to S$ we get the quotient map $R \to R/\mathfrak{p}$, so $R$ is fixed pointwise. Thus we have our result, but we also have a notion of mapping $G$ (or at least the stabilizer of $\mathfrak{P}$) to the automorphism group of $S/\mathfrak{P}$ over $R/\mathfrak{p}$.

Definition: The stabilizer of $\mathfrak{P}$ in $G$ is called the decomposition group

$D(\mathfrak{P}|\mathfrak{p}) = \{ g \in G : g\mathfrak{P} = \mathfrak{P}\}$

Theorem: Under the conditions set above, the field  extension $(S/\mathfrak{P})/(R/\mathfrak{p})$ is normal and there is a surjective homomorphism $D(\mathfrak{P}|\mathfrak{p}) \to Aut_{R/\mathfrak{p}}(S/\mathfrak{P})$.

Remark: I’ve said nothing about $S/\mathfrak{P}$ being Galois, because it need not be separable. Following Brian Conrad’s note we can construct degree $p$ Galois extensions with residue field extensions which are purely inseparable. For now I will resist the temptation to derail this post with any specifics and simply suggest the reader check out the note.

Independent of the rest of the theorem, the map described above is naturally multiplicative in $G$ and thus in the decomposition group so we have a homomorphism. To see that the field extension $(S/\mathfrak{P})/(R/\mathfrak{p})$ is normal, let $\theta \in S/\mathfrak{P}$ and $f(X) \in (R/\mathfrak{p})[X]$ be the minimal polynomial for $\theta$. If $\Theta \in S$ reduces  to $\theta$ and $F(X) \in R[X]$ is the minimal polynomial for $\Theta$ then $\theta$ is a root of the equivalence class of polynomials $F(X) + \mathfrak{p}[X]$. Thus $f(X)$ divides $F(X) +\mathfrak{p}[X]$. Since $L$ is normal, any polynomial in $K[X]$ splits into a product of linear factors when viewed as an element of $L[X]$ but a monic polynomial in $R[X]$ can only have roots in an integral extension so $F(X)$ splits into a product of linear factors when viewed as an element of $S[X]$, $F(X) + \mathfrak{P}[X]$ splits into a product of equivalence classes of linear factors and so $f(X)$ considered as an element of $(S/\mathfrak{P})[X]$ splits into a product of linear factors. Thus $S/\mathfrak{P}$ is a normal extension of $R/\mathfrak{p}$.

We are left to prove that our map is surjective. Consider the maximal separable subextension $(S/\mathfrak{P}) \supset k \supset (R/\mathfrak{p})$. Let $\theta$ be a primitive element for $k$ and $\Theta, F(X), f(X)$ correspond as before. Let $\sigma \in Gal(k/(R/\mathfrak{p})) = Aut_{R/\mathfrak{p}}(S/\mathfrak{P})$, so that $\sigma\theta$ is a root of $f(X)| (F(X) + \mathfrak{p}[X])$. Thus there is a root of  $F(X)$, say $\Theta'$ such that $\Theta' \bmod \mathfrak{P} = \sigma \theta$.

Now here comes the tricky part: we can certainly find $\Sigma \in G$ such that $\Sigma\Theta = \Theta'$. What we’d like for a map here to make sense is that ANY representative of $\theta$ to be taken to a representative of $\sigma\theta$. Thus if $p\in \mathfrak{P}$, $\Sigma(\Theta + p) = \Theta' + \Sigma(p)$ we’d like $\Sigma(p) \in \mathfrak{P}$, i.e. $\Sigma \in D(\mathfrak{P}|\mathfrak{p})$. It’s not clear that there is an overlap between $D(\mathfrak{P}|\mathfrak{p})$ and the subset of $G$ taking $\Theta$ to $\Theta'$ . To show there is an overlap, we do what reasonable people do and change the problem into something easier!

Definition: The Decomposition Field of a prime $\mathfrak{P}$ in a Galois extension $L/K$ is the fixed field $L^{D}$ of the decomposition group at $\mathfrak{P}$. By the fundamental theorem of Galois theory, $L$ is Galois over the decomposition field with Galois Group $D = D(\mathfrak{P}|\mathfrak{p})$.

Theorem: Let $\mathfrak{P}_D = \mathfrak{P} \cap L^D$. Then

• $\mathfrak{P}_D$ is a prime ideal in the integral closure $T$ of $R$ in $L^{D}$
• $\mathfrak{P}_D S = \mathfrak{P}^{e}$ for some integer $e$.
• $f(\mathfrak{P}_D |\mathfrak{p}) =e(\mathfrak{P}_D|\mathfrak{p}) = 1$, in particular $T/\mathfrak{P}_D = R/\mathfrak{p}$.

The first statement follows from the content of the post on finite separable extensions, in particular $T = S \cap L^D$ as a quick double containment argument shows.

The second statement follows from that fact that $Gal(L/L^D)$ acts transitively on the primes above $\mathfrak{P}_D$, but $\mathfrak{P}$ is one such prime and by definition the decomposition group fixes it.

If $\{\tau_i\}_{i=1}^g\subset G$ are representatives for the coset space $G/D$ , we already know that $\mathfrak{p}S = \prod_{i=1}^g \tau_i\mathfrak{P}^{e}$ where $e = e(\mathfrak{P}|\mathfrak{p})$. Then certainly $\tau_i(\mathfrak{P}_D) = \tau_i(\mathfrak{P})\cap L^D$ are all primes of $T$ lying above $\mathfrak{p}$. This is everything, as we see now. We know that $\# G = efg = [L:K]$ where $f = f(\mathfrak{P}|\mathfrak{p})$ and since $[G:D] = g$, $[L^D :K] = g$. If $\prod_i\mathfrak{q}_i^{r_i} = \mathfrak{p}T$ then since $L^D/K$ is a finite separable extension of degree $g$ $\sum_i r_i f(\mathfrak{q_i} |\mathfrak{p}) = g$, but $\prod_{i=1}^g \tau_i\mathfrak{P}_D \supset \mathfrak{p}T$, so they must be equal.  Thus $f(\mathfrak{P}_D|\mathfrak{p}) =1$. Then since $\tau_i$ are a set of coset representatives for $G/D$, if $\tau_i \ni D$, then $e(\mathfrak{P}_D|\mathfrak{p}) =1$ (although it’s not always true that if $i\ne j$ that $\tau_i\mathfrak{P}_D \ne \tau_j \mathfrak{P}_D$ because $D$ is not always a normal subgroup of $G$).

Returning to the proof that we have a surjective map $D \to Aut_{R/\mathfrak{p}}(S/\mathfrak{P})$, we consider that if $\theta$ is as above, we could instead consider the monic minimal polynomial of $\Theta$ over $T[X]$ to be $F(X)$ since $R/\mathfrak{p} = T/\mathfrak{P}_D$. If we select a root $\Theta''$ from the roots of this new $F(X)$ such that $\Theta'' \bmod \mathfrak{P} = \sigma\theta$, it need only differ from our earlier $\Theta'$ by an element of $\mathfrak{P}$. Now when we pick a $\Sigma \in Gal(L/L^D)$ such that $\Sigma\Theta = \Theta''$ we are guaranteed that $\Sigma \in D$ so we have a surjective map.

Definition: The kernel of the map $D(\mathfrak{P}|\mathfrak{p}) \to Aut_{R/\mathfrak{p}}(S/\mathfrak{P})$ is called the inertia group $I(\mathfrak{P}|\mathfrak{p})$. It is an easy observation that $\# I(\mathfrak{P}|\mathfrak{p}) = e(\mathfrak{P}|\mathfrak{p})$.

Thus we have a tower of groups

${0} \stackrel{e}{\subset} I(\mathfrak{P}|\mathfrak{p}) \stackrel{f}{\subset} D(\mathfrak{P}|\mathfrak{p}) \stackrel{g}{\subset} G$

which by Galois Theory gives the following tower of fields

$L \supset L^I \supset L^D \supset K$

and the splitting of the primes occurs as expected(up to normality of the various subgroups), i.e. the splitting of a prime occurs between $K$ and the Decomposition Field, the residue field extension occurs between the Decomposition field and the Inertia field and the ramification occurs between the Inertia field and $L$. For this reason the inertia group is sometimes called the 0-th ramification group and there exists a compatible theory of “higher ramification groups”.

An Application to Computing Galois Groups: Say $R/\mathfrak{p}$ is a finite field of size $q$, then since $f(\mathfrak{P}|\mathfrak{p}) \le [L:K]$, $S/\mathfrak{P}$ is also a finite field. Every finite extension of finite fields is Galois, and in fact cyclic, generated by the Frobenius map $x \mapsto x^q$. Considering a lift of this element to the Decomposition group (for details, consult Milne’s algebraic number theory page 136) we get an $f(\mathfrak{P}|\mathfrak{p})$ cycle in $G$.

This is particularly useful in the most commonly asked case, $K=\mathbf{Q}$ where $\mathbf{Z}/(p)$ is always finite. In fact if $L$ is generated by the roots of the monic separable polynomial $f(X)$, and$p$ does not divide the conductor of $\mathcal{O}_L$ we can determine the cycle structure of $G$ from the factorizations of $f \bmod p$. An example showing how one might do this is found in Milne’s notes.

Next time, we examine number fields a little more closely, showing in particular that the Picard group is finite and some nice things about the group of units.

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