# Galois Theory for Finite Extensions of Rings

I am learning Galois theory for schemes from Lenstra's notes, and I have a question about how this might be phrased for integral extensions with a single generator.

For fields, we have several languages for Galois theory. There is the newer language of the fundamental functor $$F : \text{ét}(K) \rightarrow \text{Set}$$, which sends a finite étale $$K$$-algebra $$A$$ to the set of maps $$[A, K^{sep}]$$. Perhaps the simplest lanuage is the case for simple extensions. While it is less general, it is explicitly and self-contained, and it is governed by polynomials and their roots.

Theorem: Let $$K(a)/K$$ be a simple extension, let $$f \in K[x]$$ be the minimal polynomial of $$K$$, and let $$G = \text{Aut}_K (K(a))$$. If $$f$$ factors completely into distinct roots in $$K(a)$$, then intermediate fields of $$K(a)/K$$ are in correspondence with subgroups of $$G$$.

For my question, let $$A$$ be a ring, and let $$B$$ be an $$A$$-algebra generated by a single element $$b \in B$$, such that $$b$$ is integral over $$A$$, and such that $$A \rightarrow B$$ is injective. Let $$f \in A[x]$$ be the unique monic polynomial which generates the kernel of the map $$A[x] \rightarrow B$$ sending $$x$$ to $$b$$. What is the relationship between the following conditions:

1. $$B$$ is étale.

2. $$f$$ splits into distinct roots in some $$A$$-algebra.

When I ask for a relationship between these, I would like to know whether $$1 \implies 2$$, and whether $$2 \implies 1$$. If at least one of these ends up being false, then I would like to know if there are some conditions you know off-hand that could be added to one to make these conditions equivalent.

• What is $f$ in (2)? A monic polynomial in $A[X]$ annihilating $b$ of minimal degree? – Alex Wertheim Apr 19 at 22:53
• Sorry, the monic generator of the kernel of the map $A[x] \rightarrow B$ sending $x$ to $b$. – Dean Young Apr 19 at 23:01
• Incidentally, I answered below predicated on the premise that you are interested in $A$-algebras of the form $A[X]/\langle f(X) \rangle$ for some $f(X)$ in $A[X]$ monic. It's not immediately clear to me why if $B$ is an $A$-algebra generated by a single $b \in B$ finite over $A$ then the kernel of the map $A[X] \to B$ sending $X$ to $b$ has to have a single generator. (Of course, I'm not saying it isn't true, I just don't see why.) – Alex Wertheim Apr 20 at 1:04
• Doesn't that follow from the Cayley-Hamilton theorem for rings: $B$ is integral by the result that finite iff integral and finitely generated as an algebra, so any element in it is integral. Therefore there the map $A[x] \rightarrow B$ sending $X$ to $b$ has kernel generated by a monic polynomial. – Dean Young Apr 20 at 1:07
• Wait, no, you were right. The map $A \rightarrow B$ needs to be an injection. There is a theorem stating that a finite faithful module is integral. – Dean Young Apr 20 at 3:13

I think (1) implies (2) should be true, if the following argument is correct. Suppose $$f \in A[X]$$ monic such that $$B = A[X]/\langle f(X) \rangle$$. Since $$A \to B$$ is étale, hence unramified, for any $$\mathfrak{p} \in \mathrm{Spec}(A)$$, the map $$k(\mathfrak{p}) \to B \otimes_{A} k(\mathfrak{p})$$ is unramified. This is the case precisely when $$B \otimes_{A} k(\mathfrak{p})$$ is a finite product of finite separable field extensions of $$k(\mathfrak{p})$$.

But $$B \otimes_{A} k(\mathfrak{p}) \cong k(\mathfrak{p})[X]/\langle \tilde{f}(X) \rangle$$, where $$\tilde{f}$$ is the image of $$f(X)$$ in $$k(\mathfrak{p})[X]$$. This implies the irreducible factors $$f_{1}, \ldots, f_{n} \in k(\mathfrak{p})[X]$$ of $$\tilde{f}$$ are distinct and are each separable over $$k(\mathfrak{p})$$, and so $$\tilde{f}$$ itself is separable over $$k(\mathfrak{p})$$. Letting $$M$$ be a splitting field for $$\tilde{f}$$ over $$k(\mathfrak{p})$$, we're done. (And of course, note that $$M$$ is an $$A$$-algebra via $$A \to k(\mathfrak{p}) \to M$$.)

I think that (2) need not imply (1), however. Take $$A = \mathbb{Z}, f(X) = X^{2}+1 \in \mathbb{Z}[X]$$, and $$B = A[X]/\langle f(X) \rangle = \mathbb{Z}[i]$$. Then the map $$\mathrm{Spec}(B) \to \mathrm{Spec}(A)$$ induced by the inclusion $$A \hookrightarrow B$$ is ramified at the prime ideal $$(1+i)\mathbb{Z}[i]$$, hence not étale, but $$f$$ splits into distinct factors over $$(\mathbb{Z}/3\mathbb{Z})[X]/\langle X^{2}+1 \rangle$$.

I have been silly. The condition you are looking for is the following:

Let $$f(X) \in A[X]$$ be a monic polynomial. Then $$B = A[X]/\langle f(X) \rangle$$ is étale over $$A$$ if and only if $$\langle f(X), f'(X) \rangle = A[X]$$. Equivalently, in local terms, $$B$$ is etale over $$A$$ if and only if the image of $$f(X)$$ is separable over $$k(\mathfrak{p})$$ for every $$\mathfrak{p} \in \mathrm{Spec}(A)$$. A reference is Example 3.4 on page 22 of Milne's "Etale Cohomology"; I can provide more details if needed.

• Thanks, Alex. Do you have any hunches on the converse? I think it's false. I'll accept this in a few days if nobody ends up knowing. – Dean Young Apr 20 at 0:55
• @DeanYoung: No problem! I think it's false. Take $A = \mathbb{Z}, f(X) = X^{2}+1 \in \mathbb{Z}[X]$. Then $B = A[X]/\langle f(X) \rangle = \mathbb{Z}[i]$ is ramified at the prime ideal $i\mathbb{Z}[i]$, hence not etale, but $f$ splits into distinct factors over $(\mathbb{Z}/3\mathbb{Z})[X]/\langle X^{2}+1 \rangle$. I'll update my answer. – Alex Wertheim Apr 20 at 0:58
• Thanks again Alex. My friend just told me that if $A$ contains a field, then we may have $2 \implies 1$, interestingly enough. – Dean Young Apr 20 at 1:39
• "Let $f(X) \in A[x]$ be a monic ..." - Oh, this is excellent! – Dean Young Apr 20 at 4:35