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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.

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  • $\begingroup$ What is $f$ in (2)? A monic polynomial in $A[X]$ annihilating $b$ of minimal degree? $\endgroup$ – Alex Wertheim Apr 19 at 22:53
  • $\begingroup$ Sorry, the monic generator of the kernel of the map $A[x] \rightarrow B$ sending $x$ to $b$. $\endgroup$ – Dean Young Apr 19 at 23:01
  • $\begingroup$ 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.) $\endgroup$ – Alex Wertheim Apr 20 at 1:04
  • $\begingroup$ 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. $\endgroup$ – Dean Young Apr 20 at 1:07
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    $\begingroup$ 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. $\endgroup$ – Dean Young Apr 20 at 3:13
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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.

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  • $\begingroup$ 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. $\endgroup$ – Dean Young Apr 20 at 0:55
  • $\begingroup$ @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. $\endgroup$ – Alex Wertheim Apr 20 at 0:58
  • $\begingroup$ Thanks again Alex. My friend just told me that if $A$ contains a field, then we may have $2 \implies 1$, interestingly enough. $\endgroup$ – Dean Young Apr 20 at 1:39
  • $\begingroup$ "Let $f(X) \in A[x]$ be a monic ..." - Oh, this is excellent! $\endgroup$ – Dean Young Apr 20 at 4:35

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