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:
$B$ is étale.
$f$ 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.
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.