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I have spent quite some time attempting to solve the following problem:

Let $A\to B$ be a finite homomorphism of Noetherian rings. We suppose that $A$ is a complete local ring, with residue field $k$. Let $k^\prime$ be the largest sub-$k$-algebra of $B\otimes_A k$ that is separable over $k$. Show that $A \to B$ decomposes into a finite étale homomorphism $A\to C$ followed by a finite homomorphism $C\to B$, with $C$ simple over $A$ (i.e., $C = A[c]$ for some $c\in C$) and $C\otimes_A k = k^\prime$.

So, as $B$ is an $A$-algebra, I want to start by writing $B = A[x_1,\cdots, x_n]/I$ for some ideal $I$, and then take $C = A[x_1, \cdots, x_n]$, hoping at the end that $n$ necessarily was $1$, but I have no idea why that might be the case. I really don't see another way to begin though, but if this is the wrong way to even start, I would appreciate any hints that you have to offer.

Thanks! :)

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I think the statement has a counterexample. Suppose $A = \mathbb{F}_2$ and $B=A^3$. Then $C = B$, which cannot be expressed as a quotient of $A[X]$ as $A[X]$ only has two $A$-valued points.

If $B$ is local, then the statement is true. In this case, $k'(\subseteq B\otimes_A k)$ is a field because $B\otimes_A k$ is local.

$k'$ is finite separable over $k$ so that we have a monic polynomial $f\in A[X]$ such that $k'\cong k[X]/(f(X))$. Here, we denote the image of $f$ in $k[X]$ also by $f$. $C:=A[X]/(f(X))$ is finite etale over $A$ (with special fiber obviously $k'$). Indeed, the finiteness is obvious while for the etaleness, we are reduced to show it at the closed point of $\operatorname{Spec} C$ because the etale locus is open. But, $A\to C$ is flat and the special fiber of $A\to C$ is the etale $k\to k'$, implying what we want by the fiber criterion. Alternatively, we can directly calculate $\Omega_{C/A}^1$ and conclude it's $0$ by NAK.

$C\to B$ is finite as $A\to B$ is.

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