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Let $A = k[x_1,\ldots,x_n]$ be a polynomial ring over a field $k$ of characteristic zero and $\{y_j\}_{1 \leq j \leq \ell}$ a family of homogeneous polynomials. Write $B$ for the subring $k[y_1,\ldots,y_\ell]$.

Assume that

  1. the quotient $A \otimes_B k$ is finite-dimensional and

  2. the tensor product $A \otimes_B A$ is a free $A$-module (the structure map is $A = A \otimes_k k \to A \otimes_B A$).

I want to conclude that either $A \otimes_B A$ or $A \otimes_B k$ is a complete intersection ring, meaning the ideal $(y_1,\ldots,y_\ell)$ in $A$ or the ideal $(1 \otimes y_j - y_j \otimes 1)_{1 \leq j \leq \ell}$ in $A \otimes_k A$ is generated by some regular subsequence.

Is either $A \otimes_B A$ or $A \otimes_B k$ a complete intersection ring?

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It will be enough to show that $B \to A$ makes $A$ a free $B$-module, by the answer to this question.

To do this pick a finite basis of $A \otimes_B A$ as a free $A$-module; it is possible to take representatives for the basis in the image of $k \otimes_B A$, say $(1 \otimes c_j)$. This choice defines a unique $A$-linear map doing what one would expect with the chosen basis (and in all but trivial cases destroying ring structure): \begin{align*}\tilde{f}\colon A \otimes_k k \otimes_B A &\overset\sim\longrightarrow A \otimes_B A,\\1 \otimes 1 \otimes c_j &\longmapsto 1 \otimes c_j.\end{align*}

Consider the restriction of this $\tilde f$ to the $B \otimes_k k \otimes_B A$; we claim this gives us a $B$-basis for $B \otimes_B A \cong A$. Indeed, the restriction factors as $$ B \otimes_k k \otimes_B A \overset f\longrightarrow B \otimes_B A \longrightarrow A \otimes_B A,$$ where the second map is the natural $(B \hookrightarrow A) \otimes_B A$ and $f$ is the unnatural $B$-linear map given by the same formula as $\tilde f$. Now, $f$ is injective because the composition is (being the inclusion followed by the bijection $\tilde f$) and surjective by inspection. Since $B \otimes_B A \cong A$ as a $B$-algebra, the composition $$B \otimes_k k \otimes_B A \underset f{\overset\sim\longrightarrow} B \otimes_B A \overset\sim\longrightarrow A$$ gives us our desired $B$-basis of $A$.

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