Polynomial rings: if $A \otimes_B A$ is free over $A$, is it a complete intersection?

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?

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 $$A$$.