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Let $K$ be a field and $\varphi: K[X_1,X_2,...,X_n] \to K[Y_1,Y_2,...,Y_m]$ a polynomial $K$-algebra morphism. Assume $n, m \ge 2$. By definition $\varphi$ endows $K[Y_1,Y_2,...,Y_m]$ with a $K[X_1,X_2,...,X_n]$-module structure.

Are there any available criteria to decide when the $K[X_1,X_2,...,X_n]$-module $K[Y_1,Y_2,...,Y_m]$ induced in this way by $\varphi$ a flat $K[X_1,X_2,...,X_n]$-module?

I want also to note that this generalizes this MathSE question.

As far as we consider the case with more than one indeterminantes. The case $n=m=1$ always has a positive answer since $K[X]$ is a PID and in this setting flat = torsion-free. For $n \ge 2$ $K[X_1,X_2,...,X_n]$ is not a PID, so the criterion is not applicable.

An approach is to use a lemma that states that if $R \to R', S \to S'$ are flat modules, then $R \otimes S \to R' \otimes S'$ is a flat $R \otimes S$-module. The point is that obviously not every polynomial map $\varphi: K[X_1,X_2,...,X_n] \to K[Y_1,Y_2,...,Y_m]$ arises from such "atomic" pieces $\varphi_i: K[X_i] \to K[Y_i]$ as tensor product $\bigotimes_i \varphi_i$.

So I'm asking if there exist approaches dealing with this problem or is it too broad?

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    $\begingroup$ The geometric way to look at it is that $\phi \colon \mathbf A^m \to \mathbf A^n$ is flat if and only if all fibres have dimension $m-n$. This follows from "miracle flatness"; see e.g. Tag 00R4. $\endgroup$
    – R. van Dobben de Bruyn
    Commented May 30, 2020 at 0:24
  • $\begingroup$ I suspect you mean "$K$-algebra homomorphism" rather than just "ring homomorphism". $\endgroup$
    – YCor
    Commented May 30, 2020 at 2:50
  • $\begingroup$ @YCor: Yes, thanks. $\endgroup$
    – user267839
    Commented Mar 20, 2021 at 15:56
  • $\begingroup$ @R.vanDobbendeBruyn: Thank you, that's exactly what I was looking for! $\endgroup$
    – user267839
    Commented Mar 20, 2021 at 15:56

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