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Fix a field $k$. Does there exist a faithfully flat extension $A\subset B$ of $k$-algebras where $B$ is finitely presented but $A$ is not?

Edit: This question has been put on hold for missing context. I am not sure what context will help the question, but here are some other places where finite presentation can be detected within a faithfully flat extension:

  • If $A\subset B$ is a faithfully flat extension of rings, $M$ is an $A$-module, and $B\otimes_A M$ is finitely presented over $B$, then $M$ is finitely presented over $A$.

  • If $A\subset B$ is an inclusion of commutative $k$-Hopf algebras and $B$ is finitely presented as a $k$-algebra, then $A$ is finitely presented as a $k$-algebra.

These situations seem to be rather different from just an inclusion of $k$-algebras, and their proofs do not seem to shed light on this situation.

For all examples that I am familiar with of extensions $A\subset B$ of $k$-algebras for $B$ is finitely presented and $A$ is not, the extension fails to be flat (or I cannot prove it is so). For example, the inclusion $k[x,xy,xy^2,\ldots]\subset k[x,y]$ is discussed in the comments below, and the examples I am familiar with are of a similar form to this. (If we work with general commutative rings rather than $k$-algebras, I still do not know an answer to the question.)

user577334
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  • Have you tried the standard example: $k[x,xy,xy^2, xy^3, \ldots ] \subset k[x,y]$? (I didn't check whether or not this extension is faithfully flat.) – Elle Najt Jul 15 '18 at 16:56
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    @Lorenzo If I write that ring as $A$, then I don't believe $0\rightarrow (x)\rightarrow A \rightarrow k\rightarrow 0$ remains exact upon base change to $k[x,y]$. – user577334 Jul 15 '18 at 17:42
  • Okay, $x \otimes y - xy \otimes 1$ is indeed a nonzero element of the kernel of the induced map. To verify this, you can use the bilinear form $B(f,g) = \pi(f) g(0,1)$, where $\pi : (x) \to (x)/(xy,x^2) \cong k$. ($x$ acts by zero on $k$.) – Elle Najt Jul 15 '18 at 21:48

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