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Let $f:R\to S$ be a ring homomorphism and M be a left S-module. We can consider M as an R-module via $ r.m := f(r)m $. I know that if M is a flat S-module and S is flat as R-module then M is a flat R-module. My question is about projectivity and finite presentation, what we can say?

Mourad Khattari
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    Consider $k[x]/(x^2) \to k$ with $k$ a field: every $k$-module is projective, but not when viewed as a module over the domain. – John Palmieri Feb 14 '22 at 22:53
  • Consider $k[x_1, x_2, x_3, \ldots] \to k$. Then the module $k$ is finitely presented as a $k$-module but not as a module over the domain. – John Palmieri Feb 14 '22 at 22:54

1 Answers1

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Consider $f\colon\mathbb{Z}\to\mathbb{Q}$. In this case we're just considering $\mathbb{Q}$-vector spaces as $\mathbb{Z}$-modules and clearly projectivity is not preserved, because no nontrivial $\mathbb{Q}$-vector space is a projective $\mathbb{Z}$-module.

Even being finitely generated is not preserved in the same situation.

egreg
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