Let $M$ be a finitely presented module over the ring $R$. Suppose that for all primes $P\subset R$ the $k(P)$ vector space $M\otimes _Rk(P)$ has a dimension $d(P)$ independent of $P$.
Can I conclude that $M$ is flat over $R$?
I am asking because I want to better understand the criterion for a family of projective schemes to be flat over some base in terms of invariance of their Hilbert polynomial.
I checked some examples. For example let $M$ be the ideal of a point $P$ of an affine curve $C$.
Here $d(P)$ is the dimension of the Zariski tangent space of $C$at $P$ so that the proposed criterion for flatness holds.

  • 2
    $\begingroup$ In general equi-dimensionality of the fibres is necessary, but not sufficient in order to guarantee flatness. However, it is also sufficient when $R$ is regular of dimension $1$, or when $R$ is regular of arbitrary dimension and $M$ is Cohen-Macaulay. See S. Kovacs' answer to the following MO question: mathoverflow.net/questions/75317/… $\endgroup$ – Francesco Polizzi Jan 14 '17 at 9:59
  • 1
    $\begingroup$ Ah, sorry, I slightly misread the question. You are asking about flatness of a (finitely presented) module over the ring $R$, so in geometric terms you want to know about the flatness of a coherent sheaf over an affine scheme (and not about the flatness of a morphism of schemes). Then the correct answer is given by abx in his comment: in the case where the fibres are of constant dimension, flatness occurs whenever $R$ is without nilpotents. $\endgroup$ – Francesco Polizzi Jan 14 '17 at 10:22

Yes if $R$ is reduced (no nilpotent elements). This is (for instance) Lemma 1 p. 51 in Mumford's Abelian Varieties (2nd edition). No in general: just take $R=k[\varepsilon ]/(\varepsilon ^2)$, $M=k$.

  • $\begingroup$ Thanks a lot abx: perfect answer! Strangely I can accept your answer but not upvote you. Anyway, in my eyes your real reputation is huge since long ago! $\endgroup$ – gregorsamsa Jan 14 '17 at 10:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.