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I have largely studied Hilbert function and Hilbert polynomial for polynomial rings over fields of characteristic zero. Is it possible to extend the theory also for polynomial rings over fields of characteristic positive?

Thank you

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Yes: for a finitely generated $k$-algebra, for instance, you can still take your additive function $\lambda$ to be $\dim_k$, and you get that the Hilbert series $\mathcal P(V,t) = \sum_{i \geq 0} \lambda(V_i)t^i$ is a rational function.

See e.g. http://tartarus.org/gareth/maths/notes/iii/Commutative_Algebra_2013.pdf, p. 31 onwards, for an approach to it.

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  • $\begingroup$ Ok, you suggest using the classic definition for Hilbert-Poincaré series. But why the majority of books and papers put the hypothesis of a field of characteristic zero? $\endgroup$ – Ella Smith May 21 '16 at 13:38
  • $\begingroup$ I haven't seen a treatment that assumes characteristic zero (though I've only seen one or two places that cover it, I think the other is Matsumara's book). The proof of the theorem that I'm thinking of doesn't need characteristic zero anywhere, so I don't see why people would want to assume it unless it made their life easier for other reasons. $\endgroup$ – Josh Hunt May 21 '16 at 22:39
  • $\begingroup$ @Ella Is the question, perhaps, about the ring where the Hilbert polynomial itself resides? Those dimensions are non-negative integers, so it is natural to view $\mathcal{P}(V,t)$ as an element of the ring of formal power series $\Bbb{Z}[[t]]\subset K[[t]]$ for any field $K$ of characteristic zero. Irrespective of the field of definition of $V$. Or,...? $\endgroup$ – Jyrki Lahtonen May 22 '16 at 7:33
  • $\begingroup$ @EllaSmith if your encounters with the Hilbert series come from invariant theory, then perhaps it is because of Molien's theorem (en.wikipedia.org/wiki/Molien_series) which requires at least the assumption that the field characteristic is not divisible by the order of the group. $\endgroup$ – Ben Blum-Smith Nov 3 '18 at 1:24

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