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Let $I \subseteq k[x_1,\dots,x_n]$ be an ideal for a field $k$ and let $A = k[x_1,\dots,x_n]/I$. For $d \geq 0$ let $$ A_{\le d} := \{f + I : f \in k[x_1,\dots,x_n], \deg{f} \leq d \}.$$ The Hilbert function of $I$ is $$ h_I: \mathbb{N}_0 \to \mathbb{N}_0, \ h_I(d) = \dim_k(A_{\leq d}). $$

This object seems quite random (at least to me) at first but after some "magic" happens, it somehow turns out to have great properties. For instance, it's a polynomial $p_I$ for all sufficiently large $d$ and we get $\deg(p_I) = \dim(A)$.

This begs the question: Why? Why should we have expected $h_I$ to be nice? How did people (probably Hilbert?) conjecture these things?

user26857
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Qi Zhu
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  • Isn't $I$ supposed to be homogeneous? – user26857 Jul 05 '19 at 19:48
  • No. At least not in our lecture. – Qi Zhu Jul 05 '19 at 20:15
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    Maybe this is why you are looking for a "demystification". Usually the Hilbert function and its generating series, that is, the Hilbert series, are defined for graded $k$-algebras. In that case, $h_R(d):=\dim_kR_d$, and if one starts with the Artinian case ($\dim R=0$) can notice that a polynomial occurs. – user26857 Jul 05 '19 at 20:46
  • Hmm, weird. We‘re currently dealing with the graded case but are still taking the definition I wrote. So that‘s interesting, thank you! Sorry, if it‘s trivial, though, but I don‘t quite see why the graded case “demystifies“ the Hilbert function. Could you elaborate a bit more on that, please? – Qi Zhu Jul 06 '19 at 05:40
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    I think that this can be useful. – Armando j18eos Aug 26 '19 at 11:20

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