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If $f$ is a homogeneous polynomial of degree $d$ in a polynomial ring in $t$ variables over a field, and it generates an ideal $I$. Then the Hilbert function of $R/I$ is $$H(R/I,n) = \binom{n+t-1}{t}-\binom{n-d+t-1}{t-d}.$$

Now i know that this is basically $H_{R/I}(t) = \dim (R) - \dim(I)$. I know that $\dim R$ gives the first binomial coefficient, but how does $\dim I = \binom{n-d+t-1}{t-d}$? Should it not be $\binom{n+d+t-1}{i}$?

Also how does the Hilbert series of $R/I$ look in this case? I mean, how does one determine the arbitrary value of $d$. Example, for $x^1$, we get $$\binom{n}{1} - \binom{n - d}{1 - d}= n - \frac{\Gamma(1-d+n)}{\Gamma(2-d)\Gamma(n)}.$$ What kind of famous series has these coefficients?

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  • $\begingroup$ @user26857, right. But how do you get the series without getting the function first (the coefficient)? $\endgroup$ – jacob smith Apr 8 '16 at 8:04
  • $\begingroup$ Well, might as well show me. I still think one needs the function before the series. $\endgroup$ – jacob smith Apr 8 '16 at 11:52
  • $\begingroup$ @user26857, why is that? $\endgroup$ – jacob smith Apr 8 '16 at 20:57
  • $\begingroup$ I figured out the first part about the functions, but I do'nt have a closed form for the series. $\endgroup$ – jacob smith Apr 8 '16 at 20:59
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Let $R=K[X_1,\dots,X_t]$, and $I\subset R$ be a homogeneous ideal. Then, $$H(R/I,n)=\dim_KR_n-\dim_KI_n,$$ where $R_n$, respectively $I_n$ is the $n$th homogeneous component of $R$, respectively $I$.

It's well known that $\dim_KR_n={{t+n-1}\choose{n}}$. On the other side, in your case $I=(f)$ with $f$ homogeneous of degree $d$, and then $\dim_KI_n={{t+n-d-1}\choose{n-d}}$. By summing up we get the Hilbert series $$H_{R/I}(z)=\frac{1}{(1-z)^t}-\frac{z^d}{(1-z)^t},$$ that is, $(1-z^d)H_R(z)$.

But you can also get this from the exact sequence $$0\to R(-d)\stackrel{f\cdot}\to R\to R/I\to0,$$ so there is no need to compute the Hilbert function of $R/I$ in order to find out its Hilbert series.

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