My question asks me to show that $\{y_1+x_1y_2+x_1^2y_3,y_1+x_2y_2+x_2^2y_3,y_1+x_3y_2+x_3^2y_3\}$ is a regular sequence in $k[y_1,y_2,y_3,x_1,x_2,x_3]$ (the order is lexico)

Going exactly in the way I was suggested in my previous question: The Grobner basis of $I=\left<f_1,f_2,f_3\right>$ that I got is:

  1. $y_1+ y_2x_3 + y_3x_3^2$

  2. $-y_2x_3 + y_2x_1 - y_3x_3^2 + y_3x_1^2$

  3. $-y_2x_3 + y_2x_2 - y_3x_3^2 + y_3x_2^2$

  4. $y_3x_1^2x_2 - y_3x_1^2x_3 - y_3x_1x_2^2 + y_3x_1x_3^2 + y_3x_2^2x_3 - y_3x_2x_3^2$

The monomial order is $y_1 > y_2 > y_3 > x_1 > x_2 > x_3$

What is a Hilbert series? How do you compute it with the help of Grobner techniques? I need some worked out example which can help me with this? I went through Stanley's paper but I think that will need a strong background in commutative algebra.

I thought it would be better to ask a new question.I am also attaching the link to my previous question

Here is the link


1 Answer 1


I will try to give a simple overview about Hilbert series, but I fear that it will not work without some background in commutative algebra.

The Hilbert Series is a formal power series that one can define for a graded $k$-algebra $R$. First of all, you therefore need a graded $k$-algebra.

You did not define exactly what $k$-algebra you want to consider, but I guess it is about the quotient ring $k[y_1, y_2, y_3, x_1, x_2, x_3] / I$. However, this is not a graded ring, as you only get a graded ring if you mod out an ideal generated by homogeneous polynomials. In Algebraic Geometry, one would now consider the homogenization of the ideal $I$, which would be \begin{equation*} \langle y_3x_3^2 + y_2x_3z +y_1z^2, y_3x_2^2 +y_2x_2z +y_1z^2, y_3x_1^2 +y_2x_1z +y_1z^2, y_2x_2^2x_3 -y_2x_2x_3^2 +y_1x_2^2z -y_1x_3^2z, y_2x_1^2x_3 -y_2x_1x_3^2 +y_1x_1^2z -y_1x_3^2z, y_2x_1^2x_2 -y_2x_1x_2^2 +y_1x_1^2z -y_1x_2^2z, y_1x_1^2x_2 -y_1x_1x_2^2 -y_1x_1^2x_3 +y_1x_2^2x_3 +y_1x_1x_3^2 -y_1x_2x_3^2 \rangle \end{equation*} I have computed that using a computer algebra system, as it obviously is quite complicated.


So for now, consider a simpler example: \begin{equation*} I = \langle xy - z^2 \rangle \subseteq k[x,y,z] \end{equation*} This ideal is generated by homogeneous polynomials (i.e. polynomials where each monomial has the same degree), so $k[x,y,z]/I$ is graded. Now the Hilbert Function $\mathrm{HF}_{k[x,y,z]/I}(n)$ is the dimension (as $k$-vector space) of the set of all homogeneous element of degree $n$.

In this example, have that all the homogeneous elements of degree $0$ are exactly $k$, so $\mathrm{HF}_{k[x,y,z]/I}(0) = 1$. Further, the homogeneous elements of degree exactly 1 are $\lambda_1 x + \lambda_2 y + \lambda_3 z$ for some $\lambda_1, \lambda_2, \lambda_3 \in k$, so $\mathrm{HF}_{k[x,y,z]/I}(1)= 3$. The case $n = 2$ is more interesting, as the homogeneous elements of degree 2 are $\lambda_1 x^2 + \lambda_2 xy + \lambda_3 xz + \lambda_4 y^2 + \lambda_5 yz$. We do not include $\lambda_6 z^2$ here, as $z^2 = xy \mod I$. Thus $\mathrm{HF}_{k[x,y,z]/I}(2) = 5$.

The Hilbert series $HS_{k[x,y,z]/I}(z)$ is now the formal power series \begin{equation*} HS_{k[x,y,z]/I}(z) = \sum_{n \in \mathbb{N}} \mathrm{HF}_{k[x,y,z]/I}(n) z^n \end{equation*} In our case, one can show \begin{equation*} HS_{k[x,y,z]/I}(z) = 1 + 3z + 5z^2 + 7z^3 + ... = \frac {1 + z} {(1 - z)^2} \end{equation*} In particular, one can always represent the Hilbert function as a fraction of polynomials in $z$.

Computing it

There is a lot of theory about Hilbert series, for example if you have a regular sequence $f_1, ..., f_m$ of homogeneous polynomials, then \begin{equation*} \mathrm{HS}_{k[x_1, ..., x_n]/\langle f_1, ..., f_m \rangle} = \frac {(1 - z^{\deg f_1}) ... (1 - z^{\deg f_m})} {(1 - z)^n} \end{equation*} Another theoretical result is that $\mathrm{HF}_{k[x_1, ..., x_n]/I} = \mathrm{HF}_{k[x_1, ..., x_n]/J}$ for the leading-term ideal $J$ of $I$ with respect to a degree-compatible term ordering (still, $I$ has to be generated by homogeneous polynomials).

If you have a Gröbner basis, then the leading term ideal is just the ideal generated by the leading terms of each polynomial in the Gröbner basis, so this can be used to compute the Hilbert Series.

In any case, you will have to work with homogeneous polynomials.

  • $\begingroup$ So to prove that the polynomials of $f_1,f_2,f_3$ are in a regular sequence we need to show that the hilbert series which is calculated is of the above form as that you have mentioned? $\endgroup$
    – Antimony
    Jul 2, 2021 at 13:48
  • $\begingroup$ is there some textbook from which I can get an idea about the hilbert series(I dont know what is graded algebra , also some exercises on hilbert series).Also if you check my previous post it was suggested that I could make the polynomials homogeneous of degree 3 by defining $deg(y_1)=3...$ $\endgroup$
    – Antimony
    Jul 2, 2021 at 13:51
  • $\begingroup$ As long as $f_1, f_2, f_3$ are not homogeneous, it does not make sense to take about the Hilbert series. If you make them homogeneous by setting $\deg(y_1) = 3, \deg(y_2) = 2$, you will get a Hilbert series. However, I am not so fluent with graded algebras not of the form $k[x_1, ..., x_n]/I$, and am unsure if the theorems I mentioned still hold. In any case, you can always compute the Hilbert series by considering the vector space dimensions (probably a lot of long computations). $\endgroup$
    – Feanor
    Jul 2, 2021 at 14:22
  • $\begingroup$ The only textbook I know on the subject is the one written by my professor (Kreuzer, Robbiano: "Computational Commutative Algebra 1" & 2), but this is not likely to be the best choice for you. For graded algebras/rings, you might look at this question math.stackexchange.com/questions/50164/…. In particular, it helps to understand how the polynomial ring $k[x_1, ..., x_n]$ becomes a graded ring. $\endgroup$
    – Feanor
    Jul 2, 2021 at 14:25

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