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Let $k$ be a field and $S=k[x_1,\dots,x_r]$ the polynomial ring in $r$ indeterminates. Let $f_1,\dots,f_n$ be a sequence of $n\le r$ forms of degrees $d_1,\dots,d_n$. If $f_1,\dots,f_n$ is a regular sequence, then it is easy to see that the Hilbert series of the quotient is \begin{align} H_{S/(f_1,\dots,f_n)}(t) = \frac{(1-t^{d_1})\cdots(1-t^{d_n})}{(1-t)^r}, \, \, \, (1). \end{align}

Question: Is the converse statement true? I.e., if (1) is true, is it the case that $f_1,\dots,f_n$ is a regular sequence? If yes, how do we see that? (I can see that if $r=n$.)

Manos
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    The Hilbert series also tells you the dimension of a ring. The order of the pole of the rational function at $(1-t)$ is the dimension. Once you have a Hilbert series as in (1), then each factor $(1-t^{d_i})$ has only one factor of $(1-t)$. Therefore, the order is $r - n$ which is $\dim S/(f_1,\dots,f_n)$. This implies that height (codimension) of $f_1, \dots, f_n$ is $n$. – Youngsu Nov 28 '15 at 08:25
  • Let me ask you Manos, how did you obtain case $r=n$? – Daniel Mar 30 '16 at 14:53
  • @SolidSnake: The formula is by induction on $n$. Can you derive it for $n=1$? – Manos Mar 30 '16 at 20:38
  • Manos, what formula? I don't see how to prove this: if the sequence has that Hilbert Series you wrote, then it is a regular sequence. Not even in the case $r=n$. The converse is clear to me. – Daniel Mar 30 '16 at 23:24
  • Sure! thanks. I will begin by stating yours: Is the converse statement true? I.e., if (1) is true, is it the case that $f_1,…,f_n$ is a regular sequence? If yes, how do we see that? (I can see that if $r=n$) ... So, my question is the same as yours, how can this implication be proven? (this implication: having the specified Hilbert series implies being regular). According to what you wrote, it seems that you could prove this implication in the case $n=r$. How could you do this? – Daniel Mar 31 '16 at 01:04
  • @SolidSnake: I see ok. For $n=r$, if (1) is true, then the Hilbert series of $S/(f_1,\dots,f_r)$ is a polynomial, and so $S/(f_1,\dots,f_r)$ has zero Krull dimension. Suppose that $f_1,\dots,f_i$ is regular, but $f_{i+1}$ is a zero divisor of $S/(f_1,\dots,f_i)$. Define $\bar{S} = S/(f_1,\dots,f_{i+1})$. Then $\dim \bar{S} = \dim S/(f_1,\dots,f_i) = r - i$. Now $\dim \bar{S}$ is equal to $height(\mathfrak{m}/(f_1,\dots,f_{i-1}))$. Since the height is $r-i$, taking the quotient with $r-i-1$ elements (the $f_{i+2},\dots,f_{r})$ can result in a height of at least $1$. But that's a contradiction. – Manos Mar 31 '16 at 01:43
  • @SolidSnake: Of course a similar argument can be used to answer the general case $n \le r$, but at the time i felt confident about the case $r=n$. – Manos Mar 31 '16 at 01:46
  • @SolidSnake: Overall, i think it is a great exercise to prove that for homogeneous polynomials $f_1,\dots,f_n$, they form a regular sequence if and only if $\dim S/(f_1,\dots,f_n) = r-n$. – Manos Mar 31 '16 at 01:53
  • Thanks for the proof, I will check it out. Also, I already tried to prove that exercise you mentioned (since it implies the original proposition). However, I was not able to do it neither. Can you give me a hint about how to solve this one? I'm studying dimension theory by myself, and I'm not still very trained with the theory. – Daniel Mar 31 '16 at 03:21
  • @SolidSnake: You need a series of Lemmas to get to that statement. For example you need everything that user26857 uses in his answer. So my suggestion would be try to study his answer and fill in the gaps. – Manos Mar 31 '16 at 12:45
  • @Manos Ok! I will. Thank you very much for your suggestions and the proof for the case $r=n$ you provided. – Daniel Apr 02 '16 at 01:51
  • @SolidSnake: You're welcome :) – Manos Apr 02 '16 at 02:16

1 Answers1

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So, the grade of the ideal generated by $f_1,\dots,f_n$ is $n$. Now the question is whether this entails that $f_1,\dots,f_n$ is a regular sequence. Since $f_i$ are homogeneous of degree $d_i\ge1$ they belong to the maximal irrelevant ideal $\mathfrak m=(x_1,\dots,x_r)$ of $S$, and $(f_1,\dots,f_n)S_{\mathfrak m}$ is an ideal of grade $n$ in $S_{\mathfrak m}$. This shows that $f_1,\dots,f_n$ form a regular sequence in $S_{\mathfrak m}$ (why?), so they form a regular sequence in $S$ (why?).

An alternative approach. Localize $S$ at $\mathfrak m=(X_1,\dots,X_r)$. The dimension equality remains the same since the dimension of a graded $k$-algebra is the height of its irrelevant maximal ideal. (Here one uses that $f_i$'s are homogeneous.) By the dimension equality the images of $f_1,\dots,f_n$ in $S_{\mathfrak m}$ form a system of parameters, and since $S_{\mathfrak m}$ is a CM local ring they form a regular sequence (in any order) (see Eisenbud, Cor. 17.12). Now we have to show the same property for $f_1,\dots,f_n$ in $S$. This follows easily since they are homogeneous.

user26857
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  • That's a great answer! however, I was wondering, is there an easy way to prove this? or, let's not say easier, but simpler in terms of mathematical tools involved. I'm dealing with this characterization problem and the only references I've seen that deal with this are Lang's Algebra and Froberg in its Groebner Basis book. I see that your approach is that of Lang, but Froberg states this as an exercise (Ex 7, Page 137) with no previous 'big' tools or anything. – Daniel Mar 30 '16 at 06:08
  • @SolidSnake When I've wrote this answer I wasn't aware of any other proof. Now that you mentioned this I remember a paper of Froberg where maybe you can find his approach; see here. – user26857 Mar 30 '16 at 08:08
  • It seems that Lang only proves the easiest direction, Theorem 6.6 in page 436. I already have seen that Froberg paper a couple of times, I think it's time to give it a deeper study! thanks a lot for the suggestion :) Anyway, I'll do my best to understand your proof, which I find elegant and sophisticated. – Daniel Mar 30 '16 at 14:18
  • There is also a "proof" (or, at least, one possible approach) in the Ph.D thesis of Magali Bardet: http://magali.bardet.free.fr/these_Bardet.pdf exactly in the page 25, where she says: "where $K$ is the kernel of multiplication bt $f_i$, then the Hilbert series of $I$ is necessarily null and therefore $K=0$", which is something I really can't get! why is this? – Daniel Mar 30 '16 at 14:52
  • I also don't get her argument. – user26857 Mar 30 '16 at 15:03