# Hilbert series characterization of regular sequences

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 $r=n$.)

• 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 8:25
• 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 1:04

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.
• There is also a "proof" (or, at least, one possible approach) in the Ph.D thesis of Magali Bardet: 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