5
$\begingroup$

I'm trying to understand the non-commutative Koszul complex, as can be found in Anick's nice paper "Non-Commutative Graded Algebras and Their Hilbert Series", J. of Algebra 78, (1982) and I'm stuck at two points, which are just where the paper "jumps" from the commutative case to the non-commutative one.

Let me set first some notation and assumptions. For the commutative situation, we have: $\mathbf{k}$ is a field, $R$ a connected commutative graded $\mathbf{k}$-algebra; that is,

$$ R = \mathbf{k} \oplus R_1 \oplus R_2 \oplus \dots $$

in which every piece $R_n$ is a finite-dimensional $\mathbf{k}$-vector space.

Let $\theta_1, \dots , \theta_r \in R$ be a regular sequence. That is, the ideal generated by $\theta_1, \dots , \theta_r$ is smaller than $R$ and each $\theta_n$ is not a zero divisor in $R/(\theta_1, \dots , \theta_{n-1})$ for all $n$. (Also the $\theta_n$ are homogeneous of positive non-zero degree.)

First. The main idea in Anick's paper seems to be replacing the notion of non zero divisors, with the help of the following characterization:

$$ \theta_1, \dots , \theta_r \quad \text{is a regular sequence}\quad \Longleftrightarrow \quad R \cong \mathbf{k}[\theta_1, \dots , \theta_r] \otimes \frac{R}{(\theta_1, \dots , \theta_r)} \ . $$

Here:

  • $\mathbf{k}[\theta_1, \dots , \theta_r]$ is the polynomial algebra on $\theta_1, \dots, \theta_r$,
  • the isomorphism is as graded $\mathbf{k}$-vector spaces,
  • the tensor product is over $\mathbf{k}$.

My first question is about this isomorphism: Anick says it's "well known" and gives a reference: Stanley, "Hilbert Functions of Graded Algebras", Adv. in Math. 28 (1978). Ok, there the closest thing looking like this result is in page 63, where Stanley says: "This is essentially a well-known property..., though an explicit statement is difficult to find in the literature." And gives in turn references for a particular case.

So, ok, I'm trying to provide myself of some proof, with the help of what Anick does for the non-commutative case. You can easily produce a morphism of graded vector spaces

$$ \mathbf{k}[\theta_1, \dots , \theta_r] \otimes \frac{R}{(\theta_1, \dots , \theta_r)} \longrightarrow R $$

by sending each $\theta_i$ to itself in $R$. For the quotient part, you choose a $\mathbf{k}$-linear section of the projection $R \longrightarrow R /(\theta_1, \dots , \theta_r)$. Anick proves that you can pick no matter which section and the resulting morphism is an epimorphism. So far, so good.

Ok so, this is my first question: how do you prove that this morphism is also a monomorphism?

In another part of the paper, Anick uses an argument on dimensions: after all, we are dealing with finite dimensional vector spaces. So I'm trying to prove (first for the case $r=1$) that, degree-wise, what we have on both sides are vector spaces of the same dimension. It's a little bit messy (I hate counting!), but I think I got it. My only doubt is: is there a more simple, direct "well-known" proof?

Second. What's the problem with the notion of (two-sided) non zero divisor in the non-commutative case that forces Anick to replace it by that isomorphism? (Then he goes on talking about "strongly free" sets, which are those $\theta_1, \dots , \theta_r \in H$, such that you have an isomorphism

$$ H \cong \mathbf{k}\langle \theta_1, \dots , \theta_r \rangle \otimes \frac{H}{H(\theta_1, \dots , \theta_r)H} \ , $$

$H$ a connected non-commutative graded $\mathbf{k}$-algebra, $\mathbf{k}\langle \theta_1, \dots , \theta_r \rangle$ is the free associative algebra...)

$\endgroup$
4
$\begingroup$

First part: Let $\theta_1,\ldots,\theta_r$ be a regular $R$-sequence. Suppose that the kernel of $\mathbb{K}[\mathrm{x}_1,\ldots,\mathrm{x}_n]\otimes R/(\theta_1,\ldots,\theta_r)\rightarrow R$ is not trivial. Then there is a non trivial polynomial $f$ that is being mapped onto $0$. The image of $f$ in $R$ would satisfy an equation $\sum_\alpha a_\alpha \theta_1^{\alpha_1}\cdots \theta_r^{\alpha_r}=0$, where $a_\alpha\notin (\theta_1,\ldots,\theta_r)$. Thus $-a_\beta \theta_1^{\beta_1}\cdots \theta_r^{\beta_r}=\sum_{\alpha\neq \beta} a_\alpha \theta_1^{\alpha_1}\cdots \theta_r^{\alpha_r}$, where $\beta\neq 0$ is the biggest multiindex with respect to the lexiographical ordering. The equation would now imply that it is possible to increase the degree in the $i$-th component by just summing and multiplying monomials by $a_\alpha\notin (\theta_1,\ldots,\theta_r)$. But that is impossible, because $\theta_1,\ldots,\theta_r$ is a regular sequence. Hence the mapping $\mathbb{K}[\theta_1,\ldots,\theta_r]\otimes R/(\theta_1,\ldots,\theta_r)\rightarrow R$ is injective.

Second part: The characterization of regular sequences does not hold in the non-commutative polynomial ring. In other words, there is a sequence $\theta_1,\ldots,\theta_r$ that is regular (by transferring the meaning of regular directly to the non-commutative case), but that fails to be regular in the other sense (Anick's definition aka. strongly free). One can see that by using Theorem 3.1. Anick defines regular (strongly free) by $R\cong \mathbb{K}\langle \theta_1,\ldots,\theta_r\rangle\ast R/R(\theta_1,\ldots,\theta_r)R$ because of the following reasons: First: Suppose that we have a commutative regular local ring $R$, then it is well known that the sequence $\theta_1,\ldots,\theta_r$ is regular if and only if $H_0(K(\theta_1,\ldots,\theta_r)\cong R/(\theta_1,\ldots,\theta_r)$ and $H_n(K(\theta_1,\ldots,\theta_r))=0$ for $n>0$, where $K(\theta_1,\ldots,\theta_r)$ is the Koszul-complex with respect to $\theta_1,\ldots,\theta_r$. We want such results in the non-commutative situation. By defining regular as above and defining an appropriate Koszul-complex in the non-commutative situation we get a similar result in form of Theorem 2.9. Second: Under certain circumstances (see Theorem 3.1) being regular (strongly free) has a combinatorial characterization, which is later used to prove the main result (Theorem 3.8).

$\endgroup$
  • $\begingroup$ Thank you very much for your help. In case I'm using this fact -or its proof- I would like to credit you with, even if it seems to be "well-known" (and nowhere to be found in the literature; its proof, I mean). Or at least acknowledge your help. How should I? $\endgroup$ – Agustí Roig Oct 6 '17 at 7:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.