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...)