Let $M$ be a finitely generated graded $k[x]$-module for some field $k$.

Let $$\cdots \rightarrow P_2 \rightarrow P_1 \rightarrow P_0 \rightarrow M \rightarrow 0$$

be a minimal graded projective resolution for $M$. In particular each $P_i$ is projective and hence free since $k[x]$ is a PID. I want to show that each map $P_{i+1} \rightarrow P_i$ maps each homogeneous $e\in P_{i+1}$ to some $(p_1, ... , p_r)$ where each $p_i$ is homogeneous and non-constant. The homogeneous part of this is by definition of a graded map, but the non-constant part is stumping me. I think I'm missing something obvious.

Suppose $e \mapsto (p_1, ... , p_r)$ where say $p_i$ is constant. Then since $(p_1, ... , p_r)$ is in the kernel of the next map, we have $p_i\partial(e_i) = -\sum p_j \partial(e_j)$, where $e_i$ is the $i$th standard basis. This should contradict minimality, and it does if $\{\partial(e_i)\}$ is the generating set of $P_{i-1}$ by which we defined the resolution, but is this necessarily so?

• Do you know that in your case, $P_n=0,n>1$? – Mohan Sep 20 '17 at 15:46
• Fair point. What if I change it to $k[x,y]$, or any other polynomial ring? – IAlreadyHaveAKey Sep 20 '17 at 22:05
• As you said, this is just minimality. Let us use the fact that projectives are free (not really necessary, since graded projectives are free easily). Then, minimal resolution will imply that the matrix corresponding to any of the maps $P_n\to P_{n-1}$ will have all entries homogeneous polynomials of positive degree. – Mohan Sep 20 '17 at 22:26
• Why though? I don't see why the matrix will have positive degree entries. – IAlreadyHaveAKey Sep 20 '17 at 22:31
• If you have a degree zero entry, it can be assumed to be a non-zero constant, since zero has any degree you wish. If it is constant and non-zero, it follows that the resolution is minimal since you can remove a free rank one part from both $P_n$ and $P_{n-1}$, contradicting minimality. – Mohan Sep 21 '17 at 2:02