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The problem is:

Let $F$ be a field, and let $R = F[x_1, \ldots, x_r]$, the polynomial ring over $F$. Consider the $R$-module $M = R/(x_1, \ldots, x_r) \cong F$. Find a resolution of $M$ by free $R$-modules.

I understand that I need to find a sequence of maps $\alpha_0,\alpha_1, \ldots$ such that im($\alpha_{i+1}) =$ ker$(\alpha_i)$. The first map $\alpha_0: R \rightarrow M$ is simply an inclusion modulo $x_1, \ldots, x_r$. An element of the kernel of $\alpha_0$ has the form $f = \sum_i^r f_ix_i$, with $f_i \in R$. The next map $\alpha_1: R^r \rightarrow R$ sends $(f_1,\ldots,f_r)$ to $x_1f_1 + \cdots +x_rf_r$. This clearly is surjective on ker($\alpha_0)$.

What I'm having trouble understanding is how to get the next map, which of course must be surjective on the kernel of $\alpha_1$. I think that the basis elements of ker($\alpha_1$) have the form $\{x_je_i - x_ie_j \mid 1 \leq i < j \leq r\}$, but I'm not sure how to prove this. Also, from what free module does $\alpha_2$ map onto $R^r$? How do I proceed and get the rest of the resolution?

share|cite|improve this question
What about the Koszul complex? – user26857 May 7 '14 at 21:11
Recommend reading Peeva's "Graded Syzygies". – Eric Towers Jul 29 '15 at 20:37

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