Consider the polynomial ring $S = C[x_1,\ldots,x_n]$. Let $$ ‎‏0 ‎‎\rightarrow‎‎‏ ‎E_{n-1} ‎‎\rightarrow ‎\cdots ‎‎‎\rightarrow ‎E_1 ‎‎‎\rightarrow ‎E_0 \rightarrow I \rightarrow 0‎ ‎,‎ $$ be a minimal free resolution of an ideal $I$ of $S$, where $$ E_p = \bigoplus _j S(-a_{pj}) . $$ The regularity of $I$ is $\max\lbrace a_{pj} - p \rbrace $. Let $m = (x_1,x_2,\ldots,x_n)$ be a maximal ideal of the ring $S = C[x_1,\ldots,x_n]$. Why is the regularity of $I$ the largest $q$ for which $\operatorname{Tor}^S_p(I,S/\mathfrak{m})_{p+q} \neq 0$ for some $p$?

  • $\begingroup$ $\mathfrak{m}=(x_1,x_2,...,x_n)$ is maximal ideal of polynomial ring $S=C[x_1 , ... ,x_n]$ $\endgroup$
    – A.B.
    Commented Jul 17, 2014 at 19:35

1 Answer 1


By the construction of derived functors, $\mathrm{Tor}(I,A)$ is the homology of the complex $$0 \to E_{n-1} \otimes A \to \cdots \to E_1 \otimes A \to E_0 \otimes A.$$ (All tensor products are $\otimes_S$.) In the particular case that $A = S/\mathfrak{m} =:k$, we have $E_p \otimes k= \oplus_j k(-a_{pj})$, so $\mathrm{Tor}_p(I, k)$ is the homology of the complex of $k$-vector spaces defined by taking the matrices in your original complex and reducing all of their elements modulo $\mathfrak{m}$. The assumption that your complex is minimal is equivalent to saying that all of the maps are $0$ modulo $m$, so you are just taking the homology of a complex where all maps are zero, and you get $$\mathrm{Tor}_p(I, k) \cong \oplus_jk(-a_{pj})$$ as a graded $k$-vector space. Now use the definition of CM regularity.


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