What makes $\textrm {Tor}^ R_ 1(R/I,I)$ free?
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- by Shiro Goto and Naoyoshi Suzuki PDF
- Proc. Amer. Math. Soc. 112 (1991), 605-611 Request permission
Abstract:
Let $I$ be a nonprincipal ideal in a Noetherian local ring $R$ and let ${H_1}(I)$ be the first homology module of the Koszul complex $K.(I)$ associated with a minimal basis of $I$. Then $T: = {\text {Tor}}_1^R(R/I,I)$ is a free $R/I$-module if and only if both the $R/I$-modules $I/{I^2}$ and ${H_1}(I)$ are free. When this is 2 2 the case, we have a canonical decomposition $T \cong {\Lambda ^2}(I/{I^2}) \oplus {H_1}(I)$ as well as the equality ${\text {rank}_{R/I}}T = {\beta _2}(R/I)$. (Here ${\beta _2}(R/I)$ denotes the second Betti number of the $R$-module $R/I$.) Some consequences are discussed too.References
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Additional Information
- © Copyright 1991 American Mathematical Society
- Journal: Proc. Amer. Math. Soc. 112 (1991), 605-611
- MSC: Primary 13D02; Secondary 13D05
- DOI: https://doi.org/10.1090/S0002-9939-1991-1052871-0
- MathSciNet review: 1052871