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What makes $ {\rm Tor}\sp R\sb 1(R/I,I)$ free?


Authors: Shiro Goto and Naoyoshi Suzuki
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
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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.


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DOI: https://doi.org/10.1090/S0002-9939-1991-1052871-0
Article copyright: © Copyright 1991 American Mathematical Society

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