Remote Access Proceedings of the American Mathematical Society
Green Open Access

Proceedings of the American Mathematical Society

ISSN 1088-6826(online) ISSN 0002-9939(print)



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
MathSciNet review: 1052871
Full-text PDF

Abstract | References | Similar Articles | Additional Information

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 [Enhancements On Off] (What's this?)

Similar Articles

Retrieve articles in Proceedings of the American Mathematical Society with MSC: 13D02, 13D05

Retrieve articles in all journals with MSC: 13D02, 13D05

Additional Information

Article copyright: © Copyright 1991 American Mathematical Society

American Mathematical Society