Timeline for Is $\mathbb{Z}_p$ flat $\mathbb{Z}_pG$-module for a finite $p$-group $G$?
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4 events
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| Jun 27, 2011 at 17:38 | comment | added | Mariano Suárez-Álvarez | @Jack, the same is true without commutativity if you are consistent about sides :) | |
| Jun 27, 2011 at 16:37 | comment | added | Jack Schmidt | $M \otimes_R R/I \cong M/IM$ for all commutative (unital, associative) rings $R$, ideals $I$ of $R$, and modules $M$ over $R$. Basically this is because $m\otimes 0 = m \otimes i = mi \otimes 1$ (so the tensor product is a quotient of M/IM, and clearly there is a bilinear map from M x R/I to M/IM given by multiplication, so the quotient is in fact an isomorphism). | |
| Jun 27, 2011 at 14:41 | comment | added | David White | Why is $I \otimes \mathbb{Z}_{(p)}G/I \cong I/I^2$? This isn't at all a question of your correctness, I'm simply new to this material and trying to learn. | |
| Jun 27, 2011 at 14:25 | history | answered | JSE | CC BY-SA 3.0 |