Timeline for Does the functor of taking invariants commute with tensor products? [closed]
Current License: CC BY-SA 3.0
14 events
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| Sep 22, 2015 at 13:31 | comment | added | David E Speyer | There are a lot of comments about confusions in the way the question is phrased, but I suspect the answer is "no" however the ambiguities are resolved. Let $R =\mathbb{C}$ with the trivial action of $G=C_2$. Let $M$ and $N$ be $\mathbb{C}$ with $G$ acting by $-1$. Then $M^G=N^G=0$ but $(M \otimes_R N)^G = M \otimes_R N$. | |
| Sep 22, 2015 at 12:15 | history | closed |
YCor Olivier Chris Godsil Stefan Kohl♦ Benjamin Steinberg |
Needs details or clarity | |
| Sep 22, 2015 at 12:05 | comment | added | Jeremy Rickard | Simple example: if $S$ is a ring, then $C_2$ acts on the ring $S\times S$ by swapping the two factors. $M=S\times\{0\}$ is an $S\times S$-module, but I think it's fairly clear that there's no action of $C_2$ on it that is "induced" in any reasonable way by the action on $S\times S$. | |
| Sep 22, 2015 at 11:49 | comment | added | Todd Trimble | Suresh, if $g$ acts as a ring automorphism and thus preserves $1 \in R$, then the composite you described in your last comment is the identity on $M$. | |
| Sep 22, 2015 at 11:47 | review | Close votes | |||
| Sep 22, 2015 at 12:19 | |||||
| Sep 22, 2015 at 11:25 | comment | added | Jason Starr | @Suresh. For an $R$-module homomorphism $\phi:N\to N$, there is an induced $R$-module homomorphism $1\otimes \phi:M\otimes_R N \to M\otimes_R N$. However, the ring automorphism induced by $g$ is not an $R$-module homomorphism of $R$. | |
| Sep 22, 2015 at 11:22 | comment | added | Jason Starr | There is no meaning to "$G$ is a group acting on a commutative ring $R$, inducing an action on each $R$-module". I advise you to read about "$G$-linearizations" in a reference such as "Geometric Invariant Theory", Mumford, Fogarty, Kirwan. For an action of $G$ on $R$, there are some $R$-modules that will admit no compatible $G$-action, e.g., if the support of the module is not a $G$-invariant subset of $\text{Spec}(R)$. On the other hand, modules may admit many different compatible $G$-actions. | |
| Sep 22, 2015 at 11:22 | comment | added | Suresh | If $g:R\longrightarrow R$ is a morphism of rings, for an $R$-module $M$, we have $g:M\cong M\otimes_RR\overset{1\otimes g}{\longrightarrow}M\otimes_RR\cong M$. Am I missing something? | |
| Sep 22, 2015 at 11:21 | comment | added | Jeremy Rickard | How does an action of $G$ on $R$ induce an action on an $R$-module? | |
| Sep 22, 2015 at 11:14 | comment | added | Suresh | Hi, I meant to say that each $g\in G$ gives a ring automorphism of $R$. | |
| Sep 22, 2015 at 11:11 | comment | added | Todd Trimble | When you say $G$ acts on $R$, do you just mean on the underlying additive group, or what? | |
| Sep 22, 2015 at 11:01 | history | edited | Suresh | CC BY-SA 3.0 |
edited title
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| Sep 22, 2015 at 10:57 | review | First posts | |||
| Sep 22, 2015 at 11:06 | |||||
| Sep 22, 2015 at 10:54 | history | asked | Suresh | CC BY-SA 3.0 |