Timeline for Can ⨁_I A be isomorphic to ∏_I A for infinite I?
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
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| Dec 16, 2010 at 18:29 | comment | added | Anton Geraschenko | @François: You're right about Maurizio's argument (unless I'm also missing something). I got ahead of myself. | |
| Dec 16, 2010 at 17:31 | comment | added | François Brunault | @Maurizio. It seems that you prove that the two objects are not isomorphic as (non-unital) rings, or am I missing some argument? Anyway, maybe this nice idea can be adapted to show the modules are not isomorphic. | |
| Dec 16, 2010 at 16:33 | comment | added | Anton Geraschenko | @Maurizio: That's a wonderful solution! Please consider posting it as an answer. | |
| Dec 16, 2010 at 16:31 | comment | added | Anton Geraschenko | @Theo: I'm failing to connect the dots. What is the analogue for modules that you're thinking of, and why does it suffice to prove it over $\mathbb Z$? | |
| Dec 16, 2010 at 16:23 | vote | accept | Anton Geraschenko | ||
| Dec 16, 2010 at 11:59 | comment | added | Maurizio Monge | Your (cardinality) proof for fields works well also for unital rings if you restrict yourself to computing the cardinality of idempotent elements (i.e. those $x$ such that $x^2$ = $x$). They are all vectors with $0-1$-components, with only a finite number of $1$'s in the direct sum. Consequently they have respactively cardinality $|I|$ and $2^{|I|}$ in the sum and in the product. | |
| Dec 15, 2010 at 19:09 | comment | added | Theo Johnson-Freyd | In the case of modules of rings, you can also try to use the following cheat: It suffices to prove the result over $\mathbb Z$. | |
| Dec 15, 2010 at 17:16 | comment | added | Anton Geraschenko | @François: Thanks for the name Erdos-Kaplansky. @darij: what is messed about about my Case 1? | |
| Dec 15, 2010 at 11:27 | comment | added | François Brunault | @darij grinberg : You're right, there could be other choice of isomorphism. So the question reduces to whether the dimensions over $k$ are always equal. | |
| Dec 15, 2010 at 11:23 | answer | added | François Brunault | timeline score: 9 | |
| Dec 15, 2010 at 11:11 | comment | added | darij grinberg | This is pretty easy: take $I$ to be non-finitely generated. The question is whether there cannot be some obscure different isomorphism. | |
| Dec 15, 2010 at 10:52 | comment | added | François Brunault | According to an exercise in Bourbaki, there exists a commutative ring $A$ together with an ideal $I$ such that $A^{\mathbf{N}} \otimes (A/I) \to (A/I)^{\mathbf{N}}$ is not injective. I haven't done the exercise though... | |
| Dec 15, 2010 at 10:38 | comment | added | François Brunault | Your argument in Case 2 also proves that the canonical map $(\prod_{i \in I} k) \otimes A \to \prod_{i \in I} A$ is injective. | |
| Dec 15, 2010 at 10:36 | comment | added | darij grinberg | Your Case 1 is pretty much messed up, I must say. Otherwise, very nice question. Wouldn't you have asked it, I would still believe that the statement of your additional question is obviously correct... | |
| Dec 15, 2010 at 10:29 | comment | added | François Brunault | If k is a field, the dimension of $k^I$ as a $k$-vector space is the same as its cardinality : this is the Erdos-Kaplansky theorem. The proof is similar to the arguments you've given, I think. See fr.wikipedia.org/wiki/… for a proof. | |
| Dec 15, 2010 at 10:27 | history | edited | Anton Geraschenko | CC BY-SA 2.5 |
fixed error
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| Dec 15, 2010 at 10:05 | history | asked | Anton Geraschenko | CC BY-SA 2.5 |