Timeline for completion of category is idempotent
Current License: CC BY-SA 2.5
17 events
| when toggle format | what | by | license | comment | |
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| Jan 4, 2010 at 0:08 | comment | added | Martin Brandenburg | yes. in fact, there is a ring homomorphism. | |
| Jan 3, 2010 at 23:05 | comment | added | Pete L. Clark | @Martin: "there exists an injective homomorphism $\prod_{p \text{prime}} \mathbb{Z}/p \to \mathbb{C}$". Certainly not: if so, it would restrict to an injection on the direct sum, and we would have an embedding of an infinite torsion abelian group into a torsionfree abelian group. (I didn't really follow the whole discussion, but maybe you meant "nontrivial" instead of "injective"?) | |
| Jan 3, 2010 at 22:38 | vote | accept | Martin Brandenburg | ||
| Jan 3, 2010 at 22:37 | comment | added | Martin Brandenburg | ah, alright. I've edited my question. perhaps the example shows that my original definition was not natural. | |
| Jan 3, 2010 at 21:54 | history | edited | Reid Barton | CC BY-SA 2.5 |
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| Jan 3, 2010 at 21:05 | comment | added | S. Carnahan♦ | @Martin: The morphism you describe does not arise from a diagram in the category A, so it is not a counterexample. The existence of morphisms in $\overline{A}$ that are not limits of maps in A is the main reason why the answer to your question is negative, and the nontriviality of W is one manifestation of that. | |
| Jan 3, 2010 at 20:23 | comment | added | Martin Brandenburg | @david: "the diagram D decomposes correspondingly as $D_{fin} \oplus D_\mathbb{Q}$." this is wrong, I gave a counterexample. | |
| Jan 3, 2010 at 20:15 | comment | added | Harry Gindi | Is there a sufficient condition for when it holds? | |
| Jan 3, 2010 at 19:11 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Jan 3, 2010 at 18:48 | comment | added | t3suji | Can I suggest the following modification of your modification :) Each group $G$ in $A$ can be written as a direct sum (=product) $G_1\oplus G_2$ where $G_1$ is finite and $G_2$ is a vector space: one of these two subgroups always vanish. Morphisms of $A$ respect such decomposition, thus a projective limit in $A$ can be written as a projective limit of products=product of projective limits. This is exactly what you do. | |
| Jan 3, 2010 at 18:08 | comment | added | David E Speyer | And, yes, Scott is right. X is isomorphic to Q_p by the obvious map. I'm going to wait until I know the argument is right before I start improving it. | |
| Jan 3, 2010 at 18:07 | comment | added | David E Speyer | I edited the answer to address Reid's objection. I don't follow what Martin is getting at; have my edits addressed your point? | |
| Jan 3, 2010 at 18:06 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Jan 3, 2010 at 17:26 | comment | added | S. Carnahan♦ | Couldn't you just use Q_p instead of defining X? | |
| Jan 3, 2010 at 17:17 | comment | added | Martin Brandenburg | I have a problem with the decomposition of your limits: it is clear that a homomorphism from a product of finite abelian groups to a vector space vanishes in the direct sum; but it does not have to vanish on the product. in fact, using transfinite methods, there exists an injective homomorphism $\prod_{p \text{prime}} \mathbb{Z}/p \to \mathbb{C}$. | |
| Jan 3, 2010 at 16:55 | comment | added | David E Speyer | For those concerned about set theoretic issues: I only need a countable sequence of finite abelian groups to get to Z_p. As a tensor product of sets, X is a set. Hom(X,Q) is a subset of 2^{X \times Q}, so it is a set, and W is a subset of Hom(X,Q). | |
| Jan 3, 2010 at 16:47 | history | answered | David E Speyer | CC BY-SA 2.5 |