Timeline for Understanding the modules of semiprimitive rings
Current License: CC BY-SA 2.5
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| Jul 4, 2010 at 1:25 | history | edited | Jamie Vicary | CC BY-SA 2.5 |
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| Jul 4, 2010 at 1:14 | comment | added | Jamie Vicary | In that case, an example of a list of primitive rings which together give rise to a subdirect product for the function ring are the values $V_x$ of the functions at a given point $x$ in [0,1], which are one-dimensional vector spaces. So we have an injective ring homomorphism $f: ([0,1],C) \to \Pi_x V_x$. Clearly in this case $([0,1],C)$ is indeed not a very big subalgebra, as you say. | |
| Jul 3, 2010 at 19:38 | comment | added | Yemon Choi | I still don't understand your revised version of the question. The algebra of continuous complex-valued functions on $[0,1]$ is semiprimitive. What are the primitive rings you're thinking of in this case? That's before we get onto examples such as the disc algebra (algebra of all continuous functions on closed unit disc which are analytic in the interior). | |
| Jul 3, 2010 at 19:21 | history | edited | Jamie Vicary | CC BY-SA 2.5 |
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| Jul 3, 2010 at 18:33 | answer | added | Bruce Westbury | timeline score: 1 | |
| Jul 3, 2010 at 18:32 | answer | added | Robin Chapman | timeline score: 3 | |
| Jul 3, 2010 at 18:21 | comment | added | Yemon Choi | just to add to my last comment: my intuition (which is gained from the Banach setting, and hence could well be in error in this setting) is that $f(R)$ might not be a very big subalgebra of $\prod_i R_i$, and so knowing the module theory of the big guy is insufficient for understanding the module theory of the subalgebra | |
| Jul 3, 2010 at 18:18 | comment | added | Yemon Choi | In the case of Banach algebras, and if we replace "module" by "Banach module" in the most obvious way, then the answer is no: when $R$ is a commutative Banach algebra then the map $f$ is just the Gelfand transform, and the problem is somehow that $R$ (or even $f(R)$) possesses many more modules than the direct product of simples. | |
| Jul 3, 2010 at 18:18 | comment | added | Robin Chapman | You want the direct sum of the simple modules for your faithful semisimple modules, not the direct product. (Think about what happens when $R=\mathbb{Z}$.) | |
| Jul 3, 2010 at 18:14 | history | asked | Jamie Vicary | CC BY-SA 2.5 |