Timeline for Geometric interpretation of a (standard) commutative algebra fact
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2013 at 15:50 | answer | added | answer_bot | timeline score: 3 | |
| Dec 11, 2013 at 20:37 | comment | added | Karl Schwede | So as already mentioned, it shows that if $R \subseteq S$ is an extension of rings such that $S$ is a finite $R$-module, then $S$ is integral over $R$. Indeed, take $S = M$ and $I = R$ and let $\phi$ be multiplication by some $x \in S$. I'm not quite sure what integral means geometrically (except for the usual going up type stuff). The other big application is Nakayama's lemma, which does have geometric interpretations certainly. Perhaps in this generality though, interpretations are harder? | |
| Dec 11, 2013 at 17:16 | comment | added | Qfwfq | @Adam: perhaps Olivier meant that once you have a geometric intuition for Caylay-Hamilton, then you also have a geometric intuition about the question in my OP. | |
| Dec 11, 2013 at 16:17 | comment | added | Adam Gal | @Olivier this is not exactly true since Cayley Hamilton says the degree can be taken at most 2, and here there is no such claim. | |
| Dec 11, 2013 at 13:08 | comment | added | Qfwfq | @Olivier: I agree | |
| Dec 11, 2013 at 13:05 | comment | added | Olivier | This amounts to finding a geometric interpretation of the Cayley-Hamilton theorem and I must say I already don't see one for $M_2(\mathbb R)$ acting on $\mathbb R^2$. | |
| Dec 11, 2013 at 11:10 | comment | added | Qfwfq | Inegral ring extensions... ok, I'll have a look to that place in Atiyah-MacDonald, thank you. | |
| Dec 11, 2013 at 10:59 | comment | added | Dietrich Burde | It plays a role for integral ring extensions - for a geometric characterization of integral ring extensions, see Atiyah-MacDonald 1969, Ch 5. Exercise 35. | |
| Dec 11, 2013 at 10:42 | history | asked | Qfwfq | CC BY-SA 3.0 |