Timeline for How to memorise (understand) Nakayama's lemma and its corollaries?
Current License: CC BY-SA 3.0
7 events
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| May 19, 2016 at 15:53 | comment | added | goblin GONE | In other words: if $I$ stabilizes $M$, then some $i \in I$ stabilizes every $m \in M.$ Very nice. | |
| Apr 13, 2011 at 10:47 | comment | added | user91132 | You're welcome. Glad to be of help. | |
| Apr 13, 2011 at 9:18 | comment | added | Georges Elencwajg | Ah yes, Konstantin, you are right. Like many algebraic geometers I tend to say and write "ring" when I mean "commutative ring". I have now edited my answer in order to prevent any misunderstanding.Thanks for your attention and apologies for my sloppiness. | |
| Apr 13, 2011 at 8:58 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
added "commutative" to ring
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| Apr 13, 2011 at 8:22 | comment | added | user91132 | You need to assume that your "arbitrary ring" $A$ is commutative. For a non-commutative counterexample, take $A$ to be any domain with a proper non-zero idempotent ideal $I$ and set $M = I$. Then $M = I = I^2 = IM$ since $I$ is idempotent so if your result were true we would get an element $i \in I$ such that $(1-i)I = 0$; since $I \neq 0$ and $A$ is a domain this forces $1 = i \in I$, contradicting the properness of $I$. For a concrete example, let $\mathfrak{g}$ be a perfect Lie algebra and let $I = \mathfrak{g}A$ be the augmentation ideal of its enveloping algebra $A = U(\mathfrak{g})$. | |
| Apr 12, 2011 at 22:56 | comment | added | aglearner | Very nice mnemonic! And also thanks for sharing the story about Rene Tom. | |
| Apr 12, 2011 at 21:46 | history | answered | Georges Elencwajg | CC BY-SA 3.0 |