Timeline for Atiyah-MacDonald, exercise 2.11
Current License: CC BY-SA 2.5
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 18, 2022 at 15:07 | comment | added | Pierre-Yves Gaillard | @JhonDoe - Yes! | |
| Feb 18, 2022 at 15:01 | comment | added | Jhon Doe | I think I get get it. The proposition proves the existence of such a polynomial. So we just pick a polynomial with smallest degree. | |
| Feb 18, 2022 at 14:54 | comment | added | Jhon Doe | @Pierre-YvesGaillard Hello could you explain the "minimal possible degree" portion? In atiyah and macdonald, the polynomial is the determinant of some matrix, so isn't the degree fixed? I don't understand why he needs to consider that condition. Is it the case it might not hold? | |
| Oct 1, 2021 at 15:14 | comment | added | Ricky | I doubt that the author will ever read this comment, but I just formalized this amazing proof in Lean. | |
| Mar 3, 2018 at 18:33 | comment | added | Asvin | Ah, extend the function instead of restricting! Brilliant, thanks. | |
| Mar 3, 2018 at 18:32 | comment | added | Pierre-Yves Gaillard | @Asvin - Here is how I understand Balazs Strenner's argument: Considering $A^n$ as a submodule of $A^m$ as explained in the post, we denote by $i:A^n\to A^m$ the inclusion and we define $\phi':A^m\to A^m$ by $\phi':=i\circ\phi$, then, abusing the notation, we drop the prime, so that the new $\phi$ is an endomorphism of $A^m$ whose image is contained in $A^n$. | |
| Mar 3, 2018 at 16:55 | comment | added | Asvin | @Pierre-YvesGaillard Sorry for replying over 7 years later but doesn't Cayley-Hamilton apply only for endomorphisms? So wouldn't the $\phi$ that satsifies the polynomial equation be the original $\phi$ restricted to the chosen copy of $A^n$? How can we evaluate this restricted $\phi$ on $(0,\dots,0,1)$ which does not lie in the chosen copy of $A^n$? | |
| Apr 17, 2016 at 14:44 | comment | added | Pierre-Yves Gaillard | @TerrenceJ - You can take $I:=A$. Thanks for your interest! | |
| Apr 17, 2016 at 12:56 | comment | added | Terrence J | @Pierre-YvesGaillard - I apologise for a rather elementary question: In applying proposition 2.4 from Atiyah-Macdonald, one requirement is $\phi(A^m) \subset IA^m$, where $I$ is an ideal of $A$. In this question, what do you take to be the ideal $I$? | |
| Jun 14, 2014 at 22:15 | comment | added | Alexey Muranov | Is Cayley-Hamilton Theorem necessary here? Is there no easier way to show that the powers of $\phi$ are linearly dependent over $A$? | |
| Sep 28, 2012 at 14:43 | comment | added | benblumsmith | +1 This is a proof from The Book. | |
| Dec 1, 2010 at 9:16 | comment | added | Pierre-Yves Gaillard | Wonderful !!!!! | |
| Dec 1, 2010 at 0:44 | history | answered | Balazs Strenner | CC BY-SA 2.5 |