Timeline for What elementary problems can you solve with schemes?
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23 events
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| Jan 10, 2023 at 7:54 | comment | added | Georges Elencwajg | @John Jiang. You can ask whatever you like. | |
| Jan 10, 2023 at 6:56 | comment | added | John Jiang | @GeorgesElencwajg are you saying I cannot ask it? | |
| Jan 9, 2023 at 17:24 | comment | added | Georges Elencwajg | @John Jiang What has your question to do with comaximal ideals?? | |
| Jan 8, 2023 at 20:44 | comment | added | John Jiang | Why doesn’t the same argument show $I = I^n$? | |
| Jul 30, 2017 at 9:43 | comment | added | Georges Elencwajg | (Continuation) With a pinch of salt one could say that Jean-Jacques Rousseau wonderfully represented that geometric point of view in a paragraph reproduced at the beginning of Fulton's Algebraic Curves. And I stand in awe at Fulton's erudition: to my knowledge no francophone algebraic geometer had ever noticed the relevance of that extract from Rousseau's Les Confessions | |
| Jul 30, 2017 at 9:35 | comment | added | Georges Elencwajg | @Noam: I knew your elementary proof, but the point is that once you know the language of elementary scheme theory the result doesn't need any proof: you just see it. It is then easy, if challenged, to translate that vision into a formal proof. (To be continued) | |
| Jul 29, 2017 at 23:29 | comment | added | Noam D. Elkies | This is a nice example, but it doesn't really answer the question, because an easy and entirely elementary proof is available: $I+J=A$ means $a+b=1$ for some $a \in I$ and $b \in J$, and then $1 = (a+b)^{m+n-1}$ is a sum of $m$ terms from $I^n$ (monomial multiples of $a^n$) and $n$ terms from $J^m$ (monomial multiples of $b^m$), so it's in $I^n + J^m$, QED. | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Apr 8, 2016 at 21:14 | history | edited | Georges Elencwajg | CC BY-SA 3.0 |
added 186 characters in body
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| Apr 12, 2012 at 19:13 | comment | added | Ostap Chervak | While I enjoyed an argument, note that it's actually much similar to Furstenberg "topological" proof of infinitude of primes, namely, if you unravel it you'll get this(schemeless proof): if $I^n$ and $J^m$ are not comaximal, we hav $I^n+J^m\not{=} A$ hence there exist a maximal $\mathbf{m}$ such that $I^n+J^m\subset\mathbf{m}$ hence $I^m \subset \mathbf{m}$, by primeness of $\mathbf{m}$ we get $I\subset \mathbf{m}$, and $J^m\subset \mathbf{m}$ so again $J\subset \mathbf{m}$ so $I+J\subset \mathbf{m}$. Contradiction. Here we see that under all algebraic manipulation there is hidden geometry. | |
| Mar 24, 2011 at 3:22 | history | made wiki | Post Made Community Wiki by Kim Morrison | ||
| Mar 22, 2011 at 16:35 | comment | added | Georges Elencwajg | Dear Sándor, you are right. Indeed a bright young undergraduate could learn this geometric language from Atiyah-MacDonald. And apparently, one of them did... | |
| Mar 22, 2011 at 15:59 | comment | added | Sándor Kovács | Hi Georges, as much as I hate arguing with you, I cannot not answer this. :). Zariski topology for arbitrary rings is introduced on page 12 of Atiyah-MacDonald. I for one read that book as a 3rd year undergrad, but did not know what schemes were until grad school. | |
| Mar 22, 2011 at 11:11 | comment | added | Georges Elencwajg | Dear Sándor, of course I understand your point and, given your remarkable command of Algebraic Geometry, I gave it the highest attention and thank you for it. Maybe you will agree with the following: it is improbable that a bright young undergraduate who had to solve this exercise on comaximal ideals would come up with such a geometric proof if he had never opened a book on elementary scheme theory. Even though you are quite right that logically the full force of nilpotents is definitely not used, it seems that you can only learn the geometric language from such books. | |
| Mar 22, 2011 at 7:55 | comment | added | Sándor Kovács | ...of $\mathrm{Spec}(A)$ not its scheme structure. | |
| Mar 22, 2011 at 7:55 | comment | added | Sándor Kovács | Georges, you are right, I buy that. I guess I was thinking of Zariski topology, but I can accept that Zariski did not consider arbitrary rings. Perhaps you should emphasize that more in your answer that this works for arbitrary rings. And of course, you are also right in your amusement: I should change my comment (if it were possible) to say "reduced scheme" instead of "variety". Finally, I would not dare challenge the fundamental importance of Grothendieck's inventions nor do I want to lessen the importance of schemes. The point I tried to make here is that this proof is using the topology | |
| Mar 22, 2011 at 7:15 | comment | added | Georges Elencwajg | Dear Sándor, the idea of associating a geometric space to an arbitrary ring is due to Grothendieck and was developed in EGA I. Nobody in the world had considered this association in such generality before: I have heard this said explicitly by Serre, Deligne and Cartier. So I don't agree with you: Spec(A) is definitely not a variety for $A=\mathbb Z$ and my proof couldn't have been written the day before EGA was published. Amusingly you implicitly acknowledge the role of schemes yourself when you define a variety as a reduced SCHEME (which is not at all the standard definition of a variety). | |
| Mar 22, 2011 at 4:37 | comment | added | Sándor Kovács | Dear Georges, while I am a big fun of doodles and a firm believer in the usefulness drawing potatoes representing all kinds of mathematical objects, and I generally enjoy reading your answers and agree with them, I feel that this is not really a scheme-y example. The same proof can be carried out by only knowing varieties. The formulae does not have anything to do with the scheme structure. In fact, the point of the proof is that the varieties (a.k.a. reduced schemes) defined by the ideals $I$ and $I^n$ are the same. | |
| Mar 21, 2011 at 20:16 | comment | added | Georges Elencwajg | Dear Thierry, what I find schemey is: a) That you can interpret that two ideals $I,J$ are comaximal as being exactly equivalent to their associated subschemes $V(I), V(J)$ being disjoint in $Spec (A)$. b) That the underlying sets $|V(I)|$ and $|V(I^n)|$ of the different subschemes $V(I)$ and $V(I^n)$ are equal. c) That there are pleasant formulae like $V(I+J)=V(I) \cap V(J)$ relating the algebra of the ring $A$ to the geometry of the scheme $Spec(A)$. | |
| Mar 21, 2011 at 19:52 | comment | added | Dustin Clausen | Related to this is the statement that every artinian ring A is a product of local artinian rings -- clear from the fact that Spec(A) is just a discrete finite set. | |
| Mar 21, 2011 at 19:43 | comment | added | Thierry Zell | For those of us who are not really into schemes, could you add something to explain what makes this an especially scheme-y idea? | |
| Mar 21, 2011 at 18:58 | comment | added | Mariano Suárez-Álvarez | Never underestimate the power of drawing potatoes... | |
| Mar 21, 2011 at 17:33 | history | answered | Georges Elencwajg | CC BY-SA 2.5 |