Timeline for the "in" function: M->gr(M) from David Eisenbud's Commutative Algebra book
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jan 16, 2021 at 1:45 | comment | added | Daniel Litt | @Mariano — Like Ravi, I think this question is absolutely within the scope of MO (Eisenbud is certainly a book that is read by people doing research in AG, and at least I’ve found myself confused by various errors therein). We’ve certainly had other well-received questions about errata. I also wanted to say — if you don’t understand a question, it’s natural to ask a clarifying question rather than dismissing it. | |
| May 18, 2014 at 11:24 | comment | added | Andrea Gagna | It works if $f, g$ have the same degree. Cfr. ex. 14.2.(ii) of Matsumura Commutative Ring Theory. | |
| May 21, 2013 at 21:01 | comment | added | Ravi Vakil | @Mariano, I'm fine with this question being asked here, as (from the faq): "MathOverflow's primary goal is for users to ask and answer research level math questions, the sorts of questions you come across when you're writing or reading articles or graduate level books." This is a reasonable question when reading a graduate level book. (The community might reasonably disagree with me of course.) And it is true that this question could work on math.stackexchange.com as well. | |
| Oct 1, 2012 at 19:52 | comment | added | Ralph | @edo: I agree. For example let $R=k[x]$ a polynomial ring, $M := (x) =: I$ with filtration $I \supseteq I^2 \supseteq ...$ and $f := x, g := x^2$. Then - to my understanding - $in(f) = x+ I^2, in(g)=x^2 + I^3, in(f+g)=in(f)$ but $in(f)+in(g)=(x+I^2) \oplus (x^2+I^3) \neq x+I^2 = in(f+g)$. | |
| Oct 1, 2012 at 19:02 | comment | added | Mariano Suárez-Álvarez | You are not exactly claiming anything, so you are not technically wrong. I do not understand what you are asking, really. | |
| Oct 1, 2012 at 18:47 | history | asked | edo arad | CC BY-SA 3.0 |