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I'm reading David Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry (3rd ed.).

in chapter 5, about filtrations and the Artin-Rees lemma, the function $in: M \to gr(M)$ was defined for a module $M$ with filtration $M=M_{0}\supset M_{1}\supset \cdots $ as follows: for $f \in M$ let $m$ be the greatest number such that $f\in M_{m}$ and set $in(f)=f$ $modulo$ $M_{m+1} \in M_{m}/M_{m+1}\subset gr(M)$, or if there is no such m: $in(f)=0$.

However, in exercise 5.1 the reader is requested to prove for $f,g\in M$ that either $in(f)+in(g)=in(f+g)$ or $in(f)+in(g)=0$. as i understand, $in(g)$ is a homogeneous element, but $in(f), in(g)$ may be of different summands.

so my question is, where am i wrong?

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  • $\begingroup$ You are not exactly claiming anything, so you are not technically wrong. I do not understand what you are asking, really. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Oct 1, 2012 at 19:02
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    $\begingroup$ @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 $in(f)+in(g)=(x+I^2) \oplus (x^2+I^3) \neq x+I^2 = in(f+g)$. $\endgroup$
    – Ralph
    Commented Oct 1, 2012 at 19:52
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    $\begingroup$ @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. $\endgroup$
    – Ravi Vakil
    Commented May 21, 2013 at 21:01
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    $\begingroup$ It works if $f, g$ have the same degree. Cfr. ex. 14.2.(ii) of Matsumura Commutative Ring Theory. $\endgroup$
    – Andrea Gagna
    Commented May 18, 2014 at 11:24
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    $\begingroup$ @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. $\endgroup$
    – Daniel Litt
    Commented Jan 16, 2021 at 1:45

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