## the “in” function: M->gr(M) from David Eisenbud’s Commutative Algebra book

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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You are not exactly claiming anything, so you are not technically wrong. I do not understand what you are asking, really. – Mariano Suárez-Alvarez Oct 1 at 19:02
In any case, MO is targeted at research-related questions mostly, and your question does not fit that description all that well. A site like math.stackexchange.com might be a much better fit for this sort of question. – Mariano Suárez-Alvarez Oct 1 at 19:03
@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 $in(f)+in(g)=(x+I^2) \oplus (x^2+I^3) \neq x+I^2 = in(f+g)$. – Ralph Oct 1 at 19:52
@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. – Ravi Vakil 2 days ago