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I'm working through the "Commutative algebra with a view toward algebraic geometry" book and stumbled onto an exercise I'm struggling to answer.

Let $R$ be a ring and let $M$ be an $R$-module. Suppose that $\{f_i\}$ is the set of elements of $R$ that generate the unit ideal, i.e $\langle f_1,...,f_n\rangle=R$. Then if $m\in M$ goes to $0$ in each $M[f_i^{-1}]$, then $m=0$. (The exercise can be found here).

First of all, $m$ goes to $0$: does this mean there are mappings $\psi_i: M\rightarrow M[f_i^{-1}]$ s.t. $\psi_i(m)=0$? If so, can can I possibly go about answering that $m$ must be $0$? Also, is there anything interesting about the mappings, are they homomorphisms, isomorphisms, etc?

The hint given in the book states that: If $m$ goes to $0$ in each $M[f_i^{-1}]$, then $m$ is killed by a power of each $i$. Show that if the set $\{f_i\}$ generates the unit ideal, then so does the set $\{f_i^{n_i}\}$ for any positive integers $n_i$.

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  • $\begingroup$ Hint: generalize this. $\endgroup$ – Bill Dubuque Feb 19 '15 at 5:05
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For any multiplicatively closed subset $S$ of $R$, there is a canonical mapping $\,\varphi\colon M\rightarrow S^{-1}M$ defined by $\,\varphi(m)=\dfrac m1$. By definition of a module of fractions, $\varphi(m)=0\iff \exists s\in S,\ sm=0$.

Here we have multiplicatively closed subsets $S_i=\bigl\{1, f_i, f_i^2,\dots, f_i^k,\dots\bigr\}$, and $m$ maps to $0$ in $M[f_i^{-1}]$ if and only if there exists a power $k_i$ of $f_i$ such that $f_i^{k_i}m=0$.

Let $K=k_1+\dots+k_n$; write $1=\lambda_1 f_1+\dots+\lambda_nf_n$. Then by the multinomial formula, $$1=(\lambda_1 f_1+\dots+\lambda_nf_n)^K=\sum_{r_1+\dots+r_n=K} \frac{N!}{r_1!\dotsm r_n!}(\lambda_1 f_1)^{r_1}\dots(\lambda_n f_n)^{r_n}. $$ Each term in this sum kills $m$. Indeed since $r_1+\dots+r_n=k_1+\dots+k_n$, at least $1$ $r_i$ is $\geq k_i$. Thus $1\cdot m = m= 0$.

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    $\begingroup$ Not much, you're right. I'm often rather slow on writing answers, and I forgot to check if any had been posted meanwhile. The only point is I'm explicit as to the power to be used on showing $1\cdot m=0$… $\endgroup$ – Bernard Feb 18 '15 at 23:50
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If $\psi_i(m) = 0$, then you know that $f_i^{n_i} m = 0$ for some $n_i \geq 0$. If you could express $1 = \sum_{i=1}^n a_i f_i^{n_i}$ with $a_i \in R$, then you'd be done, for you would know that $m = \sum_{i=1}^n a_i f_i^{n_i} m = 0$. But you know that $\langle f_1, \dots, f_n \rangle = R$, so $1 = c_1 f_1 + \dots + c_n f_n$. Raise each side of this equation to the $\sum_{i=1}^n n_i$ power and notice that when you expand by the binomial theorem, each term contains a factor of the form $f_i^{n_i}$ for at least one $i$.

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For any multiplicative set $S$ there is a canonical map $\psi_S:M\to M_S$ given by $\psi_S(m)=m/1$. This is a module homomorphism. If $\psi_S(m)=0$ then $m/1=0/1$ (in $M_S$), and by definition there is $s\in S$ such that $sm=0$. If $S=\{1,f_i,\dots,f_i^n,\dots\}$, then $\psi_S(m)=0$ means that there is $n\ge1$ such that $f_i^nm=0$.

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You can also show that the module $A^*=\prod_{i=1}^n A[f_i^{-1}]$ is faithfully $A$-flat: it is flat, since each of the terms is. Now given a maximal ideal $\mathfrak m$ of $A$, the fact that the $f_i$ generate $A$ means there must be some $f_i$ not contained in $\mathfrak m$. This means that $\mathfrak m A[f_i^{-1}]\neq A[f_i^{-1}]$, so $\mathfrak m A^*\neq A^*$.

Given $m\in M$, consider the $A$-module $Am=N$. The hypothesis means that $A^*\otimes N=0$, so that $N=0$ and $m=0$. In particular $A^*\otimes M=0$ implies $M=0$. I am not sure if Eisenbud mentions this, but note that the conditions that the $f_i$ sum to $A$ is equivalent to the $X_{f_i}$ being an open cover of ${\rm Spec}(A)$.

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