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I'm having trouble understanding the proof of theorem 8.4 in Matsumura's Commutative Algebra. It goes as follows:

Let $A$ be a ring, $I$ an ideal and $M$ an $A$-module. Suppose that $A$ is $I$-adically complete, and $M$ is separated for the $I$-adic topology. If $M/IM$ is generated over $A/I$ by $\bar{\omega}_1,\ldots \bar{\omega}_n$ and $\omega_i\in M$ is the arbitrary inverse image of $\bar{\omega}_i$, then $M$ is generated over $A$ by $\omega_1,\ldots,\omega_n$.

Now I'm having trouble understanding what it means to be "... generated over ..." some ring. In the proof he says, that by assumption we can write $M=\sum A\omega_i+IM$ - is that what he means by "If $M/IM$ is generated over $A/I$ by $\bar{\omega}_1,\ldots \bar{\omega}_n$ and $\omega_i\in M$ is the arbitrary inverse image of $\bar{\omega}_i$"? And if so, why is that the case, that we can write $M=\sum A\omega_i+IM$?

He then chooses $\xi\in M$, and then successively construct $\xi_v=\sum a_{i,v}\omega_i+\xi_{v+1}$ for $v=1,2,\ldots$ where $\xi_v\in I^vM$ and $a_{i,v}\in I^v$. Then he states that $b_i:=\sum a_{i,k}$ converges in $A$ (I guess that's because $A$ is $I$-adically complete, so $\{b_n\}_{n>0}$ is a Cauchy sequence and therefore converges), and so he says we have $\xi-\sum_{i}^n b_i\omega_i\in \cap_{v>0}I^vM$? Why is that the case?

Also, I see that this would imply that $\omega_1,\ldots,\omega_n$ generates $M$, but does that also mean they generate $M$ over $A$?

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The hypothesis is that $\;M/IM=\bigl\langle\mkern1mu\overline\omega_1,\dots,\overline\omega_n\mkern1mu\bigr\rangle$ for congruence classes $\,\overline\omega_i,\;(1\le i\le n)$ as an $A/I$-module. So, by definition of congruence classes mod. $IM$, $$M=\bigl\langle\omega_1\mkern1mu,\dots,\omega_n\mkern1mu\bigr\rangle +IM $$ This is just because $H, K$ are subgroups in a group $G$ (in additive notation), the group generated by the image of $K$ in the quotient $G/H$ is no other than $$K+(G/H)=(K+H)/H.$$ As to your other question, we have $$\xi-\sum_{i}^n b_i\omega_i=\xi-\sum_{i=1}^n \Bigl( \sum_{k\ge1} a_{i,k}\Bigr)\omega_i=\lim_{v\to\infty}\biggl(\xi-\sum_{i=1}^n \Bigl( \sum_{k=1}^{v-1} a_{i,k}\Bigr)\omega_i\biggr)=\lim_{v\to\infty}\xi_v. $$ Each $\xi_v$ lies in $I^vM$, and so does the limit since $I^vM$ is a closed subgroup of $M$.

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  • $\begingroup$ Thanks! Does the last equality follows from the fact that $lim_{v\rightarrow\infty} \xi_v=\xi$ (which is true by the succesive construction of $\xi_v$)? $\endgroup$ – jta May 2 '18 at 16:24
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    $\begingroup$ More exactly, $\xi$ is the sum of the series $\sum_v\xi_v$, if I'm not mistaken. $\endgroup$ – Bernard May 2 '18 at 16:29
  • $\begingroup$ Just to be sure, I understand it correctly, are these $\bar{\omega}$ obtained by the map $\phi:M\rightarrow M/IM$?, ie. $\phi(\omega)=\bar{\omega}$? $\endgroup$ – jta May 3 '18 at 14:06
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    $\begingroup$ @Jta: Yes, exactly. $\endgroup$ – Bernard May 3 '18 at 17:39

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