How Does a Quotient Module Relate to Its Generators in a Local Ring?

[Artusartos]

Let $R$ be a local ring with maximal ideal $J$. Let $M$ be a finitely generated $R$-module, and let $V=M/JM$. Then if $\{x_1+JM,...,x_n+JM\}$ is a basis for $V$ over $R/J$, then $\{x_1, ... , x_n\}$ is a minimal set of generators for $M$.

Proof

Let $N=\sum_{i=1}^n Rx_i$. Since $x_i + JM$ generate $V=M/JM$, we have $M=N+JM$...(the proof continues)

Question

Something in this proof is making me feel uncomfortable. Why is it true that $M=N+JM$? I understand that any element of $N+JM$ is of course an element of $M$. Also if $m \in M$, we have $m + 0 \in M+JM$. Since the $x_i +JM$ generate $M/JM$, we (obviously) have $m \in N+JM$.

But then we also have $M=M+JM$, right? Because for $m \in M$, we have $m + 0 \in JM$. Since elements of $M$ and $JM$ are obviously contained in $M$, their sum $M+JM$ must also be contained in $M$. This means that $M = M + JM$. But does this not imply that $M/JM = M$? Because elements of $M/JM$ are of the form $m+JM$ for $m \in M$, right?

This theorem (and proof) is from (0.3.4) Proposition in here [http://www.math.uiuc.edu/~r-ash/ComAlg/ComAlg0.pdf] [1]


[1]: http://www.math.uiuc.edu/~r-ash/ComAlg/ComAlg0.pdf

 

[fresh_42]

$M\supseteq N+JM$ and because $N$ contains a basis of $V$, we have equality.
You cannot conclude anything from $M=M+JM$. Factoring by $JM$ yields by the isomorphism theorem
$$
M/JM = (M+JM) /JM \cong M/(M\cap JM) = M/JM
$$
and nothing is achieved. If $JM$ is a proper nontrivial submodule, then $M/JM$ is neither $\{\,0\,\}$ nor $M$. E.g. $J=2\mathbb{Z} \subseteq \mathbb{Z} = R = M$ yields $M/JM=\mathbb{Z}_2$.