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 $M/JM=\mathbb{Z}_2$.