Quotient modules and rings

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

Proof

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

Question

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

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

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


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

  • #2
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##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##.
 

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