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Let $M$ be a $R-$module, with $R$ a commutative ring with unity. Let $I$ be a maximal ideal of $R$, and define $IM=\{xm|x\in I,m \in M\}$, prove that $IM\le M$.

Let $xm,x'm' \in IM$, so $xm-x'm'=...$ I don't know if this is related of the maximality of $I$, or what else, because the main proof, is that $M/IM$ is a $R/I$-module,and its ok, but prove that $IM$ is a subgroup of $M$ is the part I can't go on.

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It is in spanish, but the teacher put it that way, so I asume that maybe the maximality or other things can make it happen or something.

George
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    Are you sure this is the definition? $IM$ is usually defined as the set of finite sums sums $\sum_{i=1}^n a_im_i$ with $a_i\in I, m_i\in M$ and $n\geq 0$. Then it is really a submodule. The set you defined has no reason to be closed under addition. – Mark Apr 08 '23 at 16:32
  • That was the definition I was given, I'll put in and edit. – George Apr 08 '23 at 16:36
  • And because the "Product" $IM$ is a mix of the ideal of $R$ and the group $M$. – George Apr 08 '23 at 16:42
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    I believe it's a mistake then. For example, product of ideals is a special case of product of ideal by a module. Here you can find a counterexample, where not every element in $IJ$ has the form $ij$. (look at the second answer) It involves maximal ideals. https://math.stackexchange.com/questions/290229/explaining-the-product-of-two-ideals – Mark Apr 08 '23 at 16:49

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