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Let $S$ be a local ring with maximal ideal $\mathfrak{m}$. I want to prove that if $M$ is a $S-$module, then $$M\otimes_S k\cong M/\mathfrak{m}M.$$ where $k=S/\mathfrak{m}$. To prove it I defined $\psi:M\to M\otimes_Sk$ by $\psi(m)= m\otimes 1$. It was easy to prove that $\mathfrak{m}M\subseteq \ker\psi$ by seeing that for $f_j\in \mathfrak{m},m_j\in M$, $$\left(\sum_{j}m_jf_j\right)\otimes 1=\sum_jf_j(m_j\otimes 1)=\sum_j (m_j\otimes (f_j\cdot 1))=\sum_j m_j\otimes 0=0.$$

It's also easy to prove it's surjective (For an element $m\otimes (r+\mathfrak{m})$ just take $rm$ as its preimage), but I have trouble when proving that, in fact, $\mathfrak{m} M=\ker\psi$. To prove it, I need to prove that if $m\otimes 1=0$ then $m\in \mathfrak{m}M$ but I don't know how to proceed.

I needed to prove it when $M$ is a finitely generated free module over the (nonlocal) ring $S=k[x_1,\dots,x_n]$ (with maximal ideal $(x_1,\dots,x_n)$) and that's already done, but I think there must be a more general reasoning for this case, which can also be used there.

LeviathanTheEsper
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    It is more general than your situation: for any ideal $I\subset S$ (not necessarily local), one has $;M\otimes_SS/I\simeq M/IM$. It is a consequence of the tensor product being a right-exact functor when you tensor the short exact sequence $$0\to I\to S\to S/I\to 0$$ by $M$. – Bernard Apr 28 '19 at 21:53
  • Oh, I see. That's really good. – LeviathanTheEsper Apr 28 '19 at 22:19
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    You just have to check that the image of $;I\otimes M\to S\otimes M\simeq M$ is exactly $IM$. – Bernard Apr 28 '19 at 22:27

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