How to prove $M \otimes_R k\cong M/IM$?

Question

Let $R$ be a local ring with maximal ideal $I$ and residue field $k=R/I$ and $M$ finitely generated $R$-module.

My question is why $$M \otimes_R k\cong M/IM,$$ where $M \otimes_R k$ and $M/IM$ are $R/I$-modules.

Answer 1

Let $p:M\rightarrow M/IM$ the canonical projection, consider $f:M\times k\rightarrow M/IM$ defined by $f(x,k)=p(kx)$. $f$ is a $R$-bilinear map. By the universal property of the tensor product, $f$ factors by $g:M\otimes_Rk\rightarrow M/IM$.

Let $h':M\rightarrow M\otimes_Rk$ which is the composition of $h_1:M\rightarrow M\times k$ defined by $h_1(m)=(m,1)$ and the canonical projection $h_2:M\times k\rightarrow M\otimes_Rk$. For every $i\in I$, $h'(im)=h_2(im,1)=h_2(m,i)$ since $h_2$ is $R$-bilinear. Since $(m,i)=(m,0)$, we deduce that $h_2(m,i)=h'(im)=0$. This implies that $h'$ factors by $h:M/IM\rightarrow M\otimes_Rk$.

Let $x\in M, y\in k$, $h(g(x\otimes_Ry)=h(p(yx)=h_2(yx,1)=h_2(x,y)=x\otimes_Ry$.

On the other hand, let $x\in M ,g(h(p(x)))=g(h_2(x,1))=g(x\otimes_R1)=p(x)$.