Let $R$ be a commutative unital ring and $M$ an $R$-module.
I'm trying to prove $R \otimes_R M \cong M$ but I'm stuck. If $(R \otimes M, b)$ is the tensor product then I thought I could construct an isomorphism as follows:
Let $\pi: R \times M \to M$ be the map $rm$. Then there exists a unique linear map $l: R \otimes M \to M$ such that $l \circ b (r,m)= l(r \otimes m) =r l(1 \otimes m) = \pi(r,m) = rm$.
Now I need to show that $l$ is bijective. Surjectivity is clear. But I can't seem to show injectivity. In fact, by now I think it might not be injective. But I can't think of a different suitable map $\pi$.
Then I thought perhaps I should show that $l$ has a two sided inverse but for an $m$ in $M$ I can't write down its inverse. How do I finish the proof?
Best Answer
For an inverse, define $M \to R \otimes M$ by $m \mapsto 1 \otimes m$. I suppose you could also show directly that the map is injective, but point remains the same: $R$ contains $1$. If $\sum r_i \otimes m_i$ is in the kernel then $\sum r_im_i = 0$, and we can write the tensor as $\sum 1 \otimes r_im_i = 1 \otimes \sum r_im_i = 1 \otimes 0 = 0$.
Another way is to show that your map $R \times M \to M$ satisfies the universal property of the tensor product of $R$ and $M$.