Prove that $R \otimes_R M \cong M$

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 $M$.

I am following the nomenclature I learn in Dummit & Foote's Abstract Algebra.

To show that $R \otimes_R M \cong M$, let $\varphi : R \times M \to M$ defined by $\varphi(r,m) = rm$. To show that this can be extended to an $R$-module homomorphism, one can show that this map is $R$-balanced, i.e. that \begin{align} \varphi( (r_1,m) + (r_2,m) ) & = \varphi((r_1,m)) + \varphi((r_2,m)), \\\ \varphi( (r,m_1) + (r,m_2) ) & = \varphi((r,m_1)) + \varphi((r,m_2)), \\\ \varphi( (r_1 r_2,m) ) & = \varphi((r_1,r_2m)). \end{align} Once that this is done (trivially), then one can show that $\varphi : R \otimes_R M \to M$ defined by $\varphi( r \otimes m) = rm$ and extending by linearity, is well defined and is an $R$-module homomorphism.

Surjectivity is clear since $\varphi(1 \otimes m) = m$ for every $m \in M$. For injectivity, you will want to show that $\varphi$ is invertible, and $\varphi^{-1} : M \to R \otimes_R M$ would be such that $\varphi^{-1}(m) = 1 \otimes m$. Clearly $\varphi$ and $\varphi^{-1}$ are inverses of each other, so that all you need to do is to verify that $\varphi^{-1}$ is an $R$-module homomorphism. Really not that hard, since it follows from the properties of the tensor product.

Hope that helps,

As Dylan suggested, we can use the universal property of the tensor product to show that $R \otimes_R M \cong M$. Now this means we would like to show that the $R$ - module $M$ is equipped with a bilinear map $\pi : R \times M \rightarrow M$ such that for any bilinear map $B : M \times N \rightarrow P$, where $P$ is some other $R$ - module, there exists a unique linear map $L : M \rightarrow P$ such that

$$B = L \circ \pi.$$

This shouldn't be too hard. Say we are given an $R$ - bilinear map $B : M \times N \rightarrow P$. We can define the map $\pi : R \times M \rightarrow M$ by mapping the pair $(r,m)$ to $rm$. One easily checks that the map $\pi$ is well-defined and bilinear. Now define the map $L : M \rightarrow P$ on "elementary elements" by

$$L(m) = B(1,m)$$

and extend it additively. I said "elementary elements" because usually we define maps on elementary tensors - but there are none here so I just coined this term. Now your map $L$ is easily seen to be well-defined. For linearity, the fact that it is additive comes from definition of how we defined $L$. Let's check that it is compatible with scalar multiplication: For any $r \in R$, we have that

$$\begin{eqnarray*} L(rm) &=& B(1,rm) \\ &=& rB(1,m) \hspace{5mm} \text{(By definition of bilinearity)} \\ &=& r L(m) \end{eqnarray*} $$

completing the claim that $L$ was $R$ - linear. Now it remains to check that our map $B : R \times M \rightarrow P$ factors uniquely through the tensor product. The question whether $B$ factors through $M$ is equivalent to asking if first sending $(r,m)$ to $B(r,m)$ in $P$ is equal to first sending $(r,m) \mapsto rm$ in $M$, and then sending $rm$ in $M$ to $P$. But this is clear because

$$\begin{eqnarray*} L(rm) &=& B(1,rm) \\ &=& rB(1,m) \\ &=& B(r,m). \end{eqnarray*}$$

It is possible to do manipulations like that because $M$ is an $R$ module and the other guy in the direct product is $R$ itself.

It now remains to see why given our maps $B$ and $\pi$, there is a unique $R$ - linear map $L : M \longrightarrow P$. If you have any linear map $L$ out of $M$, in order for an appropriate diagram in question to commute we must have that $L(m) = L(\pi(1,m)) = B(1,m)$. There really is no choice for what $L$ is because it is defined by $B$. Hence there is only one $L$ in question for a given bilinear map $B: R \times M \longrightarrow P$ and uniqueness is proven.

We have shown that the $R$ - module $M$ satisfies the universal property of the tensor product $R \otimes_R M$, and hence must be isomorphic to $M$.

$$\hspace{6in} \square$$