For all $R$-modules $N$, $R\otimes_R N\cong N$ Why is this so? This statement is from Aluffi's book Algebra: Chapter 0 and seemingly so trivial that it deserves no proof.
So working with the universal property of tensor products, how exactly does every $R$-bilinear map $R\times N\longrightarrow P$ factor through $N$? Here $P$ is another $R$-module.
The map $R\times N\overset{\otimes}\longrightarrow N$ is given as $\otimes(r,n)=rn$.
 A: The set of $R$-bilinear maps $f: R\times N\to P$ is in natural bijection with the set of $R$-linear maps $g:N\to P$. The correspondence is obtained by setting $f(r,n) = g(rn)$.
A: Intuitively, any $R$ bilinear map $f:R \times N \to M$ is really a linear map from $N$ to $M$ in disguise, for if $g$ is a linear map from $N$ to $M$, then we may define $f$ by letting $f(1,n) = g(n)$, extended to a unique bilinear extension. The details are then trivial.
Let $f: R \times N \to M$ be a bilinear map. Given your projection map $(r,n) \mapsto rn$, if we have a factor map $g: N \to M$, then we must define
$$ g(rn) = f(r,n) $$
Certainly this defines $g$ for all $n \in N$, for $n = 1n$. Suppose $rn = r'n'$. Then, since $f$ is bilinear,
$$ f(r,n) = rf(1,n) = f(1,rn) = f(1,r'n') = r'f(1,n') = f(r',n') $$
So $g$ is well defined. That $g$ is a linear map follows because
$$ g(rn) = f(r,n) = rf(1,n) = rg(n)$$
$$ g(n + m) = f(1, n + m) = f(1, n) + f(1,m) = g(n) + g(m) $$
and $g$ is obviously a unique factor, and thus $N$ is a tensor product over $R$ for $R$ and $N$.
