David wrote:

> So I believe there is a (natural) isomorphism between L⊗M and Hom(L,M) for R-modules?

No, for example you're shown the $R$-modules $M \otimes R$ and $\mathrm{Hom}(M,R)$ are different in general: the first is isomorphic to $M$, while the second is by definition $M^*$.

The hom-tensor duality for $R$-modules says, among other things, that
for $R$-modules $M,N,P$ we have an isomorphism

$\mathrm{Hom}(M \otimes N, P) \cong \mathrm{Hom}(M,\mathrm{Hom}(N,P))$

This is easy to see. An $R$-module homomorphism from $M \otimes N$ to $P$ is the same as a bilinear map from $M \times N$ to $P$. But we can take such a bilinear map and think of it as a something that eats an element of $M$ and spits out, in a linear way, a linear map from $N$ to $P$. In other words, an element of $\mathrm{Hom}(M,\mathrm{Hom}(N,P))$

I would take your arguments that tensors work differently over a general ring work differently than tensors over a field and put a different spin on them. What's mainly true is that modules over a field are all free, while modules over a general ring aren't.

This has lots of ramifications. First, working with vector spaces makes one instantly want to grab a basis whenever there's a calculation to be done, but when working with modules you have to suppress this habit. Second, you can't identify a module with its dual. So, you shouldn't think of $M \otimes N$ as consisting of bilinear maps from $M \times N$ to $R$. Instead, it's $(M \otimes N)^*$ that consists of bilinear maps from $M\times N$ to $R$.

Once you develop the new habits, vector spaces over fields seem like a pathetically dull (and wonderfully simple) example of modules over a ring, where all the interesting (and difficult) questions become trivial.