Example of a Tensor Product of Modules with Non-Decomposable Elements 
Given a ring $R$ and $R$-modules $A_R$ and $_{R}B$, we define the tensor product $A \otimes_R B$ as the free abelian group on $A \times B$ modded out by the subgroup generated by the elements of the following form:
  $$
  (a+a',b)\!-\!(a,b)\!-\!(a',b) \quad\quad
  (a,b+b')\!-\!(a,b)\!-\!(a,b') \quad\quad
  (ar,b)\!-\!(a,rb)
$$
  Alternatively we can think of $A \otimes_R B$ as being generated by elements of the form $a \otimes b$ (which are called the decomposable elements) where $a \in A$ and $b \in B$, but not every element of $A \otimes_R B$ has to be a decomposable element that looks like $a \otimes b$.

Are there good illustrative (concrete?) examples of modules $A$ and $B$ and a ring $R$ where $A \otimes_R B$ has non-decomposable elements? 
It seems like when working with concrete modules, for all the examples I've seen so far the tensor product is isomorphic to some nice group, which I think means that there are no non-decomposable elements to consider. Alternatively when working with tensor products abstractly you can treat the decomposable elements as a basis for $A \otimes_R B$ and not bother thinking about the non-decomposable elements. Either way, I'm missing an intuitive sense of what a tensor product with non-decomposable elements "looks like."
 A: Let $\mathbb{F}$ be a field and let $V$ be a module over $\mathbb{F}$ (i.e. a vector space over $\mathbb{F}$), then $V^*$ is also a module over $\mathbb{F}$. The tensor product $V^*\otimes_{\mathbb{F}}V$ is canonically isomorphic to $\operatorname{End}_{\mathbb{F}}V$ via the map induced by the bilinear map $V^*\times V \to \operatorname{End}_{\mathbb{F}}(V)$, $(\varphi, w) \mapsto \Phi_{(\varphi, w)}$ where $\Phi_{(\varphi, w)}(v) = \varphi(v)w$. 
If $V$ is a finite-dimensional vector space over $\mathbb{F}$ of dimension $n$, choosing a basis $\{e_1, \dots, e_n\}$ for $V$ induces an isomorphism $\operatorname{End}_{\mathbb{F}}V \cong M_{n\times n}(\mathbb{F})$ by the map $a_i^je^i\otimes e_j \mapsto [a^i_j]$. Under the isomorphisms $V^*\otimes V \cong \operatorname{End}_{\mathbb{F}}V \cong M_{n\times n}(\mathbb{F})$, the decomposable elements of $V^*\otimes V$ correspond to the rank one matrices. 
To see that this last claim is true, let $\varphi \in V^*\setminus\{0\}$ and $w \in V\setminus\{0\}$, then $\varphi = x_ie^i$ and $w = y^je_j$, so $\varphi\otimes w = x_iy^j e^i\otimes e_j$ and is mapped under the isomorphism to the matrix $[x_iy^j]$. Now note that
$$[x_iy^j] = \begin{bmatrix} x_1\\ \vdots\\ x_n\end{bmatrix}[y^1\ \dots\ y^n] = xy^T$$
so $[x_iy^j]$ has rank one. Conversely, any rank one $n\times n$ matrix can be put in the form $xy^T$ for some $x, y \in \mathbb{F}^n\setminus\{0\}$ (such a product is called an outer product), and hence we can trace back the isomorphism to find a decomposable element of $V^*\otimes_{\mathbb{F}}V$.
More generally, elements of $V^*\otimes_{\mathbb{F}} V$ which can be written as a sum of $k$ decomposable elements (but no fewer) correspond to sums of $k$ outer products (but no fewer). These are precisely the matrices of rank $k$. In fact, one often defines the rank of an element in a tensor product as the smallest number of decomposable elements needed to write it as a sum, and the above simply states that the two notions of rank agree.
Note: After choosing a basis $\{e_1, \dots, e_n\}$ for $V$, one obtains the basis $\{e^i\otimes e_j \mid 1 \leq i, j \leq n\}$ for $V^*\otimes_{\mathbb{F}}V$. While the elements $e^i\otimes e_j$ (and their non-zero scalar multiples) are decomposable, they are not the only decomposable elements of $V^*\otimes_{\mathbb{F}}V$. An element of $V^*\otimes_{\mathbb{F}}V$ is decomposable if it is of the form $\varphi\otimes w$ for some non-zero $\varphi \in V^*$ and non-zero $w \in V$. For example, the element $e^1\otimes e_1 + e^2\otimes e_1$ is not of the form $ae^i\otimes e_j$, but it is decomposable as 
$$e^1\otimes e_1 + e^2\otimes e_1 = (e^1 + e^2)\otimes e_1.$$
However, what is true is that an element $L$ of $V^*\otimes_{\mathbb{F}}V$ is decomposable if and only if there is some basis $\{f_1, \dots, f_n\}$ of $V$, such that $L = f^i\otimes f_j$ for some $i$ and $j$.
A: Take $R$ to be a field $k$ and take $A = B = k^2$, with basis $e_1, e_2$. The tensor product $A \otimes_k B$ is $k^4$ with basis $e_1 \otimes e_1, e_1 \otimes e_2, e_2 \otimes e_1, e_2 \otimes e_2$, and most elements of it are indecomposable. For example, $e_1 \otimes e_1 + e_2 \otimes e_2$ is indecomposable. 
There's no more reason to expect all tensors to be decomposable than there is to expect all polynomials to factor. In fact we can be fairly explicit about which tensors decompose in the above special case: the decomposable tensors are precisely those of the form
$$(ae_1 + be_2) \otimes (ce_1 + de_2) = ac e_1 \otimes e_1 + ad e_1 \otimes e_2 + bc e_2 \otimes e_1 + bd e_1 \otimes e_2.$$
Working projectively, this defines a morphism $\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3$ (a special case of the Segre embedding), so we expect its image to be a $2$-dimensional subvariety of the $3$-dimensional variety $\mathbb{P}^3$, and in particular not the whole thing. More explicitly, an example of a nontrivial polynomial equation satisfied by the image is $x_{11} x_{22} = x_{12} x_{21}$, where $x_{ij}$ is the coefficient of $e_i \otimes e_j$ (note that this is not satisfied in our example), and this in fact generates the ideal defining the image of the Segre embedding. 
As Mariano says in the comments, another perspective is the following. If $V$ is a finite-dimensional $k$-vector space, then $V \otimes_k W$ can be identified with the space of linear maps $V^{\ast} \to W$. Among those, the decomposable tensors are the maps of rank at most $1$. If $\dim V = 2$ and $W = V^{\ast}$ then $V \otimes_k V^{\ast}$ can be identified with $\text{End}(V)$, and then an element of $\text{End}(V)$ has rank at most $1$ iff its determinant is zero; that's the equation $x_{11} x_{22} = x_{12} x_{21}$ above. In general a matrix has rank at most $1$ iff all of its $2 \times 2$ minors vanish. 
