If bimodules are important, why not bi-vector spaces?

Question

Bi-modules are a high profile math concept in abstract algebra. Why aren't bi-vector spaces not so common? I don't see them on the wikipedia article about vector space, and they yield not many results on google.

My idea: A $(K_1,K_2)$ Vector space $V$ would be the analogous definition of bimodule changing module for vector space and field for ring: $V$ is a $K_1$ vector space to the left (wich actually is the same as just "vector space" since fields are conmutative) is also a $K_2$ vector space to the right.

Compatibility: $k_1*(v*k_2)=(k_1*v)*k_2$, with $k_1 \in K_1$, $k_2 \in K_2$, $v \in V$.

Answer 1

I do think there is some non-trivial content here: even with commutative rings with identity, an $R,S$-bimodule is an $R\otimes S$-module (and without commutativity, then one ring has to be replaced by its opposite).

Even when both $R,S$ are algebras over a common field $k$, and the tensor product is over $k$, non-trivial things can happen. E.g., even when $R,S$ are fields, the tensor product is not a field unless $R,S$ are "linearly disjoint", or some other hypothesis. So $R,S$-bimodules have more structure than just "vectorspace" over the common under-field $k$.

Answer 2

Unlike the case of vector spaces, it's not particularly natural here to restrict attention to fields for bimodules. As paul garrett says, the basic problem is that if $M$ is a $(K_1, K_2)$-bimodule over a field $k$, where $K_1, K_2$ are both fields, then equivalently $M$ is a $K_1 \otimes_k K_2$-module, and $K_1 \otimes_k K_2$ need not be a field.

For a simple example, if $k = \mathbb{R}$ and $K_1 = K_2 = \mathbb{C}$, then

$$\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} \cong \mathbb{C} \times \mathbb{C}.$$

So in the nice cases, instead of a field you get a finite direct product of fields (e.g. if $K_1, K_2$ are both finite separable extensions of $k$), which is not too bad. What this means is that every $(\mathbb{C}, \mathbb{C})$-bimodule over $\mathbb{R}$ is a direct sum of copies of two simple bimodules, one given by $\mathbb{C}$ with the obvious bimodule structure and one given by $\mathbb{C}$ where one of the multiplications has been "twisted" by complex conjugation.

But worse things can happen: if $k = \mathbb{F}_p(t)$ and $K_1 = K_2 = k[x]/(x^p - t)$, then

$$K_1 \otimes_k K_2 \cong K_1[x]/(x - \sqrt[p]{t})^p$$

which has a nontrivial nilpotent, and in particular which is not semisimple.