Rings and Categories of Modules by Anderson-Fuller: Exercise 2.2.2 The following is Exercise 2.2.2 from Anderson & Fuller's book Rings and Categories of Modules:

Let $_\mathbb{C}V$ be a non-zero complex vector space. The abelian group $V$ becomes a left $\mathbb{C}$-vector space$_\mathbb{C}\overline V$ with the scalar multiplication $(\alpha,x)\mapsto \overline\alpha x$. Prove that neither of these $\mathbb{C}$-vector spaces $_\mathbb{C}V$ and $_\mathbb{C}\overline V$ is a subspace of each other and that these two $\mathbb{C}$-scalar multiplications do not form a $(\mathbb{C},\mathbb{C})$-bimodule.

The authors define a submodule of a (left) $R$-module $M$ as an abelian subgroup of $M$ closed under scalar multiplication by $R$. I'm not really sure what the exercise is asking, since both $V$ and $\overline V$ are the same when looked as sets, and therefore if one is closed under one multiplication, then both are. Some people at the webchat suggested that the authors meant that one should prove that those module structures were incompatible with each other, which makes sense, but I can't see how that connects with the definition that is given in the text.
 A: The point is that there is no $\mathbb{C}$-linear isomorphism between $_\mathbb{C}V$ and $_\mathbb{C}\overline{V}$. One way to view a subspace is: if $V$ is a vector space (over any field $k$), and $W \subseteq V$ is any subset of $V$, then $W$ is a subspace of $V$ iff $W$ has the structure of a $k$-vector space, and the inclusion map $W \hookrightarrow V$ (which is the identity on sets) is $k$-linear. This implies $W$ is an additive subgroup of $V$, and the scalar multiplications agree, for instance.
As a (perhaps silly) example, consider $\mathbb{Q}$ as a 1-dimensional vector space over itself. Then $\mathbb{N} \subseteq \mathbb{Q}$ is a subset, which has an abstract structure of a $\mathbb{Q}$-vector space (as does any countable set, by placing it in bijection with $\mathbb{Q}$, and acting via this bijection). But $\mathbb{N}$ is certainly not considered a subspace of $\mathbb{Q}$, and the inclusion $\mathbb{N} \hookrightarrow \mathbb{Q}$ is not $\mathbb{Q}$-linear.
Returning to your question, it is true that $_\mathbb{C}\overline{V}$ is a subset of $_\mathbb{C}V$,
so it is a subspace iff the inclusion $j : {}_\mathbb{C}\overline{V} \hookrightarrow {}_\mathbb{C}V$ is $\mathbb{C}$-linear. But it is not: for any nonzero $v \in _\mathbb{C}\overline{V}$, $j(i \cdot v) = j(iv) = iv$, but $i \cdot j(v) = \overline{i}j(v) = -iv$.
For the last part about bimodules, it's not completely clear what is meant: for a bimodule structure you need a left and a right action, whereas both actions seem to be defined on the left. The candidate bimodule is not explicitly mentioned either, i.e. which of $_\mathbb{C}V$ or $_\mathbb{C}\overline{V}$ is not a bimodule.
