TITLE: Does a left basis imply a right basis, without AC?
QUESTION [8 upvotes]: If $_DV_D$ is a $D$-$D$-bimodule, and we have a $D$-basis for $V_D$, do we still need AC to get a $D$-basis for $_DV$?
(The original question appears below.  But this shorter question gets at the heart of my question, and makes it clear it has more logical foundations.)

Let $D$ be a division ring and let $_D V_D$ be a $D$-$D$-bimodule.  If we temporarily forget the left module structure, and just look at the right $D$-module structure, we have $V_D= \bigoplus_{i\in I}e_iD$ for some basis $\{e_i\}_{i\in I}$.
It is a well-known fact that $E:={\rm End}(V_D)\cong {\rm CFM}_I(D)$ where ${\rm CFM}_I(D)$ is the ring of $I\times I$ column finite matrices.  These are the matrices where each column has only finitely many nonzero entries.  If we think of the elements of $V_D$ as columns of size $I\times 1$ with only finitely many nonzero entries, then ${\rm CFM}_I(D)$ acts on the left of $V_D$ simply by matrix multiplication.  (Of course, when $|I|=n$ is finite, then ${\rm CFM}_I(D)=\mathbb{M}_n(D)$ is just the usual ring of $n\times n$ matrices.)
So we have a natural bimodule structure on $V$, namely $_{E}V_D$.  Our original bimodule structure $_DV_D$ gives rise to a homomorphism $\varphi:D\to E\cong {\rm CFM}_I(D)$.  Conversely, given any such homomorphism (and a fixed basis for $V_D$ indexed by $I$) we get a $D$-$D$-module structure on $V$.
We could do all of this over again on the other side.  From the left $D$-module structure $_DV$, we can fix a basis $\{f_j\}_{j\in J}$ and corresponding decomposition $_DV=\bigoplus_{j\in J}Df_j$.  The right $D$-module structure then corresponds to a homomorphism $\psi:D\to {\rm RFM}_J(D)$.  (The ring ${\rm RFM}_J(D)$ is the ring of $J\times J$ row finite matrices.)
So given an index set $I$ and a homomorphism $\varphi:D\to {\rm CFM}_I(D)$, there is a corresponding index set $J$ and a homomorphism $\psi:D\to {\rm RFM}_J(D)$.  My question is whether there is a canonical way to describe the correspondence $(I,\varphi)\leftrightarrow(J,\psi)$.  If not a canonical way, given the information $I$ and $\varphi$, can we at least describe $|J|$ and $\psi$ explicitly from that data, after a choice of basis?

REPLY [3 votes]: This is a very incomplete answer, but maybe others can fill in the gaps (and I'll try to).
[Edit: I've not been able to make this idea work, although the ideas may lead somewhere, so I'll leave this here. Below I've added some more specific comments about the difficulties I found.]
First, a reminder of the construction in Andreas Blass' beautiful proof that in ZF "every vector space over every field has a basis" implies AC. Let $k$ be a field, and suppose that the set $X$, which is the disjoint union of subsets $X_i$, contradicts the axiom of multiple choice (i.e., it's not possible to simultaneously choose finite subsets of each $X_i$). If $K$ is the subfield of $k(X)$ consisting of elements homogeneous of degree zero in each $X_i$, and $V$ is the $K$-subspace of $k(X)$ spanned by $X$, then $V$ does not have a $K$-basis.
Here's my idea. Suppose:
(a) $k\cong K$ as fields, and
(b) $V$ has a $k$-basis.
Then $V$, considered as a $k$-bimodule with standard left action, and with right action via the isomorphism of $k$ with $K$, has a left basis but no right basis.
Condition (a) can be satisfied by taking $k$ to be the rational functions over some field in a countable union of copies of $X$, homogeneous of degree zero in each copy of each $X_i$, and $K$ the same but with one extra copy of $X$.
I don't know if (b) is necessarily true without AC, but I think I see a proof (but haven't checked the details) that it's true if every set has a total order, which is strictly weaker than AC. If I'm right, then this shows that in ZF a $k$-bimodule with a left basis may not have a right basis, but doesn't prove that AC is equivalent to the absence of a counterexample.
[Edit: I haven't been able to write down an explicit basis, but on the other hand I haven't convinced myself that it's impossible. When I tried, I kept wanting to write homogeneous rational functions in terms of homogeneous elements $y/x$ where $x,y\in X_i$ after making a choice of $x\in X_i$ for each $i$, which is not so helpful when the $X_i$ are supposed to contradict AC! Maybe this is a hint that this method can't work.
I didn't see that it's obviously easier to prove that $K$ has a $k$-basis, and if it provably doesn't, then this also answers the question, using $K$ instead of $V$.]
A simpler but related question that I don't know the answer to: Let $k$ be a field and $Y$ a set. In ZF without AC, must $k(Y)$ have a $k$-basis? If not, how strong a version of AC is necessary? As before, I think I see how to prove it if $Y$ has a total order.