Does a left basis imply a right basis, without AC?

Question

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.)

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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?

Answer 1

This isn't really an answer, but here are some thoughts. For any pair of rings $R, S$ whatsoever, the following pieces of data are all the same:

• An $(R, S)$-bimodule;
• An $R \otimes S^{op}$-module;
• A bilinear functor $R \times S^{op} \to \text{Ab}$ (where by $R$ and $S^{op}$ I mean the one-object linear categories with endomorphism rings $R$ and $S^{op}$ respectively);
• A linear functor $R \to [S^{op}, \text{Ab}]$ (where by $[-, -]$ I mean the functor category);
• A linear functor $S^{op} \to [R, \text{Ab}]$.

Now assume in addition that $R$ and $S$ are division algebras. Then the fourth bullet point gives, after a choice of basis, homomorphism of rings from $R$ to $\text{CFM}_I(S)$ for some index set $I$, and the fifth bullet point gives, after a choice of basis, a homomorphism of rings from $S^{op}$ to $\text{CFM}_J(R^{op})$ for some index set $J$. This is equivalent to the data of a homomorphism of rings from $S$ to $\text{RFM}_J(R)$ by taking opposites.

In the first approach, the underlying abelian group of the bimodule is the direct sum $\bigoplus_{i \in I} S$ regarded as a right $S$-module, whereas in the second approach, the underlying abelian group of the bimodule is the direct sum $\bigoplus_{j \in J} R$ as a left $R$-module. Any explicit answer to your question ought to generalize to this case and it ought to be equivalent to explicitly writing down an isomorphism

$$\bigoplus_{i \in I} S \cong \bigoplus_{j \in J} R$$

of abelian groups, and this seems like a potentially messy and terrible thing depending on how complicated $R$ and $S$ are; in particular if the cardinalities of $R$ and $S$ are wildly different then the cardinalities of $I$ and $J$ must also be wildly different. And note that even if $R \cong S \cong D$ it's too much to expect that we can just take $I = J$ as one might hope and then write down the simplest possible map above because most $(D, D)$-bimodules are not direct sums of copies of $_D D_D$.

So basically I don't see any hope of answering this question in a way that doesn't just involve passing through bimodules as an intermediary structure.

Answer 2

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.

Answer 3

Pace, I'm not quite sure what flavor of answer you seek, so I apologize if the following was already obvious to you. (This was just a bit too long for a comment.)

The structure ${}_E V_D$ can be "recovered" from the ring $E = \mathrm{CFM}_I(D)$ as the unique (up to isomorphism) simple module embeddable in $E$; letting $e \in C$ denote a "diagonal matrix unit" with respect to $I$, we have ${}_E V \cong Ee$ (I picture this as a "column" in $E$). This left ideal in $E$ is invariant under the "diagonal" (with respect to $I$) copy of $D \subseteq E$, and this recovers the right $D$-structure of $V$.

Thus $|J|$ is the dimension of $Ee \cong {}_D V$ as a left $D$-vector space under the left $D$-action induced by $\phi \colon D \to E$, and the morphism $\psi \colon D \to \mathrm{RFM}_I(D) \cong \mathrm{End}({}_D Ee)$ arises from the "diagonal" right $D$-action on $Ee \subseteq E$.

This at least provides a theoretical way to extract $|J|$ and $\psi$ if you have an explicit choice of $I$ and $\phi$, even if it's not particularly canonical or enlightening.