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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?

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  • $\begingroup$ Risking being silly - is not a $D$-$D$-bimodule structure equivalent to an isomorphism of $D$-vector spaces? I mean, are not the categories of left and right vector spaces equivalent? In fact, is not $D$ always isomorphic to its opposite? $\endgroup$ – მამუკა ჯიბლაძე Dec 7 '14 at 10:14
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    $\begingroup$ @მამუკაჯიბლაძე $D$ is not isomorphic to its opposite ring in general. But even when it is, there's more going on than an isomorphism of vector spaces. For example, take $D=\mathbb{C}(t)$, and make $D$ into a $D$-$D$-bimodule with $f(t)\in D$ acting on the left by multiplication by $f(t)$, and on the right by multiplication by $f(t^2)$. $\endgroup$ – Jeremy Rickard Dec 7 '14 at 11:00
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    $\begingroup$ There are old (1970--80s) results of George Bergman on weird bimodules over division rings (e.g., different dimensions on right and left), typically in connection with tensor products; have you seen these? $\endgroup$ – David Handelman Dec 7 '14 at 15:41
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    $\begingroup$ @მამუკაჯიბლაძე If $k$ is a commutative field, then the finite-dimensional division algebras over $k$ are classified by the Brauer group of $k$, with "opposite algebra" corresponding to inverse in the Brauer group. So a division algebra corresponding to an element of order greater than two in the Brauer group will not be isomorphic to its opposite algebra. The Brauer group of the rationals has such elements, and "isomorphic as $\mathbb{Q}$-algebras" is the same as "isomorphic as rings". I suspect there are $9$-dimensional examples over $\mathbb{Q}$, but I don't offhand know a reference to one. $\endgroup$ – Jeremy Rickard Dec 7 '14 at 18:09
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    $\begingroup$ I suspect the answer is "no". This could be proved by deducing AC from the statement "every vector space with a left basis has a right basis". This would strengthen @AndreasBlass's result deducing AC from "every vector space has a basis". (math.lsa.umich.edu/~ablass/bases-AC.pdf) Perhaps he or someone else can look at his old proof and rise to the challenge? $\endgroup$ – Matt F. Dec 13 '14 at 15:05
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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.

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  • $\begingroup$ I've had similar thoughts. And indeed, not only is it "too much to expect that we can just take $I=J$", even when $R\cong S\cong D$, but we can construct specific examples where $|I|\neq |J|$. (Indeed, see Jeremy's example up above.) The reason I put the "logic" tag on the question is, for all I know, the information $(I,\varphi)$ is not sufficient to give a $D$-basis on the left without using the axiom of choice. But that would be really weird to me. $\endgroup$ – Pace Nielsen Dec 7 '14 at 15:32
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    $\begingroup$ @Pace: that doesn't seem so weird to me. Already neither $I$ nor $J$ exist without appealing to the axiom of choice... $\endgroup$ – Qiaochu Yuan Dec 7 '14 at 19:30
  • $\begingroup$ Can $I$ and $\varphi$ be given explicitly, but $J$ not exist without appealing to the axiom of choice? If so, that would seem to say that the homomorphism $\varphi:D\to {\rm CFM}_I(D)$ needs some extra information to yield $J$, which would be somewhat weird. (I agree that without appealing to the axiom of choice in the first place, nothing could be said about $I$ or $J$. But given $I$ and $\varphi$, do we have enough information to get a basis on the other side?) $\endgroup$ – Pace Nielsen Dec 7 '14 at 21:24
  • $\begingroup$ @Pace: well, you do need extra information. The underlying abelian group of the bimodule has two different module structures over a division ring, and the information you're trying to relate is the information of a basis for one module structure vs. the other. For a weird enough bimodule there's no particular reason these pieces of information should be very strongly related in general. $\endgroup$ – Qiaochu Yuan Dec 8 '14 at 4:20
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    $\begingroup$ "well, you do need extra information." If you can give a proof of this statement, then that is the answer I'm looking for. $\endgroup$ – Pace Nielsen Dec 8 '14 at 4:25
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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.

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  • $\begingroup$ Jeremy, your "simpler" question is quite interesting. I'm not sure that off the top of my head I could prove (without using AC) that $k(Y)$ has a $k$-basis even in the case when $Y$ is a singleton, so I look forward to any further comments you have. $\endgroup$ – Pace Nielsen Dec 15 '14 at 17:22
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    $\begingroup$ @PaceNielsen If $Y=\{y\}$ is a singleton, then $\{y^i:0\leq i\}$ together with all $y^i/p(y)^j$ where $p(y)$ is monic and irreducible and $0\leq i<\operatorname{deg}p$ forms a basis. $\endgroup$ – Jeremy Rickard Dec 15 '14 at 17:40
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    $\begingroup$ Jeremy, if you can make the ideas in your post work, I'll put up a new bounty and accept your answer. $\endgroup$ – Pace Nielsen Dec 16 '14 at 16:55
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    $\begingroup$ @PaceNielsen No need for you to add a second bounty for my sake. Actually, since my "answer" has already been upvoted, I think the system could automatically give me half the original bounty tomorrow, which I'd feel a little guilty about if I can't fill in the gaps fairly soon. If that happens, then I'll put my own bounty on the question so that I can sleep at nights. Sorry, I've been a bit busy and haven't had a chance to sit down and think through the details, but I'll try to do so before the end of the week. $\endgroup$ – Jeremy Rickard Dec 16 '14 at 17:36
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    $\begingroup$ I've tried to complete this to a proof, but it wasn't as easy as I hoped, and I (at least temporarily) admit defeat. As promised, I've added my own bounty to encourage others to have a try. I'm not sure either way whether the ideas in my "answer" might be useful. $\endgroup$ – Jeremy Rickard Dec 18 '14 at 10:12
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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.

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  • $\begingroup$ Hi Manny! While I'm familiar with all of this, it is still good to see it phrased in a way that is slightly different than how I might have put it. To push things a littler further, I'm wondering if we can know whether or not the $D$-action induced by $\varphi:D\to E$ gives a dimension to $_DV$ without using the axiom of choice (again). I'm somewhat leaning towards "no" since everything we did can be done outside the realm of vector spaces... $\endgroup$ – Pace Nielsen Dec 7 '14 at 21:34
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    $\begingroup$ I think I'm better understanding the question. Are you asking something close to: if ${}_D V_D$ has a right $D$-basis, must it have (without assuming AC) a left $D$_basis? Since existence of bases of vector spaces over fields implies AC (math.lsa.umich.edu/~ablass/bases-AC.pdf), I wouldn't be surprised if the existence of left bases (given a right basis) also implied AC. If this is what you're asking, perhaps you should modify the question to grab the attention of more logicians. $\endgroup$ – Manny Reyes Dec 9 '14 at 0:25

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