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Inspired by other questions i have two questions about modules over division rings: given a division ring $D$ with center $Z(D)=K$. One has the notion of dimension for left modules (vector spaces) $V$ over $D$, which is well behaved like in linear algebra.

But does the following also hold: Given a left vector space $W$ of dimension $n$ and a right vector space $V$ of dimension $m$. Assume that at least one of them is even a $D$-bimodule, so that $V\otimes_D W$ has the structure of a $D$-module.

$Q1$: Do we have $dim_D(V\otimes _D W)=dim_D(V)dim_D(W)=nm$ in this case?

Now $D$ is naturally a $D$-bimodule, so is $D^{\*}=Hom_K(D,K)$.

$Q2$: Is there a $D$-bimodule isomorphism between $D$ and $D^{\*}$?

Both modules are one dimensional, so pick generators $x\in D$ and $y\in D^{\*}$. Now $x\mapsto y$ defines a map $D \rightarrow D^{\*}$, which satisfies $ax\mapsto ay$. Assume that $ax=xb$ for some $b\in D$, then we need to have $xb\mapsto yb$, which leads to $x^{-1}ax=y^{-1}ay$ for all $a\in D$. Can we find $x$ and $y$ such that the last relation is fulfilled? Or is there no such bimodule isomorphism?

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2 Answers 2

up vote 9 down vote accepted

The answer to both questions is positive if $D$ is finite-dimensional over its center $K$, and negative, in general, otherwise.

Q1. Suppose $V$ is a $D$-bimodule over $K$ (i.e., a $D\otimes_K D^{op}$-module), while $W$ is a left $D$-module of dimension $n$, as in your question. Then $W$ is isomorphic to a direct sum of $n$ copies of $D$, hence $V\otimes_D W$ as a left $D$-module is isomorphic to a direct sum of $n$ copies of $V$. So the $\dim_D (V\otimes_D W) = n\dim_D V$, where $\dim_D$ denotes the dimension of left $D$-modules. Now, to answer your original question, it remains to notice that the dimensions of $V$ as a left and a right $D$-module coincide, since both are equal to $\dim_KV/\dim_KD$.

It does not matter in this argument that $K$ is the whole center of $D$, only that $K$ is a field contained in the center of $D$ and $D$ is finite-dimensional over $K$. On the other hand, if $D$ is infinite-dimensional over $K$, it may happen that $D$ can be $K$-linearly embedded into itself as a proper subring. This would allow to define a $D$-bimodule structure on $V=D$ where $V$ is one-dimensional as a right module over $D$ and more than one-dimensional (possibly infinite-dimensional) as a left module over $D$. This would break your dimension formula.

Q2. Left $D$-module maps $D\to D^\ast$ are in one-to-one correspondence with elements of $D^\ast$, i.e., $K$-linear functions $D\to K$. A linear function $l\colon D\to K$ defines a $D$-bimodule map $D\to D^\ast$ if and only if one has $l(xy)=l(yx)$ for all elements $x$, $y\in D$. Taking $l$ to be the trace map (defined e.g. by tensoring $D$ with the separable closure of $K$ over $K$, identifying the result with matrices over the separable closure, and taking the traces of matrices), one can obtain a $D$-bimodule isomorphism between $D$ and $D^\ast$ when $D$ is finite-dimensional over its center $K$.

The above argument can be easily extended to the case when $K$ is just a subfield of the center of $D$ and $D$ is finite-dimensional over $K$ (one just composes the trace map with an arbitrary nonzero linear function from the center of $D$ to $K$). On the other hand, when $D$ is infinite-dimensional over $K$, it is no longer true that $D^\ast$ is a one-dimensional left $D$-module.

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Nice answer Leonid! –  javier Mar 21 '11 at 22:05
    
For what it's worth, while studying D-D-bimodules over K = Z(D), with D of (finite) dimension n over K, one might remark that $D\otimes_K D^\circ$ is isomorphic to $K^{n\times n}$ (e.g. Reiner, Maximal Orders, section on Brauer group). –  Matthias Künzer Mar 22 '11 at 6:24
    
A minor correction: In your answer to Q1 you need to assume that $V$ is $K$-balanced, i. e. that $kv=vk$ for all $v\in V$ and $k\in K$. If $D$ is finite-dim over a field such that this holds, then everething is well. Otherwise, one can easily contruct counterexamples (somewhat artificially, of course): Consider the function field $K=\mathbb{Q}(x)$ as $K$-$K$-bimodule, where left multiplication is as usual, and right multiplication given by $kp(x)=kp(x^2). Then the left and right dimensions are different. –  Frieder Ladisch Mar 22 '11 at 12:33
    
@F.Ladisch: I made the assumption that you mention in the first sentence of my answer to Q1: "Suppose $V$ is a $D$-bimodule over $K$ (i.e., a $D\otimes_K D^{op}$-module) ..." –  Leonid Positselski Mar 22 '11 at 14:13
    
Sorry, I did not read careful enough. I wasn't aware that the phrase "is a $D$-bimodule *over $K$*" means the same as "$K$-balanced". –  Frieder Ladisch Mar 22 '11 at 14:35

Concerning Q2, $D$ and $D^* $ are in general not isomorphic as $K$-vector spaces: suppose that $D$ has countable infinite dimension as a vector space over $K$. The dual vector space $D^*$ has then uncountable dimension over $K$. If $K$ is a countable field (say, the rational numbers) it follows that $D$ has a countable number of elements, but $D^* $ doesn't, so there cannot exist a set bijection between the two, let alone a module isomorphism.

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Countability does not matter. No infinite-dimensional vector space $D$ over a field $K$ is isomorphic to its dual vector space $D^\ast$. –  Leonid Positselski Mar 21 '11 at 22:13
    
True, the countability argument was just the easiest way I could think of to show a counterexample. Your trace argument is way more elegant :-) –  javier Mar 21 '11 at 22:24

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