2

Let $E_1$, $E_2$ be finite-dimensional division algebras over $\mathbb{Q}$. Let $X$ be a left $E_1 \otimes_\mathbb{Q} E_2^{op}-$module. In other words, $X$ is an $E_1-E_2-$bimodule, and $\mathbb{Q}$ acts in the same way from the left and from the right.

Notice that $E_1 \otimes_\mathbb{Q} E_2^{op}$ has a $\mathbb{Q}$-basis $\{f_i \otimes h_j\}$ of pure tensors, where $\{f_i\}$ is a basis of $E_1$ and $\{h_j\}$ is a basis of $E_2$. Assume moreover that $X$ is generated by a single element $M$, i.e.: $X=\{\sum_{ij} a_{ij} \ f_i M h_j \mid a_{ij} \in \mathbb{Q}\}$. Here is the question:

If I know exactly which pure tensors $f \otimes h$ fix $M$, can I deduce which sums of tensors fix $M$? In particular, if I know that no non-trivial pure tensor fixes $M$, can I show that no sum does either (and hence that the annihilator of $M$ is trivial and $X$ is freely generated of rank $1$ over $E_1 \otimes_\mathbb{Q} E_2^{op}$)?

57Jimmy
  • 6,266
  • Jimmy, I (obviously) like this question. I feel a bit bad about going with a commutative example. My guess is that the same may happen with central (over $\Bbb{Q}$) division algebras, but I'm anything but 100%. Just a hunch based on the fact that $\Bbb{Q}(\root3\of2)$ is a subfield of the 9-dimensional central division algebra I used here. It may well be that this holds for all quaternion algebras. At least in that case the subfields of the division algebras are all Galois extensions. – Jyrki Lahtonen Jun 24 '20 at 07:30
  • And thanks for this remark too. I'll try to look at it in more depth – 57Jimmy Jun 25 '20 at 08:52

1 Answers1

1

Proffering the following counterexample to get the ball rolling.

In this example $E_1=\Bbb{Q}(\root3\of2)$ and $E_2=\Bbb{Q}(\root3\of2\omega), \omega=e^{2\pi i/3}$. In other words, they are both commutative. I don't know if the conclusion is different if we add the assumption that $E_1$ and $E_2$ should be central.

Let us consider the bimodule $X=\Bbb{Q}(\root3\of2,\omega)$ with both $E_1$ and $E_2$ acting via the usual multiplication.

  • Clearly $M=1$ generates $X$ as a bimodule.
  • If $f_1\in E_1$ and $f_2\in E_2$ are such that $(f_1\otimes f_2)$ fixes $M$, then $f_1f_2=1$ implying that $f_1$ and $f_2$ are in the intersection $E_1\cap E_2=\Bbb{Q}$, and $f_1\otimes f_2=1\otimes 1$ is trivial.
  • Yet, $\dim_\Bbb{Q}X=6<9=\dim_\Bbb{Q}(E_1\otimes E_2)$ so $X$ is not freely generated by $M$.

The pair $(E_1,E_2)$ is a standard example of two extension fields of $\Bbb{Q}$ such that they intersect trivially, but are not linearly disjoint. That played a big role in finding the construction.

Jyrki Lahtonen
  • 133,153