Let $A$ and $B$ be Artin $k$-algebras for a commutative artinian ring $k$ (e.g. $A$ and $B$ are finite dimensional $k$-algebras for a field $k$). Let $M$ be an $A$-$B$-bimodule of finite length over $k$. We say that $x\in M$ generates $M$ if for all $m\in M$ there exist $a\in A$ and $b\in B$ with $m = ax+xb$. If $x\in M$ and $y\in M$ generate $M$, does it follows that $x = ayb$ for invertible elements $a\in A$ and $b\in B$? The question is motivated by If the Hom-space of finite length modules is generated by single elements, must the elements be conjugate?
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No.
Let $k = \mathbb{R}$ and $A = B = \mathbb{C}$. We can identify $A \otimes B \cong \mathbb{C} \times \mathbb{C}$. On pure tensors, this isomorphism takes $a \otimes b \mapsto (ab, a\bar{b})$. Notably, an element $(r,s) \in \mathbb{C} \times \mathbb{C}$ is a pure tensor only if $\|r\|^2 = \|s\|^2$.
Pick any impure $(r,s) \in \mathbb{C}^\times \times \mathbb{C}^\times$, and take $x = (1,1)$ and $y = (r,s)$. Then $x$ and $y$ are not conjugate in the sense that you ask, but both generate $M = A \otimes B$ as an $A$-$B$-bimodule.
answered Sep 3 at 12:17
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$\begingroup$ In the question I defined what I meant with "generated". This is not the same as being generated as a bimodule in the normal sense and I think your example does not work $\endgroup$ Commented Sep 3 at 17:42
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1$\begingroup$ Good point. Your definition does not match standard usage. My example is not singly-generated in your sense: $(1,1)$ generates-in-your-sense the set of pairs $(a+b,a+\bar{b})$, i.e. the set of pairs $(r,s)$ with $Re(r) = Re(s)$; this is not a submodule! $\endgroup$ Commented Sep 4 at 11:02