8
$\begingroup$

Let us say that an algebra $A$ over a field $k$ is Picard-surjective if the canonical map $$ \mathrm{Aut}(A) \rightarrow \mathrm{Pic}(A)$$ is surjective. Here $\mathrm{Pic}(A)$ denotes the group of isomorphism classes of invertible $A$-$A$-bimodules and the map sends an automorphism $\alpha$ to the $A$-$A$-bimodule $A_\alpha$, where the left action is the usual one and the right action is via $\alpha$.

Q: For any given finite-dimensional $k$-algebra $A$, does there exist a Morita-equivalent one that is Picard-surjective?

If not, I am interested in conditions under which this is true. I am mainly interested in the case $k=\mathbb{R}$ or $\mathbb{C}$, and for all the examples that I came up with so far, this seems to be correct, as far as I can tell.

$\endgroup$
8
$\begingroup$

Yes, the basic algebra of $A$ will be Picard-surjective.

The basic algebra is the endomorphism algebra $\operatorname{End}_A(\bigoplus_{i=1}^{n}P_i)$ of the direct sum of indecomposable projective (right) modules, one from each isomorphism class. It is Morita equivalent to $A$.

Suppose $A$ is basic. Then as a left or right module, $A$ is the direct sum of indecomposable projective modules, one from each isomorphism class. Suppose $M$ is an invertible bimodule. Since $S\otimes_AM$ is nonzero for every simple module $S$, a direct sum decomposition of $M$ as a left module must contain at least one copy of each indecomposable projective. Let $X=\bigoplus_{i=1}^nS_i$ be the direct sum of simple (right) $A$-modules, one from each isomorphism class. Since $X\otimes_AM\cong X\cong X\otimes_AA$, as a left module $X$ must contain exactly one copy of each indecomposable projective in a direct sum decomposition. So $M\cong A$ as left modules. The right $A$-module structure of $M$ is then given by an injective algebra homomorphism $A^{op}\to\operatorname{End}_A(_AA)\cong A^{op}$, which is an isomorphism by finite dimensionality: i.e., the right action is induced by an automorphism of $A$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.