1
$\begingroup$

I have $A$ and $B$ two rings and $P\in {_A\text{Mod}_B}$. What does $P_A$ mean and what does $_BP$ mean?

Also, what does $_B\text{End}(P)^{op}$ mean?

In the context of a Morita context $(B,A,P,Q,\mu,\tau)$. Then we have the following property:

$\mu$ is surjective

iff

$_BP$ is a generator, $A\cong {_B\text{End}(P)^{op}}$ and $Q\cong{_B\text{Hom}(P,B)}$

$\endgroup$
3
$\begingroup$

I would interpret $_AMod_B$ as the category of $A,B$ bimodules.

$P_A$ would mean a right $A$ module, and $_BP$ would mean a left $B$ module.

The last expression, $_B\mathrm{End}(P)^{op}$ is hard to interpret because it is unclear what $P$ is, and unclear what endomorphisms are being talked about, and unclear if it is referring to a module structure on the opposite ring, or the opposite ring of $B$ linear endomorphisms.

One simply cannot tell unless you provide more context.


From the context, it looks like $_B\mathrm{End}(P)$ is supposed to mean "the $B$ linear homomorphisms from $P$ to $P$." I think a rather clearer and more conventional way to write this is either as $\mathrm{End}_B(P)$ or better yet $\mathrm {End}(_BP)$. Writing the $B$ on the far left makes it look like notation for module structure (which is already being used) so it is a poor choice to put it there.

So in all, $\mathrm{End}(_BP)^{op}$ is the opposite ring of the ring of $B$ linear transformations on $P$.

$\endgroup$
  • $\begingroup$ sorry, I added some context $\endgroup$ – tomak Dec 19 '17 at 17:28
  • $\begingroup$ @tomak Great, I think that clears things up. $\endgroup$ – rschwieb Dec 19 '17 at 18:33

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.