# Condition for Morita equivalence

Let $$R$$ and $$S$$ be two rings. We say that $$R$$ and $$S$$ are Morita equivalent if the categories of right $$R$$-modules $$\text{Mod}_{R}$$ and right $$S$$-modules $$\text{Mod}_{S}$$ are equivalent.

We say that an $$R$$-module $$P$$ is a progenerator of $$\text{Mod}_{R}$$ if $$P$$ is a finitely generated projective module which is a generator of $$\text{Mod}_{R}$$, and we have the following known result:

Proposition. $$R$$ and $$S$$ are Morita equivalent if and only if there is a progenerator $$P$$ of $$\text{Mod}_{R}$$ such that $$\text{End}(P) \simeq S$$.

Now, it is also known that $$R$$ and $$S$$ are Morita equivalent if and only if the categories of left $$R$$-modules $$_{R}\text{Mod}$$ and left $$S$$-modules $$_{S}\text{Mod}$$ are equivalent. Therefore, my question is, what is the correspondent result of the above proposition for left modules?

The equivalent proposition is:

Proposition. $$R$$ and $$S$$ are Morita equivalent if and only if there is a progenerator $$P$$ of $$_{R}\text{Mod}$$ such that $$\text{End}(P) \simeq S^{\text{op}}$$.

While searching for the answer of this question, I found some texts in the internet which stated the above proposition putting $$\text{End}(P) \simeq S$$ instead of $$\text{End}(P) \simeq S^{\text{op}}$$, and it is wrong! I thought it was important to share this detail, because it may confuse people, as it confused me. Now let's see the explanation:

The fundamental motive for the appearence of $$S^{\text{op}}$$ instead of $$S$$ is the way we compose functions. We have the following (which is easily verified):

If we regard the ring $$S$$ as a right module ($$S_{S}$$), then $$\text{End}(S_{S}) \simeq S$$ and if we regard $$S$$ as a left module ($$_{S}S$$), then $$\text{End}(_{S}S) \simeq S^{\text{op}}$$ (those are ring isomorphisms).

Now, for more details, let us examine the necessity of those propositions closely:

If $$F : \text{Mod}_{S} \rightarrow \text{Mod}_{R}$$ is an equivalence of categories, then putting $$P_{R} = F(S_{S})$$, we may show that $$P_{R}$$ is a progenerator, and to verify that $$\text{End}(P_{R}) \simeq S$$, we proceed like this: $$\text{End}(P_{R}) = \text{End}(F(S_{S})) \simeq \text{End}(S_{S}) \simeq S.$$

And for the left case we have:

If $$F : \text{ }_{S}\text{Mod} \rightarrow \text{ }_{R}\text{Mod}$$ is an equivalence of categories, then putting $$_{R}P = F(_{S}S)$$, we may show that $$_{R}P$$ is a progenerator, and we have: $$\text{End}(_{R}P) = \text{End}(F(_{S}S)) \simeq \text{End}(_{S}S) \simeq S^{\text{op}}.$$

To prove the sufficiency of the propositions, we proceed as follows:

If $$P$$ is a progenerator of $$\text{Mod}_{R}$$ such that $$\text{End}(P) \simeq S$$, then we may show that $$\text{Hom}_{R}(P,-) : \text{Mod}_{R} \rightarrow \text{Mod}_{S}$$ is an equivalence of categories, and for $$M$$ in $$\text{Mod}_{R}$$, we regard $$\text{Hom}_{R}(P,M)$$ as a right $$\text{End}(P)$$-module in the usual way, by composing functions, so that it becomes a right $$S$$-module.

For the left case we have:

If $$P$$ is a progenerator of $$_{R}\text{Mod}$$ such that $$\text{End}(P) \simeq S^{\text{op}}$$, then we may show that $$\text{Hom}_{R}(P,-) : \text{ }_{R}\text{Mod} \rightarrow \text{ }_{S}\text{Mod}$$ is an equivalence of categories, and for $$M$$ in $$_{R}\text{Mod}$$, we regard $$\text{Hom}_{R}(P,M)$$ as a left $$\text{End}(P)^{\text{op}}$$-module in the usual way, by composing functions and making the necessary adjustments (putting the $$^{\text{op}}$$) so that everything works, and then it becomes a left $$S$$-module, since $$\text{End}(P)^{\text{op}} \simeq (S^{\text{op}})^{\text{op}} = S$$.

• I would like to add that $\operatorname{End}_R(P)\simeq S$ is not necessarily so much wrong as very understandably confusing. It is a pretty common convention when working with non-commutative rings to write $\operatorname{End}_R(P)$ for the opposite of the usual endomorphism algebra of a left module, so that $P$ is a $R$-$\operatorname{End}_R(P)$-bimodule. Commented Mar 2, 2020 at 15:46