When and .

A projective generator of (or in any abelian category) is a projective object such that if then .

is a projective generator because is projective and via .

Let be an algebra. Let be a finitely-generated projective generator of . Let . Then the functors and are inverse equivalences of categories between and . We say that and are Morita equivalent.

is an -bimodule. For an -module , is a left -module via composition by a map on the right. is already a left -module, so is a left- module for a -module .

Let and (finitely-generated and free). Then .

Alternatively, we can take , and (the space of column vectors). In fact, , where

is projective because is an idempotent. Actually, via

This tells us that is a projective generator because

Now,

So we get a Morita equivalence in the other direction.

and are Morita equivalent for any algebra (even non-commutative).

Let and . By Tensor-Hom adjunction,

There are natural transformations (the unit and counit of adjunction respectively)

To construct , by adjunction (with ), we can just take . Similarly, to construct , by adjunction (with ), we an just take . The adjunctions carry the information of the two natural transformations. Our goal is to show that (Definition (Equivalence of categories)Definition (Equivalence of categories)):

But now, for , we have

Similarly, for , we have

Then .

Let for an arbitrary indexing set . Now,

where tensor commutes with direct sum because they have the same finiteness condition. But now, we have

where the isomorphism at is because is finitely-generated. To see this, let . We then note that is determined by where generate , and each has finitely many non-zero coordinates.

Write the beginning of a free resolution

We apply to this diagram to obtain

Note that:

So the bottom row is also exact (specifically, is surjective). By the -lemma, is an isomorphism.

We begin by proving that is surjective. We have

Suppose that . Then we get a map which is non-zero as is a projective generator. It has a lift because is projective and is surjective. Note that contains every image of every map from by definition of $\varepsilon_Y$definition of . In particular, it contains , and so the composition is zero, i.e., . This is a contradiction. So is surjective.

We have from surjectivitysurjectivity a short exact sequence

Recall that is exact by projectivity of . Applying , we again obtain a short exact sequence

But we already know that is an isomorphism, and after checking the maps, we find that we have a commutative triangle

So must also be an isomorphism (-out-of- property), forcing . Since

being a projective generator forces . So is injective.

with for , and for all .

Every finite-dimensional projective -module is

for some , and the same as defined in the decomposition of above, and is a projective generator if for all .

The truth of the theorem relies on some finiteness assumptions.

Krull-Schmidt Theorem.