Here is a fact from linear algebra: if $ V$ is a finite dimensional vector space over a field $ k$ then $ V$ is naturally isomorphic to its double dual. Of course, it is also isomorphic to its dual as well, but it is the natural isomorphism that is more interesting.

Write $ V^* = \mathrm{Hom}_k(V,k)$ for the dual space of $ V$.

This is not true in general for infinite dimensional vector spaces. Before I actually talk about the subject matter, let us describe a counterexample:if we use the field $ k = \mathbb{Q}$, and $ I$ is a countable set, then $ \mathbb{Q}^{(I)}$ is also countable (here, the notation $ \mathbb{Q}^{(I)}$ refers to the direct sum of $ I$ copies of $ \mathbb{Q}$). The dual space of any vector space has cardinality at least as great as the original space, but $ \mathrm{Hom}_k(\mathbb{Q}^{(I)},\mathbb{Q})\cong\prod_{i\in I}\mathbb{Q}$, which is uncountable.

Anyways, the point is $ V\cong V^{**}$. Why is this? Of course, one can provide a short and direct proof: define $ \Phi:V\to V^{**}$ by $ \Phi(v)(f) = f(v)$. This map is injective and since $ V$ is finite dimensional, it is surjective.

Is this really satisfying though? Sometimes in mathematics, it is not completely satisfying just to prove something. Instead, one must also find the proper context for results. The proper context should be enlightening in a variety of situations. For instance, that we know $ V\cong V^{**}$ where $ V$ is a finite-dimensional vector space isn’t all that enlightening. We have a natural isomorphism, so we ought to ask: is there an equivalence of categories hiding somewhere? Indeed there is, and it’s called Morita theory!

Motivations

moritaMorita theory, created by Kiiti Morita (left) gives a detailed description of how two module categories can be equivalent. In the remainder of the post we shall see the building block theorem of Morita theory and how it explains the above phenomenon. Morita theory is described well in many places, such as Jacobson’s Basic Algebra II, so I won’t go into details with proofs.

So we start with a ring $ R$ with unity. (In the special case above, $ R$ will be a field.) Now, we want to find another ring $ S$ for which the category of (right) $ R$-modules is the same thing as (left) $ S$-modules. The reason why we switch from right to left is just a technicality. So, how can we find the ring $ S$? Whenever we have the category of $ R$-modules, there is a handy way of finding many rings that contain reams of information. That is, if $ M$ is any $ R$-module, then $ \mathrm{Hom}_R(M,M)$ is a ring, with addition being pointwise and multiplication being composition.

Of course, we cannot always expect the category of $ R$-modules to be equivalent to the category of $ S$-modules for any choices of $ M$ (take $ M= 0$). But if $ M$ is a progenerator, then we do get an equivalence! Wait, what’s a progenerator? Well, the ‘pro’ stands for finitely generated projective, and the ‘generator’ means the trace ideal is all of $ R$.

Trace Ideal

Let’s define the trace ideal. Let $ M$ be any left $ R$-module and let $ S = \mathrm{Hom}_R(M,M)$, the endomorphism ring of $ M$. Let $ M^*$ be the dual of $ M$. Now, it happens that $ M^*$ is a right $ S$-module (acting on the inside of the function!) and $ M$ is a left $ S$-module (acting as an endomorphism). We can form the map $ \theta: M^*\otimes_S M\to R$ defined by $ \sum f_i\otimes m_i = \sum f_i(m_i)$.

Definition. We call an $ R$-module $ M$ a generator if the image of the map $ \theta$ defined above is all of $ R$.

The image of the map $ \theta$ is called the trace ideal. Furthermore, if $ M$ is finitely generated and projective, then it is a short exercise to show that $ \theta$ is an isomorphism, an observation that is the basis of Morita theory itself.

The Morita Theory

Definition. We call an $ R$-module $ M$ a progenerator if it is finitely generated projective and a generator.

Building on the $ \theta$ map:

Theorem. Consider the functor $ F = -\otimes_R M:\mathcal{M}_R\to ~_S\mathcal{M}$ from right $ R$-modules to left $ S$-modules, where $ M$ is a fixed left $ R$-module. If $ M$ is a progenerator, this defines an equivalence of categories, with inverse equivalence defined by $ G = M^*\otimes_S-$, where $ M^*$ is considered as a right $ S$-module.

Now we recover that $ \theta$ is an isomorphism via $ G\circ F$.

The fact that we can consider any $ R$-module and $ S$-module allows us to do some pretty cool things: $ M$ itself is $ \mathrm{Hom}_S(S,M)$ where $ M$ is considered as an $ S$-module. Note that $ \mathrm{Hom}_S(S,M)$ is also a right $ R$-module via $ M$, and one easily verifies that this action is the same as the $ R$-action on $ \mathrm{Hom}_R(M^*\otimes_S,M^*\otimes_S M)$, which is naturally isomorphic to the $ \mathrm{Hom}_S(S,M)$ (as $ R$-modules!) via the Morita equivalence of categories, and this isomorphic to the dual space $ M^{**}$.

Of course, we needed that $ M$ be a generator – for a commutative ring, this is easy to check, and is equivalent to the annihilator of $ M$ in $ R$ being trivial. In the special case of $ R$ being commutative with no nontrivial idempotents, then this is always true.

Hence we obtain as a special case of Morita theory:

Theorem. If $ M$ is a finitely generated projective module over a commutative ring $ R$ and $ R$ has no nontrivial idempotents, then $ M\cong M^{**}$.

Of course, in the case of $ R$ being a field, this reduces to being that a finite-dimensional vector space is isomorphic to its double dual. Nonetheless, in the context of Morita theory, this phenomenon becomes clear and much more useful in a variety of situations. Hopefully this should convince the reader to look for an equivalence of categories or an adjunction whenever there is some natural homomorphism floating around.

Leave a Reply

Your email address will not be published. Required fields are marked *