User:Marty/UCSC Math 203 Fall 2008/Week 6

We have discussed categories and functors already, including additive and abelian categories. This week, we will discuss some important examples, and a bit of homological algebra.

=Category Theory=

Categories
Suppose that $C$ is a category. Remember the meaning of this. This data is required to satisfy certain axioms:
 * There is a class $Ob(C)$ of "objects" of the category $C$.
 * For any two objects $X,Y$ of $C$, there is a set $Mor(X,Y)$ of morphisms from $X$ to $Y$.
 * For any three objects $X,Y,Z$, there is a composition function $Mor(Y,Z) \times Mor(X,Y) \rightarrow Mor(X,Z)$.
 * Existence of identity morphisms, from any object to itself.
 * Associativity of composition.

We often work with additive categories. These have additional data and axioms:
 * The set of morphisms $Mor(X,Y)$ has the structure of an abelian group. It is usually called $Hom(X,Y)$.
 * Composition is $\ZZ$-bilinear on morphisms.
 * There is a zero object.
 * There are finite direct sums.

We often work with abelian categories, which are additive categories with additional axioms:
 * Every morphism has a kernel and cokernel.
 * Every monomorphism is a kernel.
 * Every epimorphism is a cokernel.

The canonical example of an abelian category is the category of (left) $A$-modules, whenever $A$ is a (not necessarily commutative, but always unital and associative) ring.

There are some operations on categories that you should be familiar with. These operations are valid for general categories, additive categories, and abelian categories:
 * The product of a finite number of categories is a category.
 * Objects are ordered tuples of objects.
 * Morphisms are ordered tuples of morphisms.
 * Composition is defined entrywise.
 * The "opposite category".
 * Objects are the same as before.
 * Morphisms are morphisms in the opposite direction.

Functors
Suppose that $C$ and $D$ are categories. A functor $F \colon C \rightarrow D$ consists of: This data is required to satisfy axioms:
 * A function $F \colon Ob(C) \rightarrow Ob(D)$.
 * For every pair of objects $X,Y$ of $C$, a function $F_{X,Y} \colon Mor_C(X,Y) \rightarrow Mor_D(F(X), F(Y))$.
 * Identity morphisms go to identity morphisms.
 * Composition is preserved.
 * If we are working with additive categories (including abelian categories), we require $F_{X,Y}$ to be a group homomorphism.

There are a few examples that are particularly important for this class:
 * Pullbacks. If $\phi \colon A \rightarrow B$ is a ring homomorphism, then every $B$ module can be "viewed" as an $A$-module.  This yields a functor $\phi^\ast$ from the abelian category $Mod_B$ to the abelian category $Mod_A$.  In particular, if $\phi \colon \ZZ \rightarrow B$ is a ring homomorphism, then this functor "forgets" the module structure, and only remembers the abelian group structure.
 * Hom functors. Suppose that $A$ is a ring, and $M$ is an $A$-module.  Then, $Hom(M, \bullet)$ is a functor from $Mod_A$ to $Ab$ (the category of abelian groups).  Also, $Hom(\bullet, M)$ is a functor from $Mod_A^{op}$ to $Ab$.  Altogether, $Hom(\bullet, \bullet)$ is a functor from $Mod_A^{op} \times Mod_A$ to $Ab$.
 * Tensor functors. Suppose that $A$ is a (commutative) ring, and $M$ is an $A$-module.  Then $\bullet \otimes_A M$ is a functor from $Mod_A$ to $Mod_A$.
 * Identity functors. There is always the identity functor $Id_C$ from any category $C$ to itself!

Natural transformations
Suppose that $C$ and $D$ are categories. Suppose that $F,G \colon C \rightarrow D$ are functors. A natural transformation from $F$ to $G$ consists of: This collection of morphisms is required to satisfy the following properties:
 * For every object $X$ of $C$, a morphism $\Phi_X \in Mor_D(F(X), G(X))$.
 * For every morphism $\phi \in Mor_C(X,Y)$ of $C$, we have: $\Phi_Y \circ F_{X,Y}(\phi) = G_{X,Y}(\phi) \circ \Phi_X$.

Two such functors $F$ and $G$ are called naturally isomorphic, if there exist natural transformations $\Phi \colon F \rightarrow G$, and $\Psi \colon G \rightarrow F$, such that $[\Psi \circ \Phi]_X = Id_{F(X)}$ and $[\Phi \circ \Psi]_X = Id_{G(X)}$ for all objects $X$ of $C$.

There is always an identity natural transformation from any functor to itself. This is a natural isomorphism. Here is another example: $$T_1(X,Y) \rightarrow T_2(X,Y), \mbox{ for every object } (X,Y) \mbox{ of } Mod_A \times Mod_A.$$
 * The tensor product is commutative. In other words, consider the following two functors from $Mod_A \times Mod_A$ to $Mod_A$ (where $A$ is a commutative ring):
 * $T_1(X,Y) = X \otimes_A Y$. (and the appropriate behavior on morphisms)
 * $T_2(X,Y) = Y \otimes_A X$. (and the appropriate behavior on morphisms)
 * There is a natural isomorphism $\Phi \colon T_1 \rightarrow T_2$, given by the canonical map:

More generally, natural transformations are used to define equivalence of categories:
 * Two categories $C$ and $D$ are called equivalent, if there exist functors $F \colon C \rightarrow D$, and $G \colon D \rightarrow C$, such that $F \circ G$ is naturally isomorphic to $Id_D$, and $G \circ F$ is naturally isomorphic to $Id_C$.

=Center of a Category=

Suppose that $C$ is an additive category. There is a commutative ring, associated to $C$, called the center of the category, or $Z(C)$. It is defined by:
 * $Z(C)$ is the ring of natural transformations from the identity functor to itself.

Understanding this definition is a long exercise in understanding the definition of functors and natural transformations. Consider an element $z \in Z(C)$. This is a natural transformation from the identity functor to itself. Therefore, for every object $X \in C$, there is a morphism: $$z_X \in Mor_C(Id_C(X), Id_C(X)) = Hom(X,X).$$ These morphisms are required to satisfy the following:
 * For all $\phi \in Hom(X,Y)$, where $X,Y$ are objects of $C$, $\phi \circ z_X = z_Y \circ \phi$.

I claim that such natural transformations -- such systems of homomorphisms $(z_X)$ -- form a commutative ring! $$\phi \circ (z_X \circ w_X) = (\phi \circ z_X) \circ w_X = (z_X \circ \phi) \circ w_X = z_X \circ (\phi \circ w_X) = z_X \circ (w_X \circ \phi) = [z_X \circ w_X] \circ \phi.$$
 * One may choose $0_X \in Hom(X,X)$, for every object $X$, for the zero element of the ring.
 * One may choose the identity $Id_X \in Hom(X,X)$, for every object $X$, for the unit element of the ring.
 * The set $Hom(X,X)$ is an abelian group, for all $X$ in the (remember, additive!) category $C$. So, define $[z + w]_X = z_X + w_X$.  This makes $Z(C)$ an abelian group for addition.  Notice that $\phi \circ [z_X + w_X] = [z_Y + w_Y] \circ \phi$, as required, since composition is $\ZZ$-bilinear.
 * The set $Hom(X,X)$ is a ring, whose "multiplication" is given by composition. This distributes over addition, since composition is $\ZZ$-bilinear.  The identity element is the unit element of the ring.  Compositions still are natural transformations, since if $z,w \in Z(C)$, then we have:
 * The ring $Z(C)$ is commutative, since for all $z,w \in Z(C)$, and all objects $X$ of $C$, we find that $w_X \in Hom(X,X)$, and thus $w_X \circ z_X = z_X \circ w_X$.

Through these verifications, we find that to every additive category $C$, there is a naturally associated commutative ring $Z(C)$, called the center of the category $C$. We can demonstrate that the center of a category depends, up to isomorphism, only upon the equivalence class of the category $C$. In other words,

Proposition: Suppose that $C$ and $D$ are equivalent categories. Then $Z(C)$ and $Z(D)$ are isomorphic rings.

Proof: A lot of diagram-following and definitions!

When $A$ is a commutative ring, one may completely recover $A$ from its additive category of modules, as follows:

Theorem: $A$ is naturally isomorphic to the center of the category $Mod_A$.

There is a natural choice of ring homomorphism from $A$ to $Z(Mod_A)$, defined as follows. For every element $a \in A$, define $\mu(a) \in Z(Mod_A)$ as follows: for every $A$-module $M$, let $\mu(a)_M \in Hom_A(M,M)$ be the endomorphism sending $m$ to $a \cdot m$. This satisfies the required properties!
 * Demonstrating that $\mu \colon A \rightarrow Z(Mod_A)$ is a ring homomorphism is not difficult -- it follows from the definitions.
 * Demonstrating that $\mu$ is injective is not difficult, by considering $\mu(a)_A$ (viewing $A$ as an $A$-module).
 * Demonstrating that $\mu$ is surjective is more difficult. One must use the following facts:
 * Suppose that $z \in Z(Mod_A)$. If $M = \bigoplus M_i$ is a direct sum of modules, then $z_M$ is uniquely determined by $z_{M_i}$ for all $i$.
 * Suppose that $z \in Z(Mod_A)$. If $M \rightarrow N$ is a surjective morphism of modules, then $z_N$ is uniquely determined by $z_M$.
 * By choosing generators, every module is a quotient of a free module.
 * The only elements of $Hom_A(A,A)$ (endomorphisms of $A$ as an $A$-module) are the "multiplication by $a$" maps.

More generally, if $A$ is a not necessarily commutative ring, $Z(Mod_A)$ is naturally isomorphic to the center of the ring $A$! This explains the terminology "center of the category".

=Central Idempotents=

Remember that idempotents in a commutative ring are elements $e$ satisfying $e^2 = e$. Such elements have the effect of "breaking up" a ring as a product of two rings: $$A \cong eA \oplus (1-e) A,$$ where $eA$ and $(1-e)A$ are rings whose unit elements are $e$ and $(1-e)$, respectively.

For noncommutative rings, this role is played by central idempotents, which are elements $e$ of the center, satisfying $e^2 = e$. These again "break up" the ring as a product of two rings.

Idempotents in a commutative ring also "break up" the spectrum into disjoint subspaces, e.g.: $$Spec(A) \cong Spec(eA) \sqcup Spec( (1-e) A).$$

Here, we observe that central idempotents break up categories of modules. Theorem: Suppose that $A$ is a (not necessarily commutative) ring. Suppose that $z$ is an idempotent in $Z(Mod_A)$, or equivalently, $z = \mu(e)$ is a central idempotent in $A$. Then $z$ breaks up the category of modules into a product of categories: $$Mod_A \cong Mod_{e A} \times Mod_{(1-e)A}.$$

We prove this, using an idempotent $z \in Z(Mod_A)$. For every object $M$ of $Mod_A$, we may consider the following two objects of $Mod_A$: $$M_1 = Ker(z_M \colon M \rightarrow M), \mbox{ and } M_2 = Ker(1 - z_M \colon M \rightarrow M).$$ Since $z$ is idempotent, we find that $z_M \circ z_M = z_M$. It follows almost directly that $M$ is the direct sum of its two subobjects: $$M = M_1 \oplus M_2.$$ Furthermore, $M_1$ is naturally an $(1-e)A$-module, and $M_2$ is naturally a $eA$-module, if $z = \mu(e)$ for a central idempotent $e$. Since $z$ is a natural transformation, it follows that the decomposition $M = M_1 \oplus M_2$ is functorial, i.e., given a morphism $\phi \colon M \rightarrow N$, one has morphisms $\phi_1 \colon M_1 \rightarrow N_1$ and $\phi_2 \colon M_2 \rightarrow N_2$, compatible with the decompositions of $M$ and $N$ and the morphism $\phi$. This yields a functor: $$Mod_A \rightarrow Mod_{(1-e)A} \times Mod_{eA}.$$ It remains to prove that this is an equivalence of categories,

For a functor in the reverse direction, suppose that $M_1, M_2$ is a $(1-e)A$-module, $eA$-module, respectively. Then, $M_1 \oplus M_2$ is naturally an $A$-module, where we define: $$a (m_1, m_2) = ((1-e)a \cdot m_1, ea \cdot m_2).$$ We leave it to the reader to demonstrate the following:
 * This defines a functor from $Mod_{(1-e)A} \times Mod_{eA}$ to $Mod_A$.
 * The categories $Mod_A$ and $Mod_{eA} \times Mod_{(1-e)A}$ are equivalent.

=Morita Equivalence=

Suppose that $A$ and $B$ are commutative rings. Then, since $A \cong Z(Mod_A)$, and $B \cong Z(Mod_B)$, we find that:

Theorem: Two commuative rings are isomorphic if and only if their categories of modules are equivalent (as additive categories).

For noncommutative rings, there is, instead, a definition:

Definition: Two rings (associative, unital, not necessarily commutative) $A$ and $B$ are called Morita equivalent if $Mod_A$ is equivalent to $Mod_B$ as additive categories.

One of the most basic and fundamental results is about "categories that look like categories of vector spaces":

Theorem: Suppose that $F$ is a (commutative) field, and $A$ is a (not necessarily commutative) ring. Then $A$ is Morita-equivalent to $F$ if and only if $A$ is isomorphic to a matrix ring over a division ring whose center is $F$. In other words, there exists a division ring $B$ whose center is $F$, and there exists a natural number $n$, such that $A$ is isomorphic to $M_n(B)$.

When $F$ is algebraically closed, there are no nontrivial division rings whose center is $F$. It follows that $A$ is Morita equivalent to $F$ if and only if $A$ is isomorphic to a matrix ring $M_n(F)$.

This is essentially the Artin-Wedderburn theorem. Indeed, if $A$ is Morita-equivalent to a field, then the center of $A$ is isomorphic to $F$, i.e., $A$ is a central $F$-algebra. One can check that $A$ is a simple $F$-algebra, i.e., that $A$ has no two-sided ideals. From this, the above theorem classifies all central simple $F$-algebras as being matrix rings over division rings.

=Projective Objects=

Suppose that $A$ is a unital associative ring. We define projective objects in the category of $A$-modules. This definition makes sense in any abelian category:

Definition: An $A$-module $P$ is projective, if one of the following equivalent conditions is satisfied:
 * $P$ is a direct summand of a free $A$-module. I.e., there exists a free $A$-module $F$, and an $A$-submodule $Q$ of $F$, such that $F = Q \oplus P$.
 * If $\pi \colon M \rightarrow N$ is a surjective $A$-module morphism, and $f \colon P \rightarrow N$ is a $A$-module homomorphism, then $f$ "lifts" to $M$, in the sense that there exists a morphism $\tilde f \colon P \rightarrow M$, such that $\pi \circ \tilde f = f$.
 * The functor $Hom(P, \bullet)$ is exact.
 * If $\pi \colon M \rightarrow P$ is a surjective $A$-module morphism, then there exists a right-inverse $\sigma$ of $\pi$, i.e., there exists $\sigma \colon P \rightarrow M$, such that $\pi \circ \sigma = Id_P$.

For example, consider the ring $A = \ZZ$. A finitely generated projective $\ZZ$-module is a direct summand of $\ZZ^r$ for some $r \geq 0$. It follows that a finitely generated projective $\ZZ$-module is torsion-free, and hence free (using the classification of modules over a PID, for example). In fact, a $\ZZ$-module (finitely-generated or not) is projective if and only if it is free.

There are important and subtle differences between the three adjectives projective, free, and flat.
 * Free implies projective implies flat.
 * Flat and finitely-presented implies projective.
 * When $A$ is commutative, flat is equivalent to "locally free".
 * For local rings and principal ideal domains, finitely generated and projective implies free.

Since every module is a quotient of a free module, every module is a quotient of a projective module as well.

=Injective Objects=

Suppose that $A$ is a unital associative ring. Suppose that $I$ is an $A$-module.

Definition: $I$ is called an injective $A$-module, if one of the following equivalent conditions is satisfied:
 * If $\pi \colon L \rightarrow M$ is an injective $A$-module homomorphism, and $f \colon L \rightarrow I$ is an $A$-module homomorphism, then $f$ extends to an $A$-module homomorphism $\tilde f \colon M \rightarrow I$, i.e. there exists $\tilde f \colon M \rightarrow I$ such that $\tilde f \circ \pi = f$.
 * If $I$ is a submodule of an $A$-module $M$, then $I$ is a direct summand of $M$.
 * The functor $Hom(\bullet, I)$ is exact.

For principal ideal domains, there is the following result: While free modules provide good examples of projective modules, injective modules are perhaps a bit less intuitive. Here is an example: Note that free modules are usually not injective! For example, $\ZZ$ is not an injective $\ZZ$-module (consider the embedding $\ZZ \rightarrow \QQ$).
 * If $A$ is a PID, and $M$ is a $A$-module, then $M$ is injective if and only if $M$ is divisible, i.e. $M = aM$ for all nonzero $a \in A$.
 * In general, if $M$ is an injective module over an integral domain, then $M$ is divisible, i.e., $M = aM$ for all nonzero $a \in A$.
 * $\QQ / \ZZ$ is an injective $\ZZ$-module.

Here is an illustrative exercise:
 * If $A$ is a local ring, and $A$ is not a field, and $M$ is a finitely-generated module, then $M$ is injective if and only if $M = 0$.

Injectivity is a local property, just like projectivity, at least for finitely-generated modules. In other words, given a finitely-generated module $M$ over a commutative ring $A$, $M$ is injective if and only if $M_p$ is injective as an $A_p$-module, for all primes $p$ of $A$. Using this property, we can construct an embedding of any $A$-module into an injective module (the Godement-construction).

=Artin-Wedderburn=

The Artin-Wedderburn theorem is the following (a slight adaptation of the statement from Dummit-Foote):

Theorem: The following conditions on an associative unital ring $A$ are equivalent:
 * Every $A$-module is projective.
 * Every $A$-module in injective.
 * Every $A$-module is completely reducible (the direct sum of irreducible modules).
 * The ring $A$ is a direct product of matrix rings over division rings.
 * The ring $A$ is Morita equivalent to a product of a finite number of fields.

Let us consider the first two conditions first.