You are currently browsing the category archive for the ‘ring theory’ category.

Let’s fix a ring $R$. A module (assumed to be a left module, I suppose, but it doesn’t really matter as long as we’re consistent) $M$ is said to be projective if the functor $Hom(M,-)$ is exact. (That is, if $0\to A\to B\to C\to 0$ is a short exact sequence of $R$-modules, then $0\to Hom(M,A)\to Hom(M,B)\to Hom(M,C)\to 0$ is also exact.) Dually, $M$ is said to be injective if the functor $Hom(-,M)$ is exact.

There are various equivalent conditions for projectives and injectives. One particularly useful result is that projective modules are exactly the direct summands of free modules. Another one is that injective modules satisfy a certain extension property: A module $J$ is injective if and only if for any map $\phi:A\to J$ and any injective map $f:A\to B$, there exists a map (not necessarily unique) $\theta:B\to J$ so that $\phi=\theta\circ f$.

Actually, we didn’t need to start with a ring $R$ at all; it would make just as much sense to allow $M$ to be an object in an arbitrary abelian category $\mathcal{C}$. We say that $\mathcal{C}$ has enough projectives if for every object $A$ of $\mathcal{C}$, there is an epimorphism (or, a surjective map, in the case of many interesting categories) $P\to A$, where $P$ is projective. Dually, $\mathcal{C}$ has enough injectives if for every object $A$ of $\mathcal{C}$, there is a monomorphism (or, an injective map, in the case of many interesting categories) $A\to J$, where $J$ is injective.

It is easy to see that the category of modules over a ring $R$ has enough projectives: if $A$ is an $R$-module, just take the free module on all the elements of $A$, and then quotient out by the submodule consisting of all relations in $A$. Hence $A$ is isomorphic to a quotient of a free (and hence projective) module.

It is also true that the category of modules over a ring $R$ has enough injectives, but this is a bit trickier to prove. To begin, we note that an arbitrary product of injective objects is injective. This follows from the isomorphism $Hom(A,\prod_{i\in I} B_i)\cong\prod_{i\in I} Hom(A,B_i)$. We also note (although I’m not going to prove it here) that in the category of abelian groups (or $\mathbb{Z}$-modules), injective modules are the same as divisible modules (i.e. modules $M$ so that the maps $m\mapsto nm$ for $n\neq 0$ are all surjective).

Let’s first show that the category of abelian groups has enough injectives. The abelian group that plays the most important role here is $\mathbb{Q}/\mathbb{Z}$. Let $A$ be any abelian group, and let $I(A)$ be the product of copies of $\mathbb{Q}/\mathbb{Z}$, indexed by the set $Hom(A,\mathbb{Q}/\mathbb{Z})$. Then $I(A)$ is injective, and there is a canonical map $e_A:A\to I(A)$. We now check that $e_A$ is actually an injective map. To do this, pick $0\neq a\in A$. It is easy to find some nontrivial map $a\mathbb{Z}\to\mathbb{Q}/\mathbb{Z}$. By the extension property for injective modules, this map extends to a map on all of $A$. This is enough to show that $e_A$ is an injective map.

Now let’s return to the category of modules over an arbitrary ring $R$. It can be shown that if $J$ is an injective abelian group, then $Hom(R,J)$ has the structure of an injective $R$-module. Now, let $M$ be an arbitrary $R$-module. Then let $I(M)$ be the product of copies of $I_0=Hom(R,\mathbb{Q}/\mathbb{Z})$, indexed by the set $Hom_R(M,I_0)$. Then, just as before, there is a canonical injective map $M\to I(M)$. This completes the proof that the category of $R$-modules has enough injectives.

Recall from the last post that if $R$ is a commutative ring, we define $K_0(R)$ to be the Grothendieck group of the isomorphism classes of finitely generated projective $R$-modules. It is natural to ask what happens if we replace finitely generated projective modules with countably generated projective modules. Let us write $\mathfrak{K}_0(R)$ for this group. It turns out that understanding $\mathfrak{K}_0(R)$ is extremely easy.

Theorem: For any commutative ring $R$, $\mathfrak{K}_0(R)=0$.

Proof: We have to show that if $A$, $B$, $C$, and $D$ are countably generated projective $R$-modules, there is some countably generated projective $R$-module $E$ so that $A\oplus D\oplus E\cong B\oplus C\oplus E$. Define $E=\bigoplus_{i=1}^\infty (A\oplus B\oplus C\oplus D)$. Hence $\mathfrak{K}_0(R)=0$.

A similar construction shows up in the theory of group rings. Here’s an exercise from T.Y. Lam’s Exercises in Classical Ring Theory:

Exercise 8.16: Let $G$ and $H$ be any two groups. Show that there is some ring $R$ so that $R[G]\cong R[H]$. (Here $R[G]$ is the ring of finite $R$-linear combinations of elements of $G$, and multiplication is defined by the group multiplication of $G$.)

Solution: Let $K=(G\times H)\times(G\times H)\times\cdots$, and set $R=\mathbb{Z}[K]$. Then $R[G]\cong R[H]$.

Lam makes the comment that, although consideration of the group rings $\mathbb{Z}[G]$ and $K[G]$ are very useful for determining properties of $G$ (for instance, the modules over these rings are the objects of study in group cohomology and representation theory, respectively), the group ring $R[G]$ for an arbitrary ring $R$ might not give us much information about $G$.

There’s an interesting article I found on more general Eilenberg swindles, but the authors don’t define progenerators, so I’ll include that here.

Let $R$ be a ring and $M$ a right $R$-module. Define $M^\ast=Hom_R(M,R)$ and $S=Hom_R(M,M)$. Then $M$ and $M^\ast$ are $S-R$ and $R-S$ bimodules, respectively. Furthermore, we can define multiplications $M^\ast M\subseteq R$ by $m^\ast m=m^\ast(m)$ and $MM\ast\subseteq S$ by $mm^\ast(m')=m(m^\ast m')$. We say that $M$ is a progenerator if $MM^\ast=S$ and $M^\ast M=R$.

In many places in mathematics, we see some variant of the following simple construction. The first time we see it is in constructing the integers from the natural numbers:

Consider pairs $(a,b)$ of elements of $\mathbb{N}$ (which includes zero for our purposes, but it doesn’t really matter this time). We form equivalence classes out of these pairs by saying that $(a,b)\sim(c,d)$ if $a+d=b+c$. We can create a group structure on these pairs by setting $(a,b)+(c,d)=(a+c,b+d)$. The resulting group is isomorphic to $\mathbb{Z}$. So that’s how to construct the integers from the natural numbers.

We see a similar construction when we discuss localizations of rings. Let $R$ be a commutative ring and $S$ a multiplicative subset containing 1. (If $R$ is noncommutative, you can still localize provided that $S$ is an Ore set, but I don’t feel like going there now.) We now consider pairs $(r,s)\in R\times S$ under the equivalence relation $(r_1,s_1)\sim(r_2,s_2)$ if there is some $t\in S$ so that $tr_1s_2=tr_2s_1$. The set of equivalence classes has the structure of a ring, called the localization of $R$ at $S$, and denoted by $R_S$. This construction is generally seen with $S=R\setminus\mathfrak{p}$, where $\mathfrak{p}$ is a prime ideal of $R$. The resulting ring is then local (meaning that it has a unique maximal ideal, namely $\mathfrak{p}R_S$. (We generally write $R_{\mathfrak{p}}$ rather than $R_S$ in this situation.) Anyway, this construction is really useful because localizations at prime ideals are frequently principal ideal domains, and we know all sorts of interesting theorems about finitely generated modules over principal ideal domains. And then we can use some Hasse principle-type result to transfer our results back to the original ring.

Notice that I allowed the multiplicative set of localization to contain zero. However, in this case, the localization becomes the trivial ring (or not a ring, if you require that $0\neq 1$ in your definition of a ring, as many people do). More generally, allowing zero divisors in the multiplicative set causes various elements in $R$ to become zero in the localization.

A similar construction shows up in $K$-theory. Suppose $A$ is any commutative semigroup. We consider pairs $(a,b)\in A\times A$ under the equivalence relation $(a,b)\sim(c,d)$ if there is some $e\in A$ so that $a+d+e=b+c+e$. (This is necessary since we do not assume that $A$ satisfies the cancellation property.) The resulting equivalence classes form a group called the Grothendieck group of $A$ and denoted by $K_0(A)$.

The Grothendieck group satisfies the following universal property. Let $\phi$ be the map sending $a\in A$ to $(a+b,b)$ for any $b\in A$. (This is easily seen to be well-defined.) Now let $G$ be any abelian group and $\psi:A\to G$ any semigroup homomorphism. Then there is a (unique) map $\theta:K_0(A)\to G$ so that $\theta\phi=\psi$.

Grothendieck groups can be very helpful for studying rings. Let $R$ be a commutative ring, and let $A$ denote the semigroup of isomorphism classes of projective $R$ modules (under the operation of direct sum). Then $K_0(A)$ (or $K_0(R)$, as people often write) is an important object of study. If $R$ is a field, for instance, then $K_0(R)\cong\mathbb{Z}$. However, if $R$ is the ring of integers of a number field $K$, then $K_0(R)\cong\mathbb{Z}\oplus C(K)$, where $C(K)$ is the ideal class group.

Perhaps more interesting is Swan’s Theorem, which relates vector bundles over a compact topological space to the projective modules over its ring of continuous functions: they have isomorphic Grothendieck groups. But that’s probably the subject of another post, especially if I can manage to understand my notes from Max Karoubi’s lecture series in Edmonton.