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## A first blog post on noncommutative rings

In this post, I’d like to record a few basic definitions and results regarding noncommutative rings. This is a subject clearly of great importance and generality, but I haven’t had much exposure to it, and I’m trying to fix that. I am working mostly from Lam’s A first course in noncommutative rings.

Some preliminary categorical remarks

Let $R$ be a ring. You probably know that a ring is a set with two operations, addition and multiplication, satisfying some compatibility relations. I prefer the following definition: a ring is a monoid object in the monoidal category $\text{Ab}$ of abelian groups under tensor product. This definition makes it clear that rings are “linearized” monoids and emphasizes that rings naturally appear as the endomorphism rings of abelian groups, just as monoids naturally appear as the endomorphism monoids of sets.

Another way to state this definition, perhaps even more categorical, is that a ring is a small $\text{Ab}$enriched category with one object, just as a monoid is a small category with one object. Among other things, this definition encourages us to think of $\text{Ab}$-enriched categories (such as abelian categories) as “rings with many objects” or ringoids, named in analogy to groupoids. It also shows that the construction of the opposite ring $R^{op}$ is a special case of the construction of the opposite category.

Definitions and examples

A left $R$module $M$ is an abelian group equipped with a ring homomorphism $R \to \text{End}(M)$. Equivalently, it is equipped with a map $R \times M \to M$ satisfying certain associativity properties, hence the term “left”; right $R$-modules are defined similarly except with a map $M \times R \to M$ (or equivalently a homomorphism $R^{op} \to \text{End}(M)$). Categorically, thinking of $R$ as a category as above, a left $R$-module is an enriched functor $R \to \text{Ab}$ (and a right $R$-module is an enriched contravariant functor). Without qualification, “module” and “$R$-module” mean “left $R$-module.”

A morphism of modules is an abelian group homomorphism $M \to N$ respecting the action of $R$ (equivalently, it is an enriched natural transformation – okay, I’ll stop now). A submodule of $M$ is a subgroup preserved by the action of $R$. Modules admit obvious notions of direct sum, kernel of a morphism, and quotient by a submodule. A module $M$ is

• simple if it is nonzero and its only submodules are $0$ and $M$ (equivalently, if its only quotient modules are $0$ and $M$),
• indecomposable if it cannot be written as a direct sum of two nonzero submodules, and
• cyclic if it is generated as a module by a single element (equivalently, if it is a quotient of $R$ as a left $R$-module).

Proposition: Any simple module is cyclic.

Proof. Let $M$ be simple and $m \in M$ a nonzero element of it. Then the map $R \ni r \mapsto rm \in M$ has image a nonzero submodule which must therefore be all of $M$.

Example. Let $R = k$ be a field. A $k$-module is just a $k$-vector space. The simple $k$-modules are precisely the one-dimensional vector spaces.

Example. Let $R = k[x]$ be a polynomial ring. A $R$-module is a $k$-vector space together with a linear operator $x$ acting on it. The simple $k$-modules are precisely the one-dimensional vector spaces where $x$ acts by a scalar and the finitely-generated indecomposable $k$-modules are precisely the vector spaces where $x$ acts by a Jordan block by the theory of Jordan normal form. This example shows that an indecomposable module need not be simple.

Example. Let $R = k \langle x, y \rangle$ be the free $k$-algebra on two generators, or the ring of noncommutative polynomials in two variables. An $R$-module is a $k$-vector space together with two linear operators acting on it, and the classification of $R$-modules is such a difficult problem that it gave rise to the notion of a wild classification problem; see also this MO question.

Example. Let $C$ be a small category. Any functor $C \to \text{Ab}$ gives rise to a left $R$-module for $R$ the category ring $\mathbb{Z}[C]$ of $C$ (and this is an equivalence of categories if $C$ has finitely many objects). This is the algebra generated by the morphisms of $C$, where the composition of two morphisms is their composition in $C$ if defined and zero otherwise.

This construction is, of course, of considerable interest when $C$ is a group regarded as a one-object category; it (or $k[C]$ for $k$ a field) gives the group algebra. It is also of considerable interest when $C$ is a free category on a graph $G$, in which case $\mathbb{Z}[C]$ (or $k[C]$ for $k$ a field) is called the path algebra or quiver algebra of the quiver $G$. When $C$ is a poset, we recover a version of the incidence algebra.

Simple modules

The example of $R = k[x]$ above shows that not every module is a direct sum of simple modules. Nevertheless, it would be nice to understand general modules in terms of simple modules since their behavior is, well, relatively simple. For example, the following is true.

Schur’s lemma: Let $M, N$ be two simple $R$-modules. Then $\text{Hom}_R(M, N)$ is empty if $M, N$ are non-isomorphic; otherwise, $\text{Hom}_R(M, M)$ is a division ring $D_M$.

Proof. Let $\phi : M \to N$ be a morphism. Then the kernel and image of $\phi$ are submodules of $M, N$ respectively, which must by hypothesis be empty or the entire module. So either $\phi = 0$ or $\phi$ is an isomorphism.

Example. Let $R = \mathbb{R}[G]$ be the real group algebra of a finite group $G$. Then an $R$-module is a real representation of $G$. The division rings that can arise as endomorphism rings of simple $R$-modules are necessarily finite-dimensional division rings over $\mathbb{R}$, hence by the Frobenius theorem can only be $\mathbb{R}, \mathbb{C}$, or the quaternions $\mathbb{H}$, and all three examples occur. See Frobenius-Schur indicator.

What can we say about understanding a general module in terms of simple modules? One naive idea is to take an arbitrary module $M$, find a simple submodule of it, consider the quotient by that submodule, find a simple submodule of that, and so forth. Unfortunately, it is false that every module contains a simple submodule! For example, consider $\mathbb{Z}$ acting on itself by left multiplication. The simple $\mathbb{Z}$-modules are the finite cyclic groups of prime order, which don’t appear as submodules of $\mathbb{Z}$.

Rather than submodules, we need to think about quotients. By Zorn’s lemma, any module $M_0$ has a maximal proper submodule $M_1$, and by maximality the quotient $M_0/M_1$ is necessarily simple. We can iterate this construction with the submodule $M_1$, but it is not guaranteed to terminate (indeed in the example of $\mathbb{Z}$ above it cannot terminate). If it does – that is, if we can find a composition series

$0 = M_n \subsetneq M_{n-1} \subsetneq ... \subsetneq M_1 \subsetneq M_0$

of submodules such that the composition factors $M_i/M_{i+1}$ are simple, we say that $M_0$ has finite length. Such a module is, in an appropriate sense, built up from the simple modules $M_i/M_{i+1}$, but if we are to take this idea seriously it would be nice if this list of modules did not depend on the choice of composition series. This is easy to see concretely for $k[x]$: here the finite length modules are precisely the finite-dimensional ones, and the list of simple modules $M_i/M_{i+1}$ can be identified with the list of eigenvalues of $x$, which only depends on the module. So it is not totally unreasonable to hope that this is true generally.

Theorem (Jordan–Hölder): Let $M$ be a finite-length module. Any two composition series have the same length $n$, the length of $M$, and the composition factors appearing in them are the same up to permutation.

Proof. Let $M_n \subsetneq ... \subsetneq M_0$ and $N_k \subsetneq ... \subsetneq N_0 = M_0$ be two composition series. We induct on $\text{max}(n, k)$. The statement is obvious if $\text{max}(n, k) = 0$. In general, if $M_1 = N_1$ then the statement follows by the inductive hypothesis. Otherwise, by maximality $M_1 + N_1 = M_0$, hence

$\displaystyle M_0/M_1 \cong N_1/(N_1 \cap M_1), M_0/N_1 \cong M_1/(N_1 \cap M_1)$

by the isomorphism theorems. WLOG $n \le k$; then intersecting $M_n \subsetneq ... \subsetneq M_1$ by $N_1$ and discarding terms, we can find a composition series for $N_1 \cap M_1$ of length at most $n-1$, and attaching such a composition series to the beginnings of the series

$\displaystyle ... \supsetneq M_1 \cap N_1 \supsetneq M_1 \supsetneq M_0, ... \supsetneq M_1 \cap N_1 \supsetneq N_1 \supsetneq M_0$

we find two series which clearly have the same length and the same composition series. On the other hand, by the inductive hypothesis the first series is equivalent to $M_n \supsetneq ... \supsetneq M_0$ and the second series is equivalent to $N_k \supsetneq ... \supsetneq N_0 = M_0$. The conclusion follows.

Semisimple modules

A module $M$ is semisimple if any submodule of $M$ is a direct summand. This turns out to be a particularly nice condition to work with on modules. For example, it is true of all simple modules, but it is also closed under direct sums, quotients, and taking submodules. In particular, every direct sum of simple modules is semisimple. What can we say about other semisimple modules?

Proposition: Any semisimple module $M$ contains a simple submodule.

Proof. Let $m \in M$ be nonzero. It suffices to reduce to the case that $M = Rm$. By Zorn’s lemma, there is a submodule $N$ of $M$ maximal with respect to the property of not containing $m$. By assumption, there is a direct sum decomposition

$\displaystyle M = N \oplus N'$.

If $N'$ contains a nonzero submodule $N''$, then by maximality $N \oplus N''$ contains $m$, hence $N \oplus N'' = M$, so $N'' = N'$. Hence $N'$ is a simple submodule of $M$.

Proposition: The following conditions on a left $R$-module are equivalent:

1. $M$ is semisimple.
2. $M$ is a direct sum of simple modules.
3. $M$ is generated by its simple submodules.

Proof. $1 \Rightarrow 3$: Let $N$ be the sum of the simple submodules of $M$. By assumption there is a direct sum decomposition $M = N \oplus N'$. Since $N'$ is semisimple, it is either zero or has a simple submodule, but by assumption the latter is not possible, so $N' = 0$.

$3 \Rightarrow 1$: Let $M = \sum M_i$ be generated by its simple submodules $M_i$. Let $N$ be a submodule of $M$. By Zorn’s lemma, there is a maximal collection of simple submodules $M_i, i \in J$ such that $\sum M_i, i \in J$ is an internal direct sum and such that $N \cap \sum M_i = 0$. Let

$M' = N + \sum M_i = N \oplus \bigoplus M_i$.

If $M' \neq M$, then there is some $M_i$ not contained in $M'$, hence $M_i \cap M' = 0$, which contradicts the maximality of $J$. So $M' = M$ as desired.

$3 \Rightarrow 2$: apply the above argument to $N = 0$.

$2 \Rightarrow 3$: obvious.

Semisimple rings

Ideally, we’d like to be able to study all the modules of a ring by studying its semisimple modules. This desire is encapsulated by the following definition.

Theorem-Definition: A (left) semisimple ring $R$ is a ring satisfying any of the following conditions, all of which are equivalent:

1. All left $R$-modules are semisimple.
2. All finitely-generated left $R$-modules are semisimple.
3. All cyclic left $R$-modules are semisimple.
4. $R$ is semisimple as a left $R$-module.

Proof. There are obvious implications $1 \Rightarrow 2 \Rightarrow 3 \Rightarrow 4$. Since we know that semisimplicity is preserved under taking quotients, we obtain $4 \Rightarrow 3$. Since any module $M$ is generated by the modules $Rm_i$ where $m_i$ is a set of generators, we obtain $3 \Rightarrow 1$.

By Maschke’s theorem, the group algebra $k[G]$ is semisimple for $G$ a finite group and $k$ a field of characteristic not dividing $|G|$, so it is clearly of interest to understand the structure of semisimple rings.

Example. Any division ring $D$ is simple, hence semisimple, as a module over itself, so any division ring is semisimple.

Example. Let $R, S$ be two semisimple rings. In the product ring $R \times S$, consider the two idempotents $e_1 = (1, 0), e_2 = (0, 1)$. Any $R \times S$-module $M$ breaks up into a direct sum $M = e_1 M \oplus e_2 M$ where the first factor is the part on which $R$ acts nontrivially and the second is the part on which $S$ acts nontrivially. In particular, $R \times S$ is semisimple as a module over itself, hence is semisimple, and moreover its simple modules are precisely the simple modules of $R$ together with the simple modules of $S$.

Example. Let $\mathcal{M}_n(D)$ denote the matrix ring of $n \times n$ matrices over a division ring $D$. As a vector space it is spanned by elements $e_{i,j}$ which satisfy the relations

$\displaystyle e_{i, j} e_{k, l} = \begin{cases} 0 \text{ if } j \neq k \\ e_{i, l} \text{ otherwise}. \end{cases}$

By inspection, as a left module over itself, $\mathcal{M}_n(D)$ breaks up into a direct sum of simple submodules $M_j = \text{span}_D(e_{i, j}, 1 \le i \le n)$. It follows that $\mathcal{M}_n(D)$ is semisimple. Moreover, since every simple module is a quotient of $\mathcal{M}_n(D)$ (as a left module over itself), it follows that all simple $\mathcal{M}_n(D)$-modules are isomorphic to some $M_j$, each of which is in turn isomorphic to $D^n$, which admits a natural left action of $\mathcal{M}_n(D)$ when thought of as a right $D$-module.

Artin-Wedderburn

The above discussion shows that any finite product of matrix rings over division rings is semisimple. In fact, this exhausts all examples. To see this, we first need the following.

Cayley’s theorem for rings: $\text{End}_R(R) \cong R^{op}$.

Proof. $R$ acts by right multiplication on $R$, and each such right multiplication is a left $R$-module homomorphism, so $\text{End}_R(R)$ certainly contains $R^{op}$. On the other hand, if $\phi : R \to R$ is a left $R$-module homomorphism, then $\phi(r) = r \phi(1)$ is already given by right multiplication.

Theorem (Artin-Wedderburn): Every semisimple ring is isomorphic to a finite product of matrix rings over division rings. Moreover, the terms in this product are uniquely determined up to permutation.

Proof. Let $R$ be semisimple. Then $R$, as a left $R$-module, admits a direct sum decomposition into simple submodules. The multiplicative unit of $R$ generates it as a module, hence generates any quotient of it, so $1$ has a nonzero image in any simple quotient of $R$; it follows that the direct sum decomposition of $R$ into simple submodules is finite, so we can write

$R \cong n_1 M_1 \oplus ... \oplus n_r M_r$

for some positive integers $n_i$ and some simple modules $M_i$. Then

$R^{op} \cong \text{End}_R(R) \cong \text{End}_R(\bigoplus n_i M_i) \cong \prod \mathcal{M}_{n_i}(D_i)$

by Schur’s lemma and the universal property of direct sums. The opposite of a matrix ring over a division ring is a matrix ring over the opposite division ring, so we conclude that $R$ is isomorphic to a finite product of matrix rings over division rings. Moreover, by the above remarks, if

$R \cong \prod \mathcal{M}_{n_i}(D_i)$

is a finite product of matrix rings over division rings, then the simple modules over $R$ are determined by the simple modules over each factor. The only simple module over $\mathcal{M}_{n_i}(D_i)$, as we have seen, is $D_i^{n_i}$, and each simple module is cyclic hence appears as a direct summand of $R$, so the product decomposition above is unique up to permutation by Jordan-Hölder.

As a corollary, a ring is left semisimple if and only if it is right semisimple, so we can drop the adjective.