The Artin-Wedderburn theorem shows that the definition of a semisimple ring is enormously restrictive. Even fails to be semisimple! A less restrictive notion, but one that still captures the notion of a ring which can be understood by how it acts on simple (left) modules, is that of a semiprimitive or Jacobson semisimple ring, one with the property that every element acts nontrivially in some simple (left) module .
Said another way, let the Jacobson radical of a ring consist of all elements of which act trivially on every simple module. By definition, this is an intersection of kernels of ring homomorphisms, hence a two-sided ideal. A ring is then semiprimitive if it has trivial Jacobson radical.
The goal of this post will be to discuss some basic properties of the Jacobson radical. I am again working mostly from Lam’s A first course in noncommutative rings.
While commutative rings only have a single notion of ideal, noncommutative rings have three. A noncommutative ring may be considered a left -module via left multiplication, a right -module via right multiplication, or an -bimodule using both. Consequently, we get three notions of ideals: a left ideal is a left submodule, a right ideal is a right submodule, and a two-sided ideal is a sub-bimodule. Quotienting by left ideals gives all cyclic left modules, quotienting by right ideals gives all cyclic right modules, and quotienting by ideals (which, without qualification, refer to two-sided ideals) gives all quotient rings.
Accordingly, a maximal left ideal is a proper left ideal which is maximal among all left ideals, and similarly we have maximal right ideals and maximal (two-sided) ideals. Quotienting by maximal left ideals gives all simple left modules, quotienting by maximal right ideals gives all simple right modules, and quotienting by maximal (two-sided) ideals gives all simple quotient rings.
When is commutative, one can think about maximal ideals geometrically as points of a space on which behaves as a space of functions. However, when is noncommutative this idea breaks down badly in general.
Example. Let be the Weyl algebra in one variable over . This algebra is isomorphic to the algebra of polynomial differential operators acting on when has characteristic zero (in characteristic we have the additional relation ), which we will assume for now.
turns out to be a simple ring; that is, it has no nontrivial two-sided ideals. To see this, first observe that the relation allows us to move all ‘s to the left of all ‘s in any element of , so we may assume WLOG that an element has the form
We want to show that the two-sided ideal generated by is all of if is nonzero. We begin by observing that it is closed under taking commutators with any element of . Now, the derivation acts trivially on for all and satisfies , so by the Leibniz rule and , hence
By repeatedly applying we obtain a nonzero polynomial (this is where we need to assume that has characteristic zero). We similarly note that the derivation acts trivially on for all and satisfies , so , hence
By repeatedly applying we obtain a nonzero constant, from which the conclusion follows (again we need to assume that has characteristic zero).
So the Weyl algebra cannot be understood as a geometric object related to its maximal ideals in a naive way, since its space of maximal ideals consists of a single point. To understand the Weyl algebra geometrically we should instead think of it as the algebra of “observables” on a “quantum particle” on the affine line over (where is position and is momentum); this is a kind of noncommutative geometry.
The Jacobson radical
Proposition: The following are equivalent for an element :
- ; that is, acts trivially in every simple (left) module.
- lies in the intersection of all maximal left ideals of .
- is left-invertible for all .
Proof. : Let be a simple (left) module. For any nonzero element , the left module homomorphism
is surjective by simplicity, so its kernel is a maximal left ideal, and all maximal left ideals occur in this way (if is such an ideal, take ). Hence iff lies in the intersection of all maximal left ideals.
: Suppose misses some maximal left ideal . Then the left ideal generated by has the property that its sum with must be the entire ring, hence there exists such that contains , hence is contained in a maximal left ideal and cannot be left-invertible. Conversely, suppose there exists such that is not left-invertible. Then it is properly contained in some maximal left ideal which cannot contain , hence cannot contain .
At this point I should still distinguish between the left and right Jacobson radicals, where the right Jacobson radical consists of elements which act trivially in all simple right modules (equivalently, which are in the intersection of all maximal right ideals or are such that is right-invertible for all ). However, the next proposition shows that this is unnecessary.
Proposition: if and only if is a unit (has a two-sided inverse) for all .
Proof. : set .
: If , then for all , hence is left-invertible, say with left inverse . This gives or . Now, , hence is also left-invertible. Since has both a left and a right inverse, they must agree, and so (and its inverse) are invertible.
The condition that is a unit is left-right symmetric, so we conclude that . One way to restate the above equivalent definition is that is the largest ideal of such that (the unit group of ).
Proposition: Let be an ideal contained in . Then .
Proof. By assumption, is contained in every maximal left ideal of , so the pullback map from maximal left ideals of to maximal left ideals of is a bijection (hence and have the same simple modules). The conclusion follows.
Corollary: is semiprimitive.
The corollary suggests a possible route to understanding a general ring : studying the semiprimitive ring first. Among other things, it may be easier to ascertain the simple modules of , which are naturally identified with the simple modules of .
For commutative rings, we do not need to distinguish between left and right ideals, and so is the intersection of all maximal ideals. In other words, is the kernel of the homomorphism
sending an element of to the “function” it defines on . Thus a commutative ring is semiprimitive if and only if it can be faithfully represented as a ring of functions on its maximal spectrum. By the Nullstellensatz, this includes for example finitely-generated algebras over a field with no nilpotents, as well as any Dedekind domain.
More generally, recall that a Jacobson ring is a commutative ring in which all prime ideals are intersections of maximal ideals. A general version of the Nullstellensatz asserts that a finitely-generated algebra over a Jacobson ring is Jacobson. In such a ring, the intersection of the maximal ideals (the Jacobson radical) is equal to the intersection of the prime ideals (the nilradical), hence a Jacobson ring is semiprimitive if and only if it is reduced (has trivial nilradical).
In general, however, the Jacobson radical need not contain any nontrivial nilpotents.
Example. Let be the ring of formal power series over a field . The maximal ideal has the property that its complement consists entirely of units, so is the unique maximal ideal, hence . More generally, the Jacobson radical of any local ring is its unique maximal left (or right) ideal.
Some more properties
Before we discuss more examples it will be helpful to discuss some additional generalities.
If is a commutative ring then it is clear that for any nilpotent we have that is also nilpotent, hence is invertible. The collection of nilpotent elements forms an ideal, the nilradical of , which consequently must lie in .
If is not commutative, then the nilpotent elements no longer form an ideal (left or right, let alone two-sided) in general, so nilpotence must be treated more carefully. We say that an ideal (left, right, or two-sided) is nil if it consists only of nilpotent elements and nilpotent if for some . (This is a strictly stronger condition: consider the ideal in the ring .)
Proposition: Let be a nil left (resp. right) ideal of . Then .
Proof. Let . Then (resp. ) is nilpotent for all , hence (resp. ) is left-invertible (resp. right-invertible).
Recall that a ring is left (resp. right) Artinian if any descending chain of left (resp. right) ideals is eventually constant. The Artinian condition generalizes the condition of being finite or finite-dimensional over a field.
Proposition: Let be left Artinian (or right Artinian). Then is a nilpotent ideal.
(This shows that in order to compute the Jacobson radical of a left Artinian ring, it suffices to find the largest nilpotent left or right ideal.)
Proof. The descending chain stabilizes by assumption, so there exists some such that . We claim that , which establishes the desired result.
Suppose otherwise. Among all the ideals such that , there is a minimal one by the descending chain condition. Pick such that . Then , hence by minimality . It follows that for some . But is a unit, so implies ; contradiction.
(If were either finite or finite-dimensional over a field then would be finitely-generated and we could appeal to an appropriate form of Nakayama’s lemma to finish this argument instead.)
Corollary: If is left Artinian, then any nil left or right ideal is nilpotent.
Corollary: If is commutative and Artinian, then is precisely the nilradical of .
It is clear that any semisimple ring is semiprimitive. The next result determines under what conditions the converse holds.
Proposition: is semisimple if and only if it is semiprimitive and left Artinian.
Proof. : By semisimplicity we have as left -modules for some left ideal . If is proper, it is contained in some maximal left ideal which does not contain ; contradiction. Hence and .
: Let be a well-ordering of the maximal left ideals in . By assumption, the descending chain stabilizes, hence the intersection of all maximal left ideals is an intersection of finitely many maximal left ideals by semiprimitivity. It follows that embeds into as a left -module. This is a direct sum of simple modules, hence semisimple, so must also be semisimple as a left -module.
Since quotients of left Artinian rings are left Artinian, we conclude the following.
Corollary: Let be left Artinian. Then is semisimple.
Thus we can use Artin-Wedderburn to gain insight into the structure of left Artinian rings; they are, roughly speaking, finite direct products of matrix rings over division rings plus some extra nilpotents. In particular, if is commutative, then is a finite direct product of fields.
Let’s turn our attention to some noncommutative examples.
Let denote the algebra of upper-triangular matrices over a field . An element of is nilpotent if and only if it is strictly upper-triangular; denote these matrices by . There is a natural epimorphism given by sending an upper-triangular matrix to its diagonal entries whose kernel is precisely , from which it follows that is an ideal. Since it is the largest nilpotent left ideal, it must be the Jacobson radical. We conclude that has simple modules, one for each diagonal entry.
This example generalizes as follows. A quiver is a directed graph in which loops and multiple edges are allowed. Given let denote the free category on ; this is the category whose objects are the vertices of and whose morphisms are paths in , with composition given by concatenation of paths. ( is left adjoint to the forgetful functor from categories to quivers which forgets the composition.) A quiver representation of is a functor for some field ; explicitly, it is a collection of vector spaces for each vertex in together with a collection of linear maps for each edge . (In the literature it seems that the are taken to be finite-dimensional, but I won’t assume this.)
The category of quiver representations is equivalent to the category of left modules over a certain algebra , the quiver algebra of , which is the -vector space spanned by the morphisms in (including the identity morphisms associated to each vertex) whose product is their composition when defined and zero otherwise.
Example. Let consist of a single vertex together with loops. Then is the free -algebra on generators.
as defined here is unital iff has finitely many vertices (in this case the identity is the sum of the elements associated to each vertex) and finite-dimensional iff has finitely many vertices and edges and no directed cycles (including loops); we summarize these conditions by saying that is finite acyclic.
Example. Let be the quiver, which consists of vertices connected by directed edges as indicated. naturally acts by left multiplication on the vector space spanned by the vertices (where, for a path , we have if is not the source of and if is a path from to ) and this action realizes an isomorphism .
For any quiver , the subspace spanned by paths of length at least is a two-sided ideal by inspection. The quotient is isomorphic to the direct sum of a copy of for every vertex of . Moreover, if is finite acyclic, then consists of nilpotent elements, so must be the largest nilpotent left ideal of ; hence . Consequently, in this case has one simple module for each vertex (corresponding to the quiver representation where the vector space associated to that vertex is one-dimensional and all others are zero-dimensional).