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The Artin-Wedderburn theorem shows that the definition of a semisimple ring is enormously restrictive. Even $\mathbb{Z}$ 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 $r \in R$ acts nontrivially in some simple (left) module $M$.

Said another way, let the Jacobson radical $J(R)$ of a ring consist of all elements of $r$ 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 $R$ 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.

Ideals

While commutative rings only have a single notion of ideal, noncommutative rings have three. A noncommutative ring $R$ may be considered a left $R$-module via left multiplication, a right $R$-module via right multiplication, or an $(R, R)$-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 $R$ is commutative, one can think about maximal ideals geometrically as points of a space on which $R$ behaves as a space of functions. However, when $R$ is noncommutative this idea breaks down badly in general.

Example. Let $R = k \langle x, y \rangle / (xy - yx - 1)$ be the Weyl algebra in one variable over $k$. This algebra is isomorphic to the algebra $k[x, \frac{\partial}{\partial x}]$ of polynomial differential operators acting on $k[x]$ when $k$ has characteristic zero (in characteristic $p$ we have the additional relation $\frac{\partial}{\partial x}^p = 0$), which we will assume for now.

$R$ turns out to be a simple ring; that is, it has no nontrivial two-sided ideals. To see this, first observe that the relation $xy = yx + 1$ allows us to move all $y$‘s to the left of all $x$‘s in any element of $R$, so we may assume WLOG that an element $f \in R$ has the form

$\displaystyle f = \sum c_{ij} x^i y^j$.

We want to show that the two-sided ideal $(f)$ generated by $f$ is all of $R$ if $f$ is nonzero. We begin by observing that it is closed under taking commutators with any element of $R$. Now, the derivation $[-, y]$ acts trivially on $y^j$ for all $j$ and satisfies $[x, y] = 1$, so by the Leibniz rule $[x^i, y] = ix^{i-1}$ and $[x^i y^j, y] = i x^{i-1} y^j$, hence

$\displaystyle [f, y] = \sum i c_{ij} x^{i-1} y^j \in (f)$.

By repeatedly applying $[-, y]$ we obtain a nonzero polynomial $g = \sum c_i y^i$ (this is where we need to assume that $k$ has characteristic zero). We similarly note that the derivation $[x, -]$ acts trivially on $x^i$ for all $i$ and satisfies $[x, y] = 1$, so $[x, y^j] = jy^{j-1}$, hence

$\displaystyle [x, g] = \sum ic_i y^{i-1} \in (f)$.

By repeatedly applying $[x, -]$ we obtain a nonzero constant, from which the conclusion follows (again we need to assume that $k$ 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 $k$ (where $x$ is position and $\frac{\partial}{\partial x}$ is momentum); this is a kind of noncommutative geometry.

Proposition: The following are equivalent for an element $x \in R$:

1. $x \in J(R)$; that is, $x$ acts trivially in every simple (left) module.
2. $x$ lies in the intersection of all maximal left ideals of $R$.
3. $1 - rx$ is left-invertible for all $r \in R$.

Proof. $1 \Leftrightarrow 2$: Let $M$ be a simple (left) module and $\phi : R \to M$ be the (left) module homomorphism given by $r \mapsto rm$ where $m \in M$ is a fixed nonzero element. By simplicity, this homomorphism is surjective, so its kernel is a maximal left ideal $I$. If $x \in J(R)$, then $x \in I$, so this proves $1 \Rightarrow 2$. Conversely, if $x$ lies in the intersection of all maximal left ideals $I$, then it maps to zero in all quotients $R/I$, hence acts trivially on $1 \in R/I$, hence acts trivially on every element of $R/I$.

$2 \Leftrightarrow 3$: Suppose $x$ misses some maximal left ideal $I$. Then the left ideal generated by $x$ has the property that its sum with $I$ must be the entire ring, hence there exists $r \in R$ such that $rx + I$ contains $1$, hence $1 - rx \in I$ is contained in a maximal left ideal and cannot be left-invertible. Conversely, suppose there exists $r$ such that $1 - rx$ is not left-invertible. Then it is properly contained in some maximal left ideal $I$ which cannot contain $rx$, hence cannot contain $x$.

At this point I should still distinguish between the left and right Jacobson radicals, where the right Jacobson radical $J(R^{op})$ 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 $1 - xr$ is right-invertible for all $r \in R$). However, the next proposition shows that this is unnecessary.

Proposition: $x \in J(R)$ if and only if $1 - rxs$ is a unit (has a two-sided inverse) for all $r, s \in R$.

Proof. $\Leftarrow$: set $s = 1$.

$\Rightarrow$: If $x \in J(R)$, then $xs \in J(R)$ for all $s \in R$, hence $1 - rxs$ is left-invertible, say with left inverse $u$. This gives $u(1 - rxs) = 1$ or $u = 1 + urxs$. Now, $rxs \in J(R)$, hence $1 + urxs = u$ is also left-invertible. Since $u$ has both a left and a right inverse, they must agree, and so $u$ (and its inverse) are invertible.

The condition that $1 - rxs$ is a unit is left-right symmetric, so we conclude that $J(R) = J(R^{op})$. One way to restate the above equivalent definition is that $J(R)$ is the largest ideal of $R$ such that $1 + J(R) \in U(R)$ (the unit group of $R$).

Proposition: Let $N$ be an ideal contained in $J(R)$. Then $J(R/N) \cong J(R)/N$.

Proof. By assumption, $N$ is contained in every maximal left ideal $I$ of $R$, so the pullback map from maximal left ideals of $R/N$ to maximal left ideals of $R$ is a bijection (hence $R$ and $R/N$ have the same simple modules). The conclusion follows.

Corollary: $R/J(R)$ is semiprimitive.

The corollary suggests a possible route to understanding a general ring $R$: studying the semiprimitive ring $R/J(R)$ first. Among other things, it may be easier to ascertain the simple modules of $R/J(R)$, which are naturally identified with the simple modules of $R$.

Commutative rings

For commutative rings, we do not need to distinguish between left and right ideals, and so $J(R)$ is the intersection of all maximal ideals. In other words, $J(R)$ is the kernel of the homomorphism

$\displaystyle R \to \prod_m R/m$

sending an element of $R$ to the “function” it defines on $\text{MaxSpec } R$. 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 $k[[x]]$ be the ring of formal power series over a field $k$. The maximal ideal $(x)$ has the property that its complement consists entirely of units, so $(x)$ is the unique maximal ideal, hence $(x) = J(k[[x]])$. 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 $R$ is a commutative ring then it is clear that for any nilpotent $n \in R$ we have that $rns$ is also nilpotent, hence $1 - rns$ is invertible. The collection of nilpotent elements forms an ideal, the nilradical of $R$, which consequently must lie in $J(R)$.

If $R$ 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) $I$ is nil if it consists only of nilpotent elements and nilpotent if $I^n = 0$ for some $n$. (This is a strictly stronger condition: consider the ideal $(x_1, x_2, x_3, ...)$ in the ring $k[x_1, x_2, ... ]/(x_1^2, x_2^3, x_3^4, ...)$.)

Proposition: Let $I$ be a nil left (resp. right) ideal of $R$. Then $I \subseteq J(R)$.

Proof. Let $x \in I$. Then $rx$ (resp. $xr$) is nilpotent for all $r \in R$, hence $1 - rx$ (resp. $1 - xr$) is left-invertible (resp. right-invertible).

Recall that a ring is left (resp. right) Artinian if any descending chain $I_1 \supseteq I_2 \supseteq I_3 ... \supseteq$ 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 $R$ be left Artinian (or right Artinian). Then $J$ 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 $J(R) \supseteq J(R)^2 \supseteq J(R)^3 \supseteq ...$ stabilizes by assumption, so there exists some $k$ such that $I = J(R)^k = J(R)^{k+1} = ...$. We claim that $I = 0$, which establishes the desired result.

Suppose otherwise. Among all the ideals $U$ such that $IU \neq 0$, there is a minimal one $U_0$ by the descending chain condition. Pick $a \in U_0$ such that $Ia \neq 0$. Then $I(Ia) = Ia \neq 0$, hence by minimality $U_0 = Ia$. It follows that $a = ya$ for some $y \in I \subseteq J(R)$. But $1 - y$ is a unit, so $(1 - y)a = 0$ implies $a = 0$; contradiction.

(If $R$ were either finite or finite-dimensional over a field then $I$ would be finitely-generated and we could appeal to an appropriate form of Nakayama’s lemma to finish this argument instead.)

Corollary: If $R$ is left Artinian, then any nil left or right ideal is nilpotent.

Corollary: If $R$ is commutative and Artinian, then $J(R)$ is precisely the nilradical of $R$.

It is clear that any semisimple ring is semiprimitive. The next result determines under what conditions the converse holds.

Proposition: $R$ is semisimple if and only if it is semiprimitive and left Artinian.

Proof. $\Rightarrow$: By semisimplicity we have $R = J(R) \oplus I$ as left $R$-modules for some left ideal $I$. If $I$ is proper, it is contained in some maximal left ideal which does not contain $J(R)$; contradiction. Hence $I = R$ and $J(R) = 0$.

$\Leftarrow$: Let $m_1, m_2, ...$ be a well-ordering of the maximal left ideals in $R$. By assumption, the descending chain $m_1 \supseteq m_1 \cap m_2 \supseteq ...$ stabilizes, hence the intersection of all maximal left ideals is an intersection of finitely many maximal left ideals $m_1 \cap ... \cap m_n = J(R) = 0$ by semiprimitivity. It follows that $R$ embeds into $\bigoplus R/m_i$ as a left $R$-module. This is a direct sum of simple modules, hence semisimple, so $R$ must also be semisimple as a left $R$-module.

Since quotients of left Artinian rings are left Artinian, we conclude the following.

Corollary: Let $R$ be left Artinian. Then $R/J(R)$ 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 $R$ is commutative, then $R/J(R)$ is a finite direct product of fields.

Let’s turn our attention to some noncommutative examples.

Quiver algebras

Let $U_n(k)$ denote the algebra of upper-triangular $n \times n$ matrices over a field $k$. An element of $U_n(k)$ is nilpotent if and only if it is strictly upper-triangular; denote these matrices by $N_n(k)$. There is a natural epimorphism $U_n(k) \to k^n$ given by sending an upper-triangular matrix to its diagonal entries whose kernel is precisely $N_n(k)$, from which it follows that $N_n(k)$ is an ideal. Since it is the largest nilpotent left ideal, it must be the Jacobson radical. We conclude that $U_n(k)$ has $n$ simple modules, one for each diagonal entry.

This example generalizes as follows. A quiver $Q$ is a directed graph in which loops and multiple edges are allowed. Given $Q$ let $F(Q)$ denote the free category on $Q$; this is the category whose objects are the vertices of $Q$ and whose morphisms are paths in $Q$, with composition given by concatenation of paths. ($F$ is left adjoint to the forgetful functor from categories to quivers which forgets the composition.) A quiver representation of $Q$ is a functor $F(Q) \to k\text{-Vect}$ for some field $k$; explicitly, it is a collection of vector spaces $V_v$ for each vertex in $Q$ together with a collection of linear maps $e : V_v \to V_w$ for each edge $e : v \to w$. (In the literature it seems that the $V_v$ 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 $k[Q]$, the quiver algebra of $Q$, which is the $k$-vector space spanned by the morphisms in $F(Q)$ (including the identity morphisms associated to each vertex) whose product is their composition when defined and zero otherwise.

Example. Let $Q$ consist of a single vertex together with $n$ loops. Then $k[Q]$ is the free $k$-algebra $k \langle e_1, ... e_n \rangle$ on $n$ generators.

$k[Q]$ as defined here is unital iff $Q$ has finitely many vertices (in this case the identity is the sum of the elements associated to each vertex) and finite-dimensional iff $Q$ has finitely many vertices and edges and no directed cycles (including loops); we summarize these conditions by saying that $Q$ is finite acyclic.

Example. Let $Q$ be the $A_n$ quiver, which consists of $n$ vertices $v_1 \to v_2 \to ... \to v_n$ connected by directed edges as indicated. $k[Q]$ naturally acts by left multiplication on the vector space spanned by the vertices $v_i$ (where, for a path $p$, we have $pv_i = 0$ if $v_i$ is not the source of $p$ and $pv_i = v_j$ if $p$ is a path from $v_i$ to $v_j$) and this action realizes an isomorphism $k[Q] \cong U_n(k)$.

For any quiver $k[Q]$, the subspace $I = k[Q]_{\ge 1}$ spanned by paths of length at least $1$ is a two-sided ideal by inspection. The quotient $k[Q]/I$ is isomorphic to the direct sum of a copy of $k$ for every vertex of $Q$. Moreover, if $Q$ is finite acyclic, then $I$ consists of nilpotent elements, so must be the largest nilpotent left ideal of $k[Q]$; hence $I = J(k[Q])$. Consequently, in this case $k[Q]$ 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).

### 3 Responses

1. on May 31, 2012 at 5:35 am | Reply Amy Szczepanski

I don’t like your definition of a Jacobson ring as “a commutative ring in which all prime ideals are intersections of maximal ideals.” If this entire post were about commutative rings, I might be OK with “a commutative ring is a Jacobson ring if ….” But to me, I prefer to think of a Jacobson ring as a not-necessarily-commutative ring where every prime ideal is the intersection of primitive ideals. Thinking this way, in the not-necessarily-commutative setting, allows for a broader interpretation of the Nullstellensatz. If R is a Jacobson ring (or, more accurately, rings with the radical property — these are not necessarily the same unless the ring’s right factor rings are all right Goldie) and an algebra over a field, then we can look at the Nullstellensatz as a question of whether the endomorphism rings of the simple R-modules are algebraic over k.

Do you have a copy of McConnell and Robson? (The AMS version is cheaper and easier to find than the Wiley version but not as pretty.) If not, you should!

• Interesting. I wasn’t aware that the Nullstellensatz had noncommutative generalizations, and I don’t have a copy of McConnell and Robson. Looks like fun.

2. […] Recall that for a left Artinian ring, the Jacobson radical is the largest nilpotent (left or right) ideal, […]