Jacobson-Hersteins Commutativity Theorem

02-23

In this article, we present Jacobson-Hersteins Commutativity Theorem, which employs a classic reduction in its proof, that is, when trying to prove certain theorems about rings, it is possible to reduce one class of rings to another, simpler, class, in the following sense.

We shall now introduce the required definitions and intermediate results. They are mainly adopted from "A First Course in Noncommutative Rings", 2nd edition, T.Y. Lam.

The Jacobson radical (or the left radical) of a ring $R$ , denoted by $rad R$ , is the intersection of all the maximal left ideals of $R$ . If $R eq 0$ , then maximal left ideals always exist by Zorns Lemma, and $rad R eq R$ . Otherwise if $R = 0$ , then there exists no maximal left ideals. In this case, the Jacobson radical is defined to be zero.

The following result characterizes the elements of $rad R$ .

(1.1) Proposition. For any $y in rad R$ , the following are equivalent.

$y in rad R$ .
$1 - xy$ is left invertible for any $x in R$ .
$yM = 0$ for any simple left $R$ -module $M$ .

Proof. 1 $Rightarrow$ 2. Suppose that $y in rad R$ and $1 - xy$ is not left invertible for some $x in R$ . Then $R cdot (1 - xy)$ is contained in a maximal left ideal $mathfrak{m}$ of $R$ . However, $1 - xy in mathfrak{m}$ and $y in mathfrak{m}$ imply that $1 in mathfrak{m}$ , a contradiction.

2 $Rightarrow$ 3. Suppose that $1 - xy$ is left invertible for any $x in R$ and $y$ does not kill $m$ for some $m in M$ . Then it follows from the simplicity of $M$ that $R cdot ym = M$ . In particular, $x cdot ym = m$ for some $x in R$ . Therefore $(1 - xy)m = 0$ . By 2, $m = 0$ , a contradiction.

3 $Rightarrow$ 1. For any maximal left ideal $mathfrak{m}$ , $R/mathfrak{m}$ is a simple left $R$ -module. By 3, $y cdot R/mathfrak{m} = 0$ . It follows that $y in mathfrak{m}$ . By definition, $y in rad R$ .

It is not hard to see that $rad R = igcap Ann(M)$ , where $M$ ranges over all the simple left $R$ -modules. In particular, $rad R$ is an ideal of $R$ .

Also, it is not hard to see that a fourth condition can be added to (1.1) to strengthen the second condition.

(1.2) Proposition. For any $y in rad R$ , the following are equivalent.

$1 - xy$ is left invertible for any $x in R$ .
$1 - xyz$ is invertible for any $x, z in R$ .

Proof. 1 $Rightarrow$ 2. Since $rad R$ is an ideal, $yz in rad R$ . By (1.1), $u(1 - xyz) = 1$ for some $u in R$ . Since $rad R$ is an ideal, $xyz in rad R$ . Again by (1.1), $u = 1 + u(xyz)$ is left invertible. Since $u$ is also right invertible, $u$ is invertible and hence $1 - xyz$ is invertible.

2 $Rightarrow$ 1. Clear.

Consequently, we have the following corollary.

(1.3) Corollary. The left radical of $R$ agrees with its right radical.

The Jacobson radical leads to a new notion, which we now introduce.

Semiprimitivity. A ring $R$ is called semiprimitive if $rad R = 0$ .

Observe that $R/rad R$ is always semiprimitive, as $rad (R/mathfrak{A}) cong (rad R)/mathfrak{A}$ for any ideal $mathfrak{A}$ of $R$ .

Alternatively, semiprimivity can be characterized in the following way.

(1.4) Proposition. A ring $R$ is semiprimitive if and only if $R$ has a faithful semisimple left module $M$ .

Proof. Suppose that $M$ exists. Since $rad R$ acts as zero on all left simple $R$ -modules, $(rad R) cdot M = 0$ . Then the faithfulness of $M$ implies that that $rad R = 0$ . In other words, $R$ is semiprimitive. Conversely, suppose that $rad R = 0$ . Let ${M_i}$ be a complete set of mutually nonisomorphic simple left $R$ -modules. Then $M = igoplus M_i$ is semisimple, and

$Ann(M) = igcap Ann(M_i) = rad R$ . Since $rad R = 0$ , $M$ is a faithful $R$ -module.

The notion of a left primitive ring can be introduced in a similar manner.

Primitivity. A ring $R$ is called left primitive if $R$ has a faithful simple left module.

Although semiprimitivity is left-right symmetric, primitivity, on the other hand, is not. The first example was constructed by G. Bergman in 1964. See Errata: A Ring Primitive on the Right but Not on the Left for more information.

Before proceeding, we extend the notion of primitivity to ideals.

Primitivity. An ideal $mathfrak{A} subseteq R$ is called left primitive if $R/mathfrak{A}$ is left primitive.

And we have the following characterization of left primitive ideals.

(1.5) Proposition. An ideal $mathfrak{A}$ in $R$ is left primitive if and only if $mathfrak{A}$ is the annihilator of a simple left $R$ -module.

Proof. Suppose that $mathfrak{A} = Ann(M)$ , where $M$ is a simple left $R$ -module. Then the natural $R/mathfrak{A}$ -action on $M$ is faithful. In other words, $R/mathfrak{A}$ is a left primitive ring. Conversely, suppose that $R/mathfrak{A}$ is left primitive and $M$ is a faithful simple left $R/mathfrak{A}$ -module. When considered as an $R$ -module, $M$ preserves simplicity. Moreover, $Ann(M) = mathfrak{A}$ .

(1.5) implies that the Jacobson radical $rad R$ is the intersection of all the left primitive ideals in $R$ .

By definition, a left primitive ring is always semiprimitive. Conversely, it is also possible to reduce a semiprimitive ring to a left primitive ring, in the following procedure.

Let ${R_i}$ be rings, and $epsilon: R o prod R_i$ be an injective ring homomorphism. $epsilon$ represents $R$ as a subdirect product of the $R_i$ s if each of the maps $R o R_i$ obtained by composing $epsilon$ with the coordinate projections is surjective. Moreover, a subdirect product representation is trivial if one of the maps $R o R_i$ is an isomorphism.

(1.6) Theorem. A nonzero ring $R$ is semiprimitive if and only if $R$ is a subdirect product of left primitive rings.

Proof. Suppose that $R$ is semiprimitive. Let ${mathfrak{A}_i}$ be the family of left primitive ideals in $R$ . Then $igcap mathfrak{A}_i = 0$ and $R$ is a subdirect product of left primitive rings ${R/mathfrak{A}_i}$ . Conversely, suppose that $epsilon: R o prod R_i$ is a subdirect product representation, where $R_i$ s are left primitive rings. Then $mathfrak{A}_i = ker(R o R_i)$ is a left primitive ideal, and $igcap mathfrak{A}_i = 0$ . Since $rad R$ is contained in $igcap mathfrak{A}_i$ , $rad R = 0$ .

We shall now define the notion of density, which will play an important role in the structure theorem of left primitive rings.

Acts Densely. Let $R, k$ be rings, and $V$ be an $(R,k)$ -bimodule. $R$ acts densely on $V_k$ if for any $f in End(V_k)$ and $v_1, ..., v_n in V$ , there exists $r in R$ such that $rv_i = f(v_i)$ for all $1 leq i leq n$ .

(1.7) Lemma. In the above notation, if $_RV$ is a semisimple left $R$ -module and $k = End(_RV)$ . Then any $R$ -submodule $W$ of $V$ is an $End(V_k)$ -submodule.

Proof. Write $V = W oplus W$ for some $R$ -submodule $W$ of $V$ . Let $e in End(_RV)$ be the projection of $V$ on $W$ with respect to this decomposition. Then for any $f in End(_RV)$ , $f(W) = f(We) = (fW)e subseteq W$ , implying that $W$ is an $End(_RV)$ -submodule of $V$ .

(1.7) prepares us for the following fundamental result.

(1.8) Density Theorem. Let $R$ be a ring, and $V$ be a semisimple left $R$ -module. Then $R$ acts densely on $V_k$ , where $k = End(_RV)$ .

Proof. For any $f in End(_RV)$ and any $v_1, ..., v_n in V$ , define $ilde{f}: ilde{V} o ilde{V}$ by $ilde{f} = (f, f, ..., f)$ , where $ilde{V} = V^n$ . It is routine to verify that $ilde{f} in End( ilde{V}_{ ilde{k}})$ , where $ilde{k} = End(_R ilde{V}) cong M_n(End(_RV)) = M_n(k)$ . Now consider the cyclic $R$ -submodule $ilde{W}$ of $ilde{V}$ generated by $(v_1,...,v_n) in ilde{V}$ . By (1.7), $ilde{W}$ is stabilized by $End( ilde{V}_{ ilde{k}})$ , in particular, by $ilde{f}$ . Therefore $ilde{f}(v_1,...,v_n) = (f(v_1),...,f(v_n))$ , implying the existence of a desired $r in R$ .

(1.9) Corollary. In the above notation, if $V_k$ is finitely generated, then the natural map $ho: R o End(V_k)$ is surjective.

Proof. Let $v_1, ..., v_n in V$ be a finite set of generators of $V_k$ . Then for any $f in End(V_k)$ , there exists $r in R$ such that $rv_i = f(v_i)$ for all $1 leq i leq n$ . Therefore $rv = sum r(v_ia_i) = sum (rv_i)a_i = sum (f(v_i))a_i = f(sum{v_ia_i}) = f(v)$ for all $v in V$ . By definition, $f = ho(r)$ .

Let $k$ be a division ring, and $V$ be a right vector space over $k$ . A subset $S subseteq End(V_k)$ is called $m$ -transitive if for any set of $n leq m$ linearly indepedent vectors $v_1, ..., v_n$ and any other set of $n$ vectors $v_1, ..., v_n$ , there exists $s in S$ such that $s(v_i) = v_i$ for all $1 leq i leq n$ . If $S$ is $m$ -transitive for all (finite) $m$ , then $S$ is called a dense set of linear transformations on $V_k$ . The following result ensures that this definition is consistent with the one we introduced.

(1.10) Proposition. Let $V$ be an $(R,k)$ -bimodule, where $R$ is a ring and $k$ is a division ring. Then $R$ acts densely on $V_k$ if and only if $ho(R)$ is a dense ring of linear transformations on $V$ , where $ho: R o End(V_k)$ is the natural ring homomorphism.

Proof. One direction is clear. For the other direction, assume that $ho(R)$ is dense, and let $f in End(V_k)$ , $v_1, ..., v_n in V$ . After a possible permutation, $v_1, ..., v_m$ are linearly independent over $k$ for some $m leq n$ , and each $v_i$ is a linear combination of $v_1, ..., v_m$ for all $m leq i leq n$ . Since $ho(R)$ is $m$ -transitive, there exists $r in R$ such that $rv_j = f(v_j)$ for all $1 leq j leq m$ . It follows that this is true when $j$ is extended to $1, ..., n$ .

(1.8), (1.9), and (1.10) yield the following result, which characterizes the structure of left primitive rings completely. (1.11) is sometimes also referred as the Density Theorem.

(1.11) Theorem. Let $R$ be a left primitive ring, $V$ be a faithful simple left $R$ -module, and $k$ be the division ring $End(_RV)$ (by Schurs Lemma). Then $R$ is isomorphic to a dense ring of linear transformations on $V_k$ . Moreover:

If $R$ is left Artinian, then $n = dim_k{V}$ is finite and $R cong M_n(k)$ .
If $R$ is not left Artinian, then $dim_k{V}$ is infinite, and for any $n > 0$ , there exists a subring $R_n$ of $R$ which admits a ring homomorphism onto $M_n(k)$ .

1 recovers the Artin-Wedderburns Theorem for left Artinian simple rings, as these are precisely the left Artinian left primitive rings, by a moments thought. A more generalized version of Artin-Wedderburns Theorem will be presented at the end of this article.

Proof. The first conclusion (before "moreover") follows easily from the faithfulness of $_RV$ and the density of $ho(R)$ . For the second conclusion (after "moreover"), assume that $dim_k{V} = n < infty$ . By (1.9), $ho(R) woheadrightarrow End(V_k)$ . And by faithfulness of $_RV$ , $ho(R) hookrightarrow End(V_k)$ . In conclusion, $R cong M_n(k)$ . Since $R$ is semisimple, $R$ is clearly left Artinian. Next assume that $dim_k{V}$ is infinite, and let $v_1, ..., v_n$ be $n$ linearly independent vectors, $V_n = sum_{i = 1}^n{v_ik}$ for all $1 leq n < infty$ . Finally, let $R_n = {r in R: r(V_n) subseteq V_n}$ , and $mathfrak{A_n} = {r in R: r(V_n) = 0}$ . Then $R_n/mathfrak{A}_n$ acts faithfully on the $k$ -vector space $V_n$ . By $n$ -transitivity of $R$ , any $k$ -endomorphism of $V_n$ can be realized as the action of some $r in R_n$ . Therefore the natural map $R_n/mathfrak{A}_n o End((V_n)_k)$ is an isomorphism, yielding $R_n/mathfrak{A}_n cong M_n(k)$ . Furthermore, by $(n+1)$ -transitivity, there exists $r in R$ such that $rv_1 = ... = rv_n = 0$ but $rv_{n+1} eq 0$ . This induces a strictly descending sequence of left ideals $mathfrak{A}_1 supset mathfrak{A}_2 supset mathfrak{A}_3 supset ...$ , so $R$ is not left Artinian.

We shall now present the Jacobson-Hersteins Theorem.

Jacobson-Hersteins Theorem. A ring $R$ is commutative if and only if for any $a, b in R$ , there exists an integer $n(a,b) > 1$ such that $(ab - ba)^{n(a,b)} = ab - ba$ .

Proof. In this proof, we assume the following theorem, which will be proved later.

(1.12) Theorem. Let $R$ be a division ring such that for any $a in R$ , $a^{n(a)} = a$ for some integer $n(a) > 1$ . Then $R$ is commutative.

(1.12) is true for any left primitive ring $R$ . By (1.11), there exists a division ring $k$ such that either $R cong M_n(k)$ or for any $m$ , there exists a subring $R_m subseteq R$ which admits a ring homomorphism onto $M_m(k)$ . However, for $m geq 2$ , $M_m(k)$ cannot satisfy the required property "for any $a, b in R$ , there exists an integer $n(a,b) > 1$ such that $(ab - ba)^{n(a,b)} = ab - ba$ " by a routine check. Therefore $R cong k$ . (1.12) applied.
(1.12) is true for any semiprimitive ring $R$ . By (1.6), there exists a subdirect produce representation $R o prod{R_i}$ , where $R_i$ s are left primitive rings. Since each $R_i$ is a homomorphic image of $R$ , $R_i$ is commutative by 1 and (1.12). Since $R$ is isomorphic to a subring of $prod{R_i}$ , $R$ is commutative.
(1.12) is true for any ring $R$ . By 2, $R/rad R$ is commutative. Therefore for any $a, b in R$ , $ab - ba in rad R$ . Since $(ab - ba)^{n(a,b)} = ab - ba$ for some $n(a,b) > 1$ , $(ab - ba)(1 - (ab - ba)^{n(a,b) - 1}) = 0$ . However, $(1 - (ab - ba)^{n(a,b) - 1}) in 1 + rad R$ is a unit, by a moments reflection on (1.2), $ab - ba$ must equal to zero.

Proof of (1.12).

Claim 1. A finite division ring $R$ is a field.

Proof. Let $F$ be the center of $R$ . Then $F$ is finite field or order $q$ for some prime power $q$ . Assume $R$ is not field, or in other words, $dim_F{D} = n > 1$ . Write down the class equation for the finite group $D^{ imes}$ : $|D^{ imes}| = q^n - 1 = q - 1 + sum |D^{ imes}: C(a)^{ imes}|$ , and $r = r(a) = dim_{F}C(a)$ . Then $r | n$ and it is possible to rewrite the class equation: $|D^{ imes}| = q^n - 1 = q - 1 + frac{q^n -1}{q^r - 1}$ . Since $r|n$ , $q^n - 1 = Phi_n(q)(q^r - 1)h(q)$ , where $Phi_n(q)$ is the $n$ -th cyclotomic polynomial, and $h(q) in mathbb{Z}[q]$ . The class equation implies that $frac{q^n-1}{q^r-1}$ is divisible by $Phi_n(q)$ and hence $q - 1$ is divisible by $Phi_n(q)$ . In particular, $q - 1 geq |Phi_n(q)| = prod|q - zeta|$ , where $zeta$ ranges over all primitive $n$ -th roots of unity. A contradiction.

Claim 2. A finite subgroup $G$ of a division ring $R$ of characteristic $p > 0$ is cyclic.

Proof. Let $F$ be the prime field of $R$ , and $K = {sum a_ig_i: a_i in F, g_i in G}$ . Then $K$ is a finite subring of $R$ , which is a field by Claim 1. Since $G$ is a subgroup of $K^{ imes}$ , $G$ is cyclic.

Claim 3. If all additive commutators are central in a division ring $R$ , then $R$ is a field.

Proof. It suffices to show that if $y in R$ commutes with all additive commutators, then $y in Z(R)$ . If $y otin Z(D)$ , $xy eq yx$ for some $x in D$ . Consider $x(xy) - (xy)x = x(xy - yx)$ . Since $y$ commutes with $x(xy) - (xy)x$ and $xy - yx eq 0$ , $y$ commutes with $x$ , a contradiction.

Claim 4. Let $R$ be a division ring of characteristic $p > 0$ . If $a in R^{ imes}$ is noncentral and torsion, then there exists $y in R^{ imes}$ such that $yay^{-1} = a^i eq a$ for some $i > 0$ . Moreover, $y$ can be chosen to be an additive commutator in $R$ .

Proof. This proof is removed from this article because of its length and my laziness.

By the assumption in (1.12), any nonzero additive commutator has finite order in the multiplicative group $D^{ imes}$ . If $R eq F = Z(R)$ , then there exists some additive commutator $a = bb - bb otin F$ by Claim 3. For any $c in F^{ imes}$ , $ca = (cb)b - b(cb)$ is also a nonzero additive commutator. Since $a$ and $ca$ both have finite order, there exists an integer $k > 0$ such that $1 = a^k = (ca)^k = c^ka^k$ . This implies that $char F = char R > 0$ . Also, since $a$ is non central and torsion, Claim 4 yields an additive commutator $d in R^{ imes}$ such that $dad^{-1} = a^i eq a$ , where $i > 0$ . As $y$ normalizes $langle a angle$ , $langle a angle cdot langle y angle$ is a finite subgroup of $R^{ imes}$ . By Claim 2, $langle a angle cdot langle y angle$ is commutative, a contradiction.

Proof of Artin-Wedderburns Theorem.

Statement. Let $R$ be a semisimple ring. Then $R cong M_{n_1}(D_1) imes ... imes M_{n_r}(D_r)$ for some division rings $D_i$ and natural numbers $n_i$ . Moreover, $(n_1,D_1),...,(n_r,D_r)$ is uniquely determined, up to permutation.

Proof. Decompose $_RR$ into a finite direct sum of minimal left ideals and group the minimal left ideals according to isomorphism types. Then $_RR cong n_1V_1 oplus ... oplus n_rV_r$ , where $V_1, ..., V_r$ are mutually nonisormophic simple left $R$ -modules. Since $R$ -endomorphism of $_RR$ is determined by right multiplication by elements of $R$ , $R cong End(_RR)$ . Therefore $R cong End(n_1V_1 oplus ... oplus n_rV_r) = End(n_1V_1) imes End(n_rV_r)$ . By Schurs Lemma, each $D_i = End(V_i)$ is a division ring. And by a routine verification, each $End(n_iV_i) cong M_{n_i}(D_i)$ . In conclusion, we have the desired decomposition $R cong M_{n_1}(D_1) imes ... imes M_{n_r}(D_r)$ . The uniqueness of this decomposition follows easily from Jordan Holders Theorem.

Reference.

"A First Course in Noncommutative Rings", 2nd edition, T.Y. Lam.