The Jacobson radical; nil & nilpotency

Posted: August 16, 2022 in Jacobson Radical, Noncommutative Ring Theory Notes
Tags: algebraic algebra, Artinian rings, Jacobson radical, nil ideal, nill, nilpotent

See part (1) and part (2) of this post here and here. All rings, as usual, are assumed to have identity.

So far, we have defined the Jacobson radical of a ring and we have given some basic results and examples. In this post, we go a little deeper and look at some cases where the Jacobson radical is nil or nilpotent.

Let $I$ be a left ideal of a ring. Recall that we say $I$ is nil if for every $a \in I,$ there exists some positive integer $n$ such that $a^n=0.$ We say that $I$ is nilpotent if $I^n=(0)$ for some positive integer $n,$ i.e. $a_1a_2 \cdots a_n=0$ for all $a_i \in I.$ Clearly if $I$ is nilpotent, then $I$ is also nil.

Fact 1. The Jacobson radical of a ring contains every nil left ideal of the ring.

Proof. Let $I$ be a nil left ideal of a ring $R.$ Let $x \in I, \ r \in R.$ Then $rx \in I$ and so $rx$ is nilpotent. Hence $(rx)^n=0$ for some positive integer $n$ and so $1+rx+ \cdots + (rx)^{n-1}$ is the inverse of $1-rx.$ Thus $1-rx$ is invertible for all $r \in R$ and so $x \in J(R). \ \Box$

The next fact is about the Jacobson radical of a left Artinian ring. I give two versions of the fact: the first one is weak and it says the Jacobson radical is nil but the second one is strong and it says the Jacobson radical is in fact nilpotent. The reason for doing that is because sometimes we only need the weak version, which is much easier to prove.

Fact 2. Let $R$ be a left Artinian ring. Then

i) $J(R)$ is nil,

ii) $J(R)$ is in fact nilpotent.

Proof. i) Let $x \in J(R)$ and consider the descending chain of left ideals $Rx \supseteq Rx^2 \supseteq \cdots .$ Since $R$ is left Artinian, there exists a positive integer $n$ such that $Rx^n=Rx^{n+1}$ and so $x^n=rx^{n+1}$ for some $r \in R.$ Hence $(1-rx)x^n=0$ and so $x^n=0$ because $x \in J(R)$ and so $1-rx$ is invertible.

ii) Let $J:=J(R)$ and consider the descending chain of ideas $J \supseteq J^2 \supseteq J^3 \supseteq \cdots.$ Since $R$ is (left) Artinian, there exists an integer $m \ge 1$ such that $J^n=J^m$ for all $n \ge m.$ In particular, $J^{2m}=J^m.$ We show that $J^m=(0).$ So suppose, to the contrary, that $J^m \ne (0)$ and consider the set

$S:=\{\text{left ideals} \ I \ \text{of} \ R \ \text{such that} \ J^mI \ne (0)\}.$

Since $J^m \ne (0),$ we have $R \in S$ and so $S \ne \emptyset.$ Since $R$ is left Artinian, $(S, \subseteq)$ has a minimal element, say $K.$ Since $K \in S,$ we have $J^mK \ne (0)$ and therefore there exists $x \in K$ such that $J^mx \ne (0).$ In particular, $x \ne 0.$ Let $K_1:=J^mx, \ K_2:=Rx.$ Then both $K_1,K_2$ are in $S$ and both are contained in $K.$ Thus, by minimality of $K,$ we must have $K_1=K_2=K$ and so $J^mx=Rx.$ Hence $x=ax$ for some $a \in J^m$ and that gives $(1-a)x=0.$ But $1-a$ is invertible because $a \in J,$ and so $(1-a)x=0$ gives $x=0,$ which is a contradiction. $\ \Box$

Now suppose that $k$ is a field, and $R$ is a $k$ -algebra. So $R$ is a vector space over $k$ and thus we can talk about $\dim_k R.$ If $\dim_k R$ is finite, then $R$ is Artinian and hence, by the first part of Fact 2, $J(R)$ is nil. Amitsur extended this result to $k$ -algebras $R$ for which $\dim_k R < \text{card}(k).$

Fact 3. (Amitsur) Let $k$ be a field, and let $R$ be a $k$ -algebra. If $\dim_k R < |k|,$ as cardinal numbers, then $J(R)$ is nil.

Proof. See this post. $\Box$

Again, suppose that $k$ is a field, and $R$ is a $k$ -algebra. Recall that we say $r \in R$ is algebraic (over $k$ ) if it is a root of some nonzero polynomial in $k[x].$ We say that $R$ is an algebraic algebra if every element of $R$ is algebraic. Obviously, if $\dim_k R$ is finite, then $R$ is algebraic.

Fact 4. Let $k$ be a field, and let $R$ be a $k$ -algebra. Let $r \in J(R).$ If $r$ is algebraic, then $r$ is nilpotent. In particular, the Jacobson radical of every algebraic algebra is nil.

Proof. Let $p(x)=a_mx^m+a_{m+1}x^{m+1} + \cdots + a_nx^n, \ a_i \in k, \ a_m \ne 0,$ be a polynomial such that $p(r)=0.$ So $a_mr^m+a_{m+1}r^{m+1}+ \cdots + a_nr^n=0$ and hence

$r^m(1+b)=0, \ \ \ b:=a_m^{-1}a_{m+1}r + \cdots + a_m^{-1}a_nr^{n-m} \in J(R).$

Now, $1+b$ is invertible because $b \in J(R),$ and so $r^m(1+b)=0$ gives $r^m=0. \ \Box$

Remark. Since, by Fact 1, the Jacobson radical contains every nil (left) ideal, saying the Jacobson radical is nil is equivalent to saying the Jacobson radical is the largest nil ideal of the ring.

Next, we give examples of rings whose Jacobson radical is zero, see it here.

Abstract Algebra

Blogroll

Top Posts & Pages

Categories