2
$\begingroup$

In Herstein's book "Noncommutative rings" it is said that it is a simple exercise to show that the Jacobson radical of an algebraic algebra is nil. But I can't see why.

Let $A$ be an algebraic algebra. And take $a\in J(A)$. Since $A$ is algebraic there is a polynomial $f(x)=x^n+\alpha_{1}x^{n-1}+\dots+\alpha_{n-1}x+\alpha_n$ s.t. $f(a)=0$.

Also, $Ma=(0)$ for every irreductible $A$-module $M$, since $a\in J(A)$.

(I don't know if the following is helpful) We also have that for any $b\in A$, $$[f(a),b]= [a^n,b]+\alpha_1[a^{n-1},b]+\dots+\alpha_{n-1}[a,b]=0$$ Commute this with $[a,b]$ and we get $$[[a^n,b],[a,b]]+\alpha_1[[a^{n-1},b],[a,b]]+\dots+\alpha_{n-2}[[a^2,b],[a,b]]=0$$ Commute this with $[[a^2,b],[a,b]]$ and so on, $n$ times, and we get a polynomial involving only commutators of $[a^i,b]$ and none of the coefficients $\alpha_i$.

How does that leads to the fact that $J(A)$ is nil?

$\endgroup$
4
$\begingroup$

You can write $f(x) = x^i g(x)$ for some $i$ such that $g(0) \neq 0$. Note that $g(a) \in g(0) 1 + Aa$ is invertible since $a \in J(A)$. It follows that $0 = f(a) = a^i g(a)$ and multiplying by $g(a)^{-1}$ shows $a^i = 0$.

$\endgroup$
  • $\begingroup$ Klupsch But that can be the case that g=f. Then g(a)=0, right? $\endgroup$ – Andre Gomes Apr 18 '17 at 13:21
  • $\begingroup$ And even if $g(a)\neq 0$, i can't see why the fact that $a\in J(A)$ implies that $g(a)$ is invertible. $\endgroup$ – Andre Gomes Apr 18 '17 at 13:31
  • $\begingroup$ @AndréGomes It can't be $i=0$: if $f(0)\ne0$, then $f(a)$ is invertible. $\endgroup$ – egreg Apr 18 '17 at 14:29
  • $\begingroup$ @egreg i can't see that. $\endgroup$ – Andre Gomes Apr 18 '17 at 14:36
  • 4
    $\begingroup$ @AndréGomes Yes, sorry, it is $c=f(0)$. An element $a\in A$ belongs to $J(A)$ if and only if, for every $b\in A$, $1+ab$ is invertible: it's a basic characterization of the Jacobson radical. $\endgroup$ – egreg Apr 18 '17 at 17:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.