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?


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 $a^i = 0$.

  • $\begingroup$ 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

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