## Jacobson radical of algebras; a theorem of Amitsur

Posted: February 7, 2011 in Jacobson Radical, Noncommutative Ring Theory Notes
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Let $k$ be a field and let $A$ be a $k$-algebra. Let $J(A)$ be the Jacobson radical of $A.$

Theorem. (Amitsur) If $\dim_k A < |k|,$ as cardinal numbers, then $J(A)$ is nil.

Proof. If $k$ is finite, then $\dim_k A < \infty$ and so $A$ is Artinian. Thus $J(A)$ is nilpotent and therefore nil. So we may assume that $k$ is infinite. Let $a \in J(A).$ So $1-ca$ is invertible in $A$ for all $c \in A.$ In particular, if  $0 \neq \gamma \in k,$ then $1 - \gamma^{-1}a$ is invertible and thus $a - \gamma = -\gamma(1- \gamma^{-1}a)$ is invertible. Now consider the set

$S = \{(a-\gamma)^{-1}: \ 0 \neq \gamma \in k \}.$

Clearly for any non-zero elements $\beta, \gamma \in k,$ we have $(a-\gamma)^{-1}=(a-\beta)^{-1}$ if and only if $\gamma=\beta.$ Thus, since $k$ is infinite, we have $|S|=|k| > \dim_k A.$ Therefore the elements of $S$ cannot be $k$-linearly independent. So there exist an integer $n \geq 1$ and non-zero elements $\beta_i, \gamma_i \in k$ such that

$\sum_{i=1}^n \beta_i(a-\gamma_i)^{-1}=0. \ \ \ \ \ \ \ \ \ \ (1).$

Obviously all $a - \gamma_i$ commute with each other and with all $\beta_i$ because $\beta_i$ and $\gamma_i$ are in $k.$ So multilpying $(1)$ through by $\prod_{i=1}^n (a - \gamma_i)$ will clear denominators and we will get

$\sum_{i=1}^n \beta_i \prod_{j \neq i} (a - \gamma_j) = 0. \ \ \ \ \ \ \ \ \ \ \ (2)$

Define the polynomial $p(x) = \sum_{i=1}^n \beta_i \prod_{j \neq i} (x - \gamma_j) \in k[x].$ Then $p(a) = 0$ by $(2).$ Besides $p(x)$ is not identically zero because for example $p(\gamma_1) = \beta_1(\gamma_1 - \gamma_2)(\gamma_1 - \gamma_3) \cdots (\gamma_1 - \gamma_n) \neq 0.$ So we have proved that $a$ is algebraic over $k.$ Thus

$a^m + \alpha_1 a^{m-1} + \cdots + \alpha_m = 0, \ \ \ \ \ \ \ \ \ \ \ (3)$

for some integer $m \geq 1$ and some $\alpha_i \in k.$ Now, $a$ is not invertible because $a \in J(A).$ Thus $\alpha_m = 0.$ On the other hand, if $\alpha_i = 0,$ for all $1 \leq i \leq m,$ then $(3)$ becomes $a^m = 0$ and we are done. So we may assume that there exists $1 \leq \ell < m$ such that $\alpha_{\ell} \neq 0$ and

$a^m + \alpha_1 a^{m-1} + \cdots + \alpha_{\ell}a^{m- \ell} = 0. \ \ \ \ \ \ \ \ \ \ \ (4)$

If we let $c = - \alpha_{\ell}^{-1}(a^{\ell - 1} + \alpha_1 a^{\ell - 2} + \cdots + \alpha_{\ell - 1}),$ then we can re-write $(4)$ as

$\alpha_{\ell} (1 - ca)a^{m- \ell} = 0. \ \ \ \ \ \ \ \ \ \ \ \ (5)$

Since $a \in J(A),$ the element $1-ca$ is invertible and thus $(5)$  gives us $a^{m- \ell} = 0. \ \Box$

An interesting case of the theorem is when $A$ is finitely generated and $k$ is uncountable. Then clearly $\dim_k A$ is countable, because $A$ is finitely generated, and thus $\dim_k A < |k|.$ So $J(A)$ is nil.