Archive for the ‘Jacobson Radical’ Category

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.