Proposition 10

Let A be a K-algebra and L an extension of K. Let M be an A-module.

a) We make one of the following two assumptions:

(i) The A-module M is finitely generated and L is algebraic over K.

(ii) The ring A is left Artinian.

Then we have the inclusion

b) If L is a separable extension of K, then we have the inclusion

Let us prove assertion a). Consider case (i). First suppose that L has finite degree over K. Then the A-module $M_{(L)}$ is finitely generated. Denote the canonical homomorphism from A to $A_{(L)}$ by $f$; the ring $A_{(L)}$ is generated by the union of its center and $f(A)$. We can therefore apply Proposition 3 of VIII, p. 174 to the $A_{(L)}$-module $M_{(L)}$. This gives the inclusion $\mathfrak{R}_A(M_{(L)})\subset \mathfrak{R}_{A_{(L)}}(M_{(L)})$ and a fortiori $\mathfrak{R}_A(M)_{(L)}\subset \mathfrak{R}_{A_{(L)}}(M_{(L)})$ (VIII, p. 152, Corollary 1).

Let us now treat the general case. Let $x_1, . . . , x_n$ be elements that generate the A-module M and $x$ an element of $\mathfrak{R}_A(M)$. Let $a_1, . . . , a_n$ be elements of $A_{(L)}$; since L is algebraic over K, there exists a finite extension $L'$ of K contained in L such that the $a_i$ belong to $A_{(L')}$. By the above, $x$ belongs to the radical of the $A_{(L')}$-module $M_{(L')}$; it follows from the corollary of VIII, p. 153 that the elements $x_i+a_ix$ for $1\leqslant i\leqslant n$ generate the $A_{(L')}$-module $M_{(L')}$ and therefore the $A_{(L)}$-module $M_{(L)}$. By the same corollary, $x$ belongs to the radical of the $A_{(L)}$-module $M_{(L)}$.

Let us now consider case (ii). Let $\mathfrak{r}$ be the radical of A, so that the radical of the A-module M is equal to $\mathfrak{r}M$ (VIII, p. 174, Corollary). The radical $\mathfrak{r}$ of A is a nilpotent two-sided ideal of A (VIII, p. 173, Proposition 1); therefore, $\mathfrak{r}_{(L)}$ is a nilpotent two-sided ideal of $A_{(L)}$. It follows that $\mathfrak{r}_{(L)}\subset \mathfrak{R}(A_{(L)})$ (VIII, p. 156, Theorem 1), and Proposition 6 of VIII, p. 158 implies $\mathfrak{R}(A_{(L)})M_{(L)}\subset \mathfrak{R}_{A_{(L)}}(M_{(L)})$. We therefore have $\mathfrak{R}_A(M) =\mathfrak{r}M\subset$ $\mathfrak{R}_{A_{(L)}}(M_{(L)})$, which concludes the proof of assertion a).

The module $M/\mathfrak{R}_A(M)$ is without radical. If L is a separable extension of K, then it follows from Proposition 9 of VIII, p. 223 that the $A_{(L)}$-module $(M/\mathfrak{R}_A(M))_{(L)}$ is without radical. Consequently, we have the inclusion