Proposition 4

For $i\in \{1,2\}$, let $A_i$ be a K-algebra, $M_i$ a semisimple $A_i$-module, and $Z_i$ the center of the commutant of $M_i$. Set $A = A_1\otimes A_2$, $M = M_1\otimes M_2$, and $Z = Z_1\otimes Z_2$. We have $\mathfrak{R}_A(M) =\mathfrak{R}(Z)M$. If one of the algebras $Z_1$ and $Z_2$ is separable over K, in particular if the field K is perfect, then the A-module M is without radical.

For $i\in \{1,2\}$, let $S_i$ be a simple $A_i$-module, $D_i$ its commutant, and $I(i)$ a set. We begin by treating the case when $M_i$ is the $A_i$-module $S_i^{(I(i))}$. We identify the center $Z_i$ of its commutant with the center of $D_i$. Set $D = D_1\otimes D_2$. We have $\mathfrak{R}(D) =\mathfrak{R}(Z)D$ (Proposition 3 of VIII, p. 217) and $\mathfrak{R}_A(S_1\otimes S_2) =$ $\mathfrak{R}(D)(S_1\otimes S_2)$ (VIII, p. 215, Corollary 2), hence $\mathfrak{R}_A(S_1\otimes S_2) =\mathfrak{R}(Z)(S_1\otimes S_2)$. The A-module M is the direct sum of a family of A-modules isomorphic to $S_1\otimes S_2$, and the radical of the direct sum of a family of modules is the direct sum of the radicals (VIII, p. 152, Corollary 2). We therefore have $\mathfrak{R}_A(M) =\mathfrak{R}(Z)M$.

Now consider the general case. For $i\in \{1,2\}$, denote the support of the $A_i$-module $M_i$ by $\mathscr{S}_{M_i}$. For $\lambda \in \mathscr{S}_{M_i}$, denote the isotypical component of $M_i$ of type $\lambda$ by $M_{i;\lambda}$ and the center of its commutant by $Z_{i;\lambda}$. We identify the ring $Z_i$ with the product of the rings $Z_{i;\lambda}$ for $\lambda \in \mathscr{S}_{M_i}$. Let $\lambda \in \mathscr{S}_{M_1}$ and $\mu\in \mathscr{S}_{M_2}$. Denote by $i_{\lambda}: Z_{1;\lambda}\rightarrow Z_1$ the unique K-linear mapping such that pr$_{\lambda}\circ i_{\lambda}$ is the identity mapping on $Z_{1;\lambda}$ and pr$_{\lambda'}\circ i_{\lambda}$ is the zero mapping for $\lambda '\in \mathscr{S}_{M_1}-\{\lambda \}$. Define $i_\mu: Z_{2;\mu}\rightarrow Z_2$ likewise. Set $Z_{\lambda ,\mu}= Z_{1;\lambda}\otimes Z_{2;\mu}$ and $i_{\lambda ,\mu}=i_{\lambda}\otimes i_\mu$, and denote the mapping pr$_{\lambda}\otimes$ pr$_\mu$ from Z to $Z_{\lambda ,\mu}$ by $\pi_{\lambda ,\mu}$. The mapping $\pi_{\lambda ,\mu}$ is a surjective ring homomorphism; we therefore have $\pi_{\lambda ,\mu}(\mathfrak{R}(Z))\subset \mathfrak{R}(Z_{\lambda ,\mu})$ (VIII, p. 155, Proposition 5, b)). Let us prove the reverse inclusion. Let $z$ be an element of $\mathfrak{R}(Z_{\lambda ,\mu})$; since $Z_{1;\lambda}$ and $Z_{2;\mu}$ are fields, $z$ is nilpotent (Theorem 3 of VIII, p. 215). We have $i_{\lambda ,\mu}(xy) =i_{\lambda ,\mu}(x)\cdot i_{\lambda ,\mu}(y)$ for $x, y$ in $Z_{\lambda ,\mu}$; consequently, $i_{\lambda ,\mu}(z)$ is nilpotent and therefore belongs to $\mathfrak{R}(Z)$. Since $\pi_{\lambda ,\mu}\circ i_{\lambda ,\mu}$ is the identity mapping on $Z_{\lambda ,\mu}$, the element $z$ belongs to $\pi_{\lambda ,\mu}(\mathfrak{R}(Z))$. We have thus proved the equality $\pi_{\lambda ,\mu}(\mathfrak{R}(Z)) =\mathfrak{R}(Z_{\lambda ,\mu})$.

Set $M_{\lambda ,\mu}= M_{1;\lambda}\otimes M_{2;\mu}$; this is a submodule of M that is stable under Z. For $z\in Z$ and $m\in M_{\lambda ,\mu}$, we have $zm=\pi_{\lambda ,\mu}(z)m$. Consequently, $\mathfrak{R}(Z)M_{\lambda ,\mu}$ is equal to $\mathfrak{R}(Z_{\lambda ,\mu})M_{\lambda ,\mu}$ and therefore to $\mathfrak{R}_A(M_{\lambda ,\mu})$ by the isotypical case. Since the radical of a direct sum is the direct sum of the radicals (VIII, p. 152, Corollary 2) and M is the direct sum of the submodules $M_{\lambda ,\mu}$ for $(\lambda , \mu)\in \mathscr{S}_{M_1}\times \mathscr{S}_{M_2}$, the equality $\mathfrak{R}_A(M) =\mathfrak{R}(Z)M$ is proved. The last assertion then follows from Proposition 3 of VIII, p. 217.