Proposition 3

Suppose that the algebras $A_1$ and $A_2$ are semisimple, with respective centers $Z_1$ and $Z_2$. Set $Z = Z_1\otimes Z_2$.

a) The mapping $\mathfrak{a}\mapsto \mathfrak{a}A$ is an isomorphism from the set of ideals of Z, ordered by inclusion, to the set of two-sided ideals of A, ordered by inclusion.

b) The radical of A is equal to the intersection of the maximal two-sided ideals of A and equals $\mathfrak{R}(Z)A$.

c) If one of the K-algebras $Z_1$ and $Z_2$ is separable, in particular if the field K is perfect, then the radicals of the rings Z and A are reduced to 0.

Each algebra $A_i$ is the product of finitely many simple algebras. Now, the center of a product of rings is the product of the centers, and we have the analogous assertions for radicals (VIII, p. 156, Corollary 3) and for two-sided ideals (I, §8, No. 10, p. 109, Proposition 8). It therefore suffices to prove Proposition 3 under the assumption that $A_1$ and $A_2$ are simple algebras.

For $i\in \{1,2\}$, set $B_i= A_i\otimes A^o_i$, and view $A_i$ as a $B_i$-module, where the homotheties are given by the formula

for $x,y$, and $z$ in $A_i$. The commutant of the $B_i$-module $A_i$ is $(Z_i)_{A_i}$, the set of homotheties by elements of $Z_i$; we identify it with $Z_i$. Moreover, the $B_i$-submodules of $A_i$ are the two-sided ideals of $A_i$, and since the ring $A_i$ is simple, the only two-sided ideals are 0 and $A_i$. Hence $A_i$ is a simple $B_i$-module. Moreover, the $(B_1\otimes B_2$)-submodules of $A_1\otimes A_2$ are the two-sided ideals of the ring $A_1\otimes A_2$.

Assertion a) therefore follows from VIII, p. 214, Theorem 2, b) applied to the simple $B_1$-module $A_1$, with commutant $Z_1$, and to the simple $B_2$-module $A_2$, with commutant $Z_2$.

Let us prove assertion b). The intersection of the maximal two-sided ideals of A is the radical of the $(B_1\otimes B_2$)-module $A_1\otimes A_2$; by Corollary 2 of VIII, p. 215, this intersection coincides with $\mathfrak{R}(Z)A$. The algebras $Z_1$ and $Z_2$ are commutative and semisimple. The radical of the ring Z therefore consists of nilpotent elements (VIII, p. 215, Theorem 3), and the two-sided ideal $\mathfrak{R}(Z)A$ of the ring A is contained in the radical $\mathfrak{R}(A)$ (VIII, p. 157, Remark 1). However, the intersection of the maximal two-sided ideals of A contains $\mathfrak{R}(A)$ (VIII, p. 155, Proposition 5, d)). This proves assertion b).

The tensor product of a separable commutative algebra and a reduced commutative algebra is a reduced ring (V, §15, No. 2, p. 120, Proposition 5). The algebras $Z_1$ and $Z_2$ are commutative and semisimple, hence reduced. If one of the algebras $Z_1$ and $Z_2$ is separable, then the algebra Z is reduced; it is therefore without radical by Theorem 3 of VIII, p. 215, and we have $\mathfrak{R}(A) =\mathfrak{R}(Z)A = 0$. This is, in particular, the case when the field K is perfect because every reduced commutative algebra over a perfect field is separable (V, §15, No. 5, p. 125, Theorem 3).