Theorem 3

Let $Z_1$ and $Z_2$ be semisimple commutative algebras over K. The radical of the ring $Z_1\otimes Z_2$ is equal to the set of its nilpotent elements.

Let us first treat the case when $Z_1$ and $Z_2$ are extensions $L_1$ and $L_2$ of the field K. By interchanging $L_1$ and $L_2$ if necessary, we reduce to the case when the transcendent degree of $L_1$ over K is less than that of $L_2$ over K. Choose an algebraic closure Ω of $L_2$; by Corollary 1 of Theorem 5 of V, §14, No. 6, p. 114, we may assume that $L_1$ is a subextension of Ω.

A) Let us first prove that the radical of $L_1\otimes L_2$ is contained in that of $L_1\otimes \Omega$. Set $\mathfrak{a}=\mathfrak{R}(L_1\otimes L_2)(L_1\otimes \Omega )$; it is an ideal of the commutative ring $L_1\otimes \Omega$, and we must prove that $\mathfrak{a}$ is contained in the radical of $L_1\otimes \Omega$. In other words (VIII, p. 156, Theorem 1), we must prove that for $x\in \mathfrak{a}$, the element $1 +x$ is invertible in $L_1\otimes \Omega$. Now, since Ω is an algebraic extension of $L_2$, there exists an extension $L_3$ of $L_2$, of finite degree, such that $x$ belongs to $\mathfrak{R}(L_1\otimes L_2)(L_1\otimes L_3)$. It obviously suffices to prove that $1 +x$ is invertible in $L_1\otimes L_3$. Now, $C = L_1\otimes L_3$ is a finitely generated module over the ring $B = L_1\otimes L_2$. By the corollary of VIII, p. 175, we have $\mathfrak{R}(B)C\subset \mathfrak{R}$(C), so $x$ belongs to the radical of C and $1 +x$ is invertible in C.

B) Let us prove that the radical of $L_1\otimes \Omega$ consists of nilpotent elements. Denote the characteristic exponent of K by $p$ and the relative $p$-radical (that is, purely inseparable) closure of K in Ω (V, §5, No. 2, p. 25) by P; the field P is perfect. Since P is an algebraic extension of K, we have $L_1(P) = L_1[P]$ (V, §3, No. 2, p. 18, Corollary 1). Let $\mathfrak{b}$ be the kernel of the canonical homomorphism from $L_1\otimes P$ to the field $P_1= L_1[P]$. Let $x\in \mathfrak{b}$; there exist elements $y_1, . . . , y_n$ of $L_1$ and elements $z_1, . . . , z_n$ of P such that $x=\sum^n_{i=1}y_i\otimes z_i$ and $\sum_{i=1}^ny_iz_i= 0$. Since P is $p$-radical over K, there exists a power $q$ of $p$ such that $z^q_1, . . . , z_n^q$ belong to K. We then have

So $\mathfrak{b}$ consists of nilpotent elements.

Set $\mathfrak{c}=\mathfrak{b}\otimes_P\Omega$; it is the kernel of the canonical homomorphism from $(L_1\otimes P)\otimes_P\Omega$ to $P_1\otimes_P\Omega$, and it consists of nilpotent elements by the above. Now, Ω is an algebraically closed extension of P, and $P_1$ is a subextension of Ω. Since the field P is perfect, $P_1$ is a separable extension of P (V, §15, No. 5, p. 125, Theorem 3). By Theorem 4 of V, p. 120, the intersection of the maximal ideals of the commutative ring $P_1\otimes_P\Omega$ is reduced to 0. In other words, the ring $P_1\otimes_P\Omega$, which is isomorphic to $((L_1\otimes P)\otimes_P\Omega )/\mathfrak{c}$, is without radical. This proves (VIII, p. 155, Proposition 5) that $\mathfrak{c}$ contains the radical of the ring $(L_1\otimes P)\otimes_P\Omega$. Now, this ring is isomorphic to $L_1\otimes \Omega$, and $\mathfrak{c}$ consists of nilpotent elements. Therefore, the radical of $L_1\otimes \Omega$ consists of nilpotent elements.

C) End of the proof in the specific case. By A) and B), the radical $\mathfrak{r}$ of $L_1\otimes L_2$ is contained in the set $\mathfrak{n}$ of nilpotent elements of this commutative ring. We moreover know that $\mathfrak{n}$ is contained in $\mathfrak{r}$ (VIII, p. 157, Remark 2).

Now consider the general case. Since a semisimple commutative K-algebra is the product of finitely many extensions of the field K (VIII, p. 137, Proposition 3) and the radical of a product of rings is the product of the radicals (VIII, p. 156, Corollary 3), the radical of $Z_1\otimes Z_2$ is the set of nilpotent elements of this ring.