If X and Y are Semi-Sim. Com. Ban. Algs. then X⊗Y is Semi-Simple

# If X and Y are Semi-Simple Commutative Banach Algebras then X⊗Y is Semi-Simple

 Proposition 1: Let $X$ and $Y$ be semi-simple commutative Banach algebras over $\mathbf{C}$. Then $X \otimes Y$ is semi-simple.
(1)
\begin{align} \quad [f \square g](x \otimes y) = f(x)g(y) \end{align}
• To show that $X \otimes Y$ is semi-simple we need to show that:
(2)
\begin{align} \quad \mathrm{Rad}(X \otimes Y) = \bigcap_{T \in \Phi_{X \otimes Y}} \ker (T) = \{ 0 \} \end{align}
• Let $u \in X \otimes Y$ and suppose that $[f \square g](u) = 0$ for all $f \in \Phi_X$ and for all $g \in \Phi_Y$. We aim to show that $u = 0$ to conclude that $\mathrm{Rad}(X \otimes Y) = \{ 0 \}$.
• Now since $X$ and $Y$ are semisimple commutative Banach algebras we have that $\mathrm{Rad}(X) = \{ 0 \}$ and $\mathrm{Rad}(Y) = \{ 0 \}$, i.e., $\bigcap_{f \in \Phi_X} \ker (f) = \{ 0 \}$ and $\bigcap_{g \in \Phi_Y} \ker (g) = \{ 0 \}$. By the assumption that $[f \square g](u) = 0$ for all $f \in \Phi_X$ and all $g \in \Phi_Y$ we have that:
(3)
\begin{align} \quad 0 &= [f \square g](u) \\ &= [f \square g] \left ( \sum_{i=1}^{m} x_i \otimes y_i \right ) \\ &= \sum_{i=1}^{m} [f \square g] (x_i \otimes y_i) \\ &= \sum_{i=1}^{m} f(x_i) g(y_i) \\ &= g \left ( \sum_{i=1}^{m} f(x_i)y_i \right ) \end{align}
• Since the above equality holds for all $g \in \Phi_Y$, we have that $\sum_{i=1}^{m} f(x_i)y_i \in \mathrm{Rad}(Y) = \{ 0 \}$, so:
(4)
\begin{align} \quad 0 &= \sum_{i=1}^{m} f(x_i)y_i \\ \end{align}
• But since $\{ y_1, y_2, ..., y_m \}$ is linearly independent, the above equation implies that for each $f \in \Phi_X$ and for each $1 \leq i \leq m$ that $f(x_i) = 0$. So for each $1 \leq i \leq m$, $x_i \in \mathrm{Rad}(X) = \{ 0 \}$, so each $x_i = 0$.
• Similarly we have that:
(5)
\begin{align} \quad [f \square g](u) &= [f \square g] \left ( \sum_{i=1}^{m} x_i \otimes y_i \right ) \\ &= \sum_{i=1}^{m} [f \square g] (x_i \otimes y_i) \\ &= \sum_{i=1}^{m} f(x_i)g(y_i) \\ &= f \left ( \sum_{i=1}^{m} g(y_i)x_i \right ) \end{align}
• Since the above equality holds true for all $f \in \Phi_X$ we have that $\sum_{i=1}^{m} g(y_i)x_i \in \mathrm{Rad}(X) = \{ 0 \}$, so:
(6)
\begin{align} \quad 0 &= \sum_{i=1}^{m} g(y_i)x_i \end{align}
• Since $\{ x_1, x_2, ..., x_m \}$ is linearly independent, the above equation implies that for each $g \in \Phi_Y$ and for each $1 \leq i \leq m$ that $g(y_i) = 0$. So for each $1 \leq i \leq m$, $y_i \in \mathrm{Rad}(Y) = \{ 0 \}$, so each $y_i = 0$.
• Thus we have that $u = 0$, and so $\mathrm{Rad}(X \otimes Y) = \{ 0 \}$, i.e., $X \otimes Y$ is semi-simple. $\blacksquare$