Semi-Simple Com. Banach Algebras A and B - A⊗B is Semi-Simple
For Semi-Simple Commutative Banach Algebras A and B - A⊗B is Semi-Simple
| Proposition 1: Let $\mathfrak{A}$ and $\mathfrak{B}$ be semi-simple commutative Banach algebras over $\mathbb{C}$. Then $\mathfrak{A} \otimes \mathfrak{B}$ is semi-simple. |
- Proof: Let $f \in \Phi_{\mathfrak{A}}$ and let $g \in \Phi_{\mathfrak{B}}$. By the theorem on The Existence of a Linear Map σ on X⊗Y to Z that Matches a Bilinear Map on X×Y to Z page there exists a linear functional $f \square g$ on $\Phi_{\mathfrak{A}} \otimes \Phi_{\mathfrak{B}}$ such that for all $x \in \mathfrak{A}$ and for all $y \in \mathfrak{B}$:
\begin{align} \quad [f \square g](x \otimes y) = f(x)g(y) \end{align}
- To show that $\mathfrak{A} \otimes \mathfrak{B}$ is semi-simple we need to show that:
\begin{align} \quad \mathrm{Rad}(\mathfrak{A} \otimes \mathfrak{B}) = \bigcap_{T \in \Phi_{\mathfrak{A} \otimes \mathfrak{B}}} \ker (T) = \{ 0 \} \end{align}
- Since $\mathfrak{A}$ and $\mathfrak{B}$ are commutative Banach algebras over $\mathbb{C}$, we have by the For Commutative Banach Algebras A and B over C - ΦA×ΦB is Homeomorphic to ΦA⊗pB page we have that that $\Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}}$ is homeomorphic to $\Phi_{\mathfrak{A}\otimes_p \mathfrak{B}}$. In particular, every multiplicative linear functional on $\Phi_{\mathfrak{A} \otimes_p \mathfrak{B}}$ corresponds to a multiplicative linear functional on $\Phi_{\mathfrak{A}} \times \Phi_{\mathfrak{B}}$.
- Let $u \in \mathfrak{A} \otimes \mathfrak{B}$ and suppose that $[f \square g](u) = 0$ for all $f \in \Phi_{\mathfrak{A}}$ and for all $g \in \Phi_{\mathfrak{B}}$. We aim to show that $u = 0$ to conclude that $\mathrm{Rad}(\mathfrak{A} \otimes \mathfrak{B}) = \{ 0 \}$.
- By one of the theorems on the Basic Theorems Regarding the Algebraic Tensor Product of Two Normed Spaces page, we can assume that $u = \sum_{i=1}^{m} x_i \otimes y_i$ where $\{ x_1, x_2, ..., x_m \}$ is linearly independent in $\mathfrak{A}$ and $\{ y_1, y_2, ..., y_m \}$ is linearly independent in $\mathfrak{B}$.
- Now since $\mathfrak{A}$ and $\mathfrak{B}$ are semisimple commutative Banach algebras we have that $\mathrm{Rad}(\mathfrak{A}) = \{ 0 \}$ and $\mathrm{Rad}(\mathfrak{B}) = \{ 0 \}$, i.e., $\bigcap_{f \in \Phi_{\mathfrak{A}}} \ker (f) = \{ 0 \}$ and $\bigcap_{g \in \Phi_{\mathfrak{B}}} \ker (g) = \{ 0 \}$. By the assumption that $[f \square g](u) = 0$ for all $f \in \Phi_{\mathfrak{A}}$ and all $g \in \Phi_{\mathfrak{B}}$ we have that:
\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 $g \in \Phi_{\mathfrak{B}}$, we have that $\sum_{i=1}^{m} f(x_i)y_i \in \mathrm{Rad}(\mathfrak{B}) = \{ 0 \}$, so:
\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 $f \in \Phi_{\mathfrak{A}}$ and for each $1 \leq i \leq m$ that $f(x_i) = 0$. So for each $1 \leq i \leq m$, $x_i \in \mathrm{Rad}(\mathfrak{A}) = \{ 0 \}$, so each $x_i = 0$.
- Similarly we have that:
\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 $f \in \Phi_{\mathfrak{A}}$ we have that $\sum_{i=1}^{m} g(y_i)x_i \in \mathrm{Rad}(\mathfrak{A}) = \{ 0 \}$, so:
\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 $g \in \Phi_{\mathfrak{B}}$ and for each $1 \leq i \leq m$ that $g(y_i) = 0$. So for each $1 \leq i \leq m$, $y_i \in \mathrm{Rad}(\mathfrak{B}) = \{ 0 \}$, so each $y_i = 0$.
- Thus we have that $u = 0$, and so $\mathrm{Rad}(\mathfrak{A} \otimes \mathfrak{B}) = \{ 0 \}$, i.e., $\mathfrak{A} \otimes \mathfrak{B}$ is semi-simple. $\blacksquare$
