Let $A$ and $B$ be commutative Banach algebra. I have proven that if $A$ and $B$ have identity $e_A$ and $e_B$ respectivly , then $e_A\hat\otimes e_B$ is identity for $A\hat\otimes B$ (the projective tensor product of $A$ with $B$)
I want to find a proof for the converse
Any help will be appreciated.
The answer is NO, if you count $0$ as a unital Banach algebra. Otherwise, it's YES. Let $e \in A \hat\otimes B$ be a unit and take $b_0\in B$ and $g\in B^*$ with $g(b_0)=1$. Let $g\cdot b_0\in B^*$ be defined by $g\cdot b_0\colon x\mapsto g(b_0x)$. Then $e_A:=(\mathrm{id}\otimes (g\cdot b_0))(e) \in A$ is a left unit for $A$. Indeed, $$ae_A=(\mathrm{id}\otimes g)((a\otimes b_0)(e))=(\mathrm{id}\otimes g)(a\otimes b_0)=a$$ holds for every $a\in A$. One can equally prove that $A$ has a right unit and hence a two-sided unit. Likewise for $B$.
