The well-known Banach–Mazur theorem says that $C([0, 1])$ is a universal separable Banach space, in the sense that if $X$ is any separable Banach space then there is a map $f : X \to C([0, 1])$ which is both linear and isometric. Note that $C([0, 1])$ also has the structure of a Banach algebra.
My question is this: Is $C([0, 1])$ universal for separable commutative Banach algebras? Of course a separable Banach algebra is a separable Banach space, so there is a linear isometry into $C([0, 1])$, but I'm asking if that map can also be taken to preserve the multiplication operation.
If $C([0, 1])$ is not universal for separable commutative Banach algebras, does there exist such a universal object? I'm interested mostly in ZFC results, but would also not mind hearing consistent answers (especially if they are consistent with $\neg CH$).
The short answer to the first question is no. A simple counterexample is $\mathbb{C}^2$, or indeed any Banach algebra containing a nontrivial idempotent, since $C([0,1])$ contains no nontrivial idempotents.
Indeed if a universal algebra of the sort you ask about exists, it cannot be semisimple, and hence cannot be of the form $C(X)$. For if $U$ is such a universal algebra and $f\colon A\to U$ is a homomorphism and $\phi$ is a character on $U$, then $\phi\circ f$ is a character on $A$. If $x\in A$ and $f(x)\ne0$, you could pick $\phi$ so that $\phi(f(x))\ne0$, so $x$ is not in the Jacobson ideal of $A$.