Let $A$ be a complex, unital and commutative Banach-algebra. 


**Question:** Is the maximal spectrum $Max(A)$ of $A$ endowed with the topology induced by the prime spectrum $Spec(A)$ of $A$, Hausdorff? 


**Background:** 
By Gel'fand-Mazur, there is a continuous bijection from the spectrum $Sp(A)$ of $A$ (defined as the set of characters of $A$ endowed with the weak\*-topology) onto $Max(A)$. 
The map sends a character to its kernel.
I would like to know whether this bijection is a homeomorphism and this is equivalent to asking whether $Max(A)$ is Hausdorff.
If $A$ is a $C^\*$-algebra, then the answer is 'yes', because then $A$ is the ring of complex, continuous functions on $Sp(A)$ (by Gel'fand-Naimark) and one can use partition of unity to get the Hausdorff property of $Max(A)$ 



Maximal spectrum of a complex, unital and commutative Banach-algebra
Well, it's the Zariski topology, so why should it be Hausdorff in general? For a specific example, take the Banach algebra $H(D)$ of functions that are holomorphic inside the unit disk and continuous on its closure. Its maximal spectrum is the closed disc, and the closed sets are locally finite in its interior, so it's certainly not Hausdorff.


