## Maximal spectrum of a complex, unital and commutative Banach-algebra

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 $Max(A)$

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 Could you remind us of the definition of the topology on Spec(A)? – Yemon Choi Nov 3 at 17:35 Is it the same as what analysts call the hull-kernel topology? If so, then the answer to your question is negative in general – Yemon Choi Nov 3 at 17:37 @Yemon The topology on $Spec(A)$ has the sets $D(f)=\{p\vert f\notin p\}$, where $f\in A$, as open (sub-)basis. I think it is indeed called the hull-kernel topology in functional analysis. If you have an example or a reference for the negative answer, could you post it as an answer? Thanks. – Marcus Nov 3 at 18:46

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.