# Does the non-emptiness of the spectrum of an element of a Banach algebra depend on the Axiom of Choice?

One of the most basic results in functional analysis states that the spectrum of any element of a Banach algebra is non-empty. The proof, as most people might have seen, makes use of Liouville's Theorem in complex analysis. However, in order to use Liouville's Theorem at all, one has to transform the resolvent function for the given element into some holomorphic function on some domain of $\mathbb{C}$, and this particular step requires the Hahn-Banach Theorem, which is a weak variant of the Axiom of Choice (AC). Another approach is to first develop the holomorphic functional calculus on Banach spaces, but I think the Krein-Milman Theorem, which is another weak variant of AC, is needed to define such a functional calculus (this is explained in Rudin's Functional Analysis).

Hence, my question is whether or not there exists anything in the literature that says that when you assume the negation of AC, it is possible for the spectrum of an element of a Banach algebra to be empty. More specifically, under the assumption of $\neg$ AC, does there exist a Banach algebra $\mathcal{A}$ and an $a \in \mathcal{A}$ such that $\sigma(a) = \varnothing$?

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Krein-Milman is not equivalent to the axiom of choice. However the assumption Krein-Milman+Ultrafilter Theorem (which is stronger than Hahn-Banach) is equivalent to full choice. See more here: math.stackexchange.com/questions/145271 –  Asaf Karagila Oct 19 '12 at 14:31
Also, could you write in slightly more details what exactly is needed to be found, e.g. "a Banach algebra with an element $T$ such that the set $\{c\in\Bbb C\mid\ldots\}$ is empty"? –  Asaf Karagila Oct 19 '12 at 14:34
Related: math.stackexchange.com/questions/157217/…. The resolvent function of $a$ is a holomorphic function on $\mathbb C\setminus\sigma(a)$ with values in the Banach algebra. –  Jonas Meyer Oct 26 '12 at 4:31