# Maximal Ideal of a Banach Algebra is closed?

Let $$A$$ be a commutative Banach Algebra with unit and $$M$$ a maximal Ideal of $$A$$. Then $$M$$ is obviously a subspace.
But why must $$M$$ also be closed?

I don't see in which way a maximal ideal is connected to a closed subspace.

The closure $$\operatorname{cl}(M)$$ of $$M$$ is also an ideal, as can be checked easily. As $$M$$ is maximal, we must have either $$\operatorname{cl}(M) = M$$ or $$\operatorname{cl}(M) = A.$$ We need show that the later doesn't happen; specifically, $$1\not\in \operatorname{cl}(M).$$
This follows from the fact that the set of invertible elements in a Banach algebra is open (see Prove that the set of invertible elements in a Banach algebra is open). Note that for a commutative Banach algebra we have that any element $$x$$ is not invertible exactly if it is contained in a maximal ideal.
• Potentially $\operatorname{cl}(M)$ could be $A$. To argue that this doesn't happen one needs to prove that any element sufficiently close to $1$ is invertible, and therefore, not in $M$. Jan 3, 2020 at 16:42