# Maximal ideals and maximal subspaces of normed algebras

This is a kind of "prove or give a counter-example" question, and I'm having some difficults with it: By a maximal ideal $I$ of an algebra $A$, we mean an ideal $I\neq A$ which is not properly contained in any other proper ideal of $A$ (possibly, $I=\left\{0\right\}$. Let's write $\mathbf{1}$ for the unit of $A$ (if it exists).

Question. Let $A$ be a unital commutative Banach Algebra and $I$ a maximal ideal of $A$. Prove that $I$ is a maximal subspace of $A$. Is this result still valid if $A$ is not Banach or commutative or unital?

The first part is pretty easy: Maximal ideals are of the form $\ker\tau$ for some character $\tau:A\to\mathbb{C}$, and then $A=\ker\tau+\mathbb{C}\mathbf{1}$, so $\ker\tau$ is a maximal subspace.

If $A$ is not commutative, we have a counter-example: The algebra $M_2(\mathbb{C})$ is simple, so the unique maximal ideal of $M_2(\mathbb{C})$ is $\left\{0\right\}$, which is not a maximal subspace.

I couldn't solve the rest of the question: If $A$ is Banach and commutative but non-unital, maybe I could consider its unitization and associate maximal ideals of $A$ and $\widetilde{A}$, but I couldn't do that.

The only example of commutative, unital, non-Banach normed algebra I can think of is $\mathbb{C}[x]$, the polynomial algebra in one variable (with any of the usual norms), but the only maximal ideals of $\mathbb{C}[x]$ are of the form $p(x)\mathbb{C}[x]$ with $\operatorname{degree}(p)=1$, and these are maximal subspaces.

Any hints are appreciated. Thank you.

• Natural examples of commutative unital non banach normed algebras are $RG$ for some commutative ring $R$ and abelian group $G$. For non-unital algebra, take any proper ideal of some algebra. – ougao Jul 9 '14 at 18:29

Of course the result is not valid if we drop commutativity. For instance the algebra of all bounded operators on $\ell_2$ can serve as a counterexample.

Concerning having a unit,

Maximal ideals of commutative Banach algebras have codimension one.

Here is a proof under an extra assumption that a maximal ideal is closed.

According to Gamelin

Th. Gamelin, Uniform Algebras, Prentice-Hall, Englewood Cliffs, N. J., 1969.

p. 22, closed maximal non-modular ideals are precisely those subspaces of codimension 1 which contain the linear span of all $xy$ where $x,y\in A$ (let us denote this set by $A\cdot A$). Indeed, a proper closed ideal $J$ is modular if and only if $A\cdot A\not\subseteq J$. Note that any subspace $M$ containing $A\cdot A$ is an ideal, so $M$ is a closed maximal ideal if and only if it has codimension 1. Since maximal modular ideals are kernels of characters the result follows.

The case where $M$ is not closed. (Edit 19.08.2014) H. Garth Dales, basing on some observations from this discussion managed to answer the question in the remaining case where the maximal ideal is not closed. He kindly uploaded his answer to the arXiv, so that you can find it here.

Non-complete algebras. There exist non-complete, commutative and unital normed algebras with maximal ideals having infinite codimension, though. For instance the algebra of all entire functions on the complex plane endowed with the sup norm on the closed unit disc is such an example. (I learnt this example from H. G. Dales.) To see this, endow this algebra with (non-normable) topology by specifying a family of semi-norms:

$$p_r(f) = \sup_{|z|\leqslant r}|f(z)|\;\;\;(r>0).$$

Now, take the ideal $$J = \{f\mbox{ entire}\colon (\exists{n_0\in \mathbb{N}})(\forall n\geqslant n_0) (f(n)=0)\}.$$ This is a proper ideal not contained in the kernel of any point evaluation. One can prove that kernels of point evaluations are the only maximal ideals which are closed in this topology. Since this algebra is unital, $J$ has to be contained in some dense maximal ideal $M$. Consequently, $M$ has infinite codimension.

• $L^1$ of course has maximal ideals, for example $I=\{ f: \int f = 0\}$. – user138530 Jul 18 '14 at 23:48
• The example of entire functions is correct, but not the choice of ideal. The ideal of functions vanishing at a point has codimension 1. – Jonas Meyer Jul 23 '14 at 14:14
• @Tomek: Yes, the functional is discontinuous, but that doesn't change the fact that the ideal has codimension one. The quotient is $\mathbb C$. Given an entire $f$, $f-f(5)$ is in the ideal $I$ of functions vanishing at $5$, thus $f=(f-f(5))+f(5)$ is in $I+\mathbb C1$. However, although I am not very knowledgeable in them, searching reveals references to infinite codimensional maximal ideals, which necessarily are not kernels of homomorphisms to $\mathbb C$. – Jonas Meyer Jul 23 '14 at 19:55
• @Tomek: The part about closedness and denseless isn't quite relevant I don't think. You were correct to say that the kernels of point evaluations at points outside the disk are dense, but they are also one codimensional, so I don't follow your reasoning in the update. (Unless you are talking about closedness in some other topology, e.g. compact uniform convergence?) – Jonas Meyer Jul 23 '14 at 21:13
• @Tomek: By the way, in case you aren't aware, you can put "@username" in your comment to notify other users in the same comment thread. Otherwise they might miss comments. You get notified of comments on your answers regardless. – Jonas Meyer Jul 23 '14 at 22:31