Maximal Ideals and Maximal Subspaces in normed algebras

[Murtuza Tipu]

Homework Statement

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?

Homework Equations

The first part is pretty easy: Maximal ideals are of the form kerτ for some character τ:A→C, and then A=kerτ+C1, so kerτ is a maximal subspace.

If A is not commutative, we have a counter-example: The algebra M2(C) is simple, so the unique maximal ideal of M2(C) is {0}, which is not a maximal subspace.

The Attempt at a Solution

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 A˜, but I couldn't do that.

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

Any hints are appreciated. Thank you.

 

[mighty2000]

One approach to solving this problem is to use the fact that every non-unital commutative Banach algebra A can be embedded into a unital commutative Banach algebra B. This embedding is given by taking the unitization of A, which is defined as A˜ = A ⊕ C1, where C1 is a copy of the complex numbers with multiplication given by (a, λ)(b, μ) = (ab + aμ + bλ, λμ).

Now, let I be a maximal ideal of A. We want to show that I is a maximal subspace of A. Since A is embedded into B, we can also view I as a subset of B. Since I is a maximal ideal of A, it is also a maximal ideal of B. This means that B/I is a field.

Now, consider any subspace J of B that contains I, i.e. I ⊆ J. We want to show that either J = B or J = I. Since B/I is a field, J/I is a vector space over B/I. Since J/I contains I/I, which is the zero element of B/I, J/I must contain all elements of B/I. This means that J/I = B/I, or equivalently, J = B. Therefore, I is a maximal subspace of B, and hence of A.

This result is still valid if A is not Banach, commutative, or unital. In the non-unital case, we can use the same argument as above, but instead of considering the unitization of A, we consider the unitization of the non-unital algebra. In the non-commutative case, we can use the same argument, but instead of considering maximal ideals, we consider maximal left ideals. In the non-Banach case, we can use the same argument, but instead of considering maximal ideals, we consider maximal subspaces that are also closed under the norm.