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.
Maximal Ideal of a Banach Algebra is closed
banach-algebrasideals
Related Solutions
If $A$ is not abelian, then clearly $B\not=A$ for any abelian subalgebra $B\subseteq A$. If $A$ is abelian, then $B=A$ is the only maximal abelian subalgebra of $A$ (and all the claims are trivially true).
If $A$ itself is a unital Banach algebra (as stated in Murphy's exercise), then any maximal abelian subalgebra $B\subseteq A$ is closed and unital (because $B_1=\{\lambda 1 + b;\lambda\in C, b\in B\}$ ($C$ the field of complex numbers) and $\overline{B}$ are abelian subalgebras of $A$ and they contain $B$, so by maximality of $B$ we get $B=\overline{B}=B_1$).
We have $\sigma_A(b)\subset\sigma_B(b)$ for all $b\in B$, because if $\lambda 1-b$ has an inverse in $B$, then it also has an inverse in $A$ (so if it doesn't have an inverse in $A$, then it can't have an inverse in $B$, either).
But note that the converse is true also: if for an element $b\in B$, $\lambda 1-b$ has an inverse in $A$, say $c$, then $c$ belongs to $B$. This is because the relation $(\lambda 1-b)b'=b'(\lambda 1-b)$ (satisfied for all $b,b'\in B$ because $B$ is abelian) implies $cb'=b'c$ for all $b'\in B$. Then $B_c=\{p(c)b';p(c)$ a polynomial in $c, b'\in B\}$ is abelian, contains $B$, so since $B$ is maximal abelian we have $B_c=B$ and $c\in B$, i.e. $\lambda 1-b$ has an inverse in $B$. Therefore $\sigma_B(b)\subset\sigma_A(b)$.
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.
Best Answer
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.