 Solution

The Gelfand-Mazur theorem says that a nonzero complex unital Banach algebra in which every nonzero element is invertible is the complex field itself: $a\mapsto \lambda$ where $a=\lambda 1$. This is an isometric isomorphism under the usual normalization $\Vert 1\Vert =1$; without that normalization its norm is simply $\Vert \lambda 1\Vert =|\lambda |\Vert 1\Vert$. We first prove the required spectral and resolvent facts rather than assume them.

For any element $a$ of a complex unital Banach algebra $A$, define its spectrum of an element byIf $\Vert b\Vert <1$, the partial sums of ${\sum }_{n\ge 0}{b}^n$ converge in the complete algebra. Multiplying a partial sum by $1-b$ on either side gives $1-b^{N+1}\to 1$, proving the Neumann series inverse. For $|\lambda |>\Vert a\Vert$ this givesThus the spectrum is bounded. Under $\Vert 1\Vert =1$, $\Vert R(\lambda ,a)\Vert \le (|\lambda |-\Vert a\Vert {)}^{-1}$; for an unnormalized identity the same estimate with numerator $\Vert 1\Vert$ suffices. Also $\lambda R(\lambda ,a)\to 1$ as $|\lambda |\to \infty$.

If ${\lambda }_0$ is outside the spectrum and $R_0=R({\lambda }_0,a)$, then for $|h|\Vert R_0\Vert <1$,This follows by factoring $({\lambda }_0+h)1-a=({\lambda }_{0}1-a)(1+h{R}_0)$ and applying the just-proved series. The series converges locally in norm, proving that the resolvent set is open, the spectrum is closed, and the resolvent of an element is holomorphic. Subtracting inverse equations also gives the resolvent identityconsistent with the derivative $R^{\prime }(\lambda ,a)=-R(\lambda ,a{)}^2$ from the local series.

The spectrum is nonempty. If it were empty, the resolvent would be an entire algebra-valued function tending to zero at infinity. Choose a bounded linear functional $\ell :A\to \mathbb{C}$ with $\ell (1)=1$, extending the functional on $\mathbb{C}1$ by the complex Hahn-Banach theorem. Then $\ell (R(\lambda ,a))$ is an entire function, bounded outside a disk by the Neumann estimate and bounded inside by continuity and compactness. The Liouville theorem makes it constant, and its decay makes that constant zero. But $\lambda \ell (R(\lambda ,a))\to \ell (1)=1$, a contradiction. We have proved that every spectrum is a nonempty compact subset of the disk $|\lambda |\le \Vert a\Vert$.

Now assume every nonzero element of $A$ is invertible. For any $a$, choose $\lambda \in {\sigma }_A(a)$. The noninvertible element $a-\lambda 1$ must be zero. Thus every element is a scalar multiple of the identity, uniquely because $1\ne 0$, andNo commutativity assumption was needed for this part; it is forced by the conclusion. Complex scalars are essential to the spectral argument.

For the maximal ideal space and its correspondence with characters of an algebra, now take $A$ to be a commutative complex unital Banach algebra. Every proper ideal lies in an algebraic maximal ideal by Zorn's lemma: a chain's union is an ideal not containing $1$. Every proper ideal also has proper closure. Otherwise it contains an element $m$ with $\Vert 1-m\Vert <1$, which is invertible by the Neumann series, forcing the ideal to contain $1$. A maximal ideal $I$ is therefore closed, since its closure is a proper ideal containing it.

The quotient $A/I$, with its quotient norm, is a complex Banach algebra. Completeness follows from closedness of $I$, and submultiplicativity follows by taking infima over representatives. It is a division algebra: if $a\notin I$, maximality gives $I+Aa=A$, so $ba+i=1$ for some $b\in A$, $i\in I$. Hence $a+I$ is invertible. By the proved Gelfand-Mazur theorem, $A/I\cong \mathbb{C}$, with the identity mapping to $1$. Composing this isomorphism with the quotient map produces a nonzero multiplicative complex linear functional ${\varphi }_I$ with kernel $I$.

Conversely let $\varphi :A\to \mathbb{C}$ be a nonzero multiplicative complex linear functional, without assuming continuity. Nonzeroness and multiplicativity give $\varphi (1)=1$, hence $\varphi (\lambda 1)=\lambda$ and surjectivity onto $\mathbb{C}$. Its kernel is a maximal ideal. If $a-\varphi (a)1$ were invertible, applying $\varphi$ to its product with an inverse would give $0=1$. Therefore $\varphi (a)\in {\sigma }_A(a)$, soThis proves automatic continuity of characters; with normalized identity their operator norm is one. Finally a kernel determines its character uniquely: for every $a$, its class in $A/I=\mathbb{C}1$ is a unique scalar multiple of the unit, and every character with kernel $I$ must take that scalar value. We have obtained the bijectionThe commutativity in this correspondence cannot be dropped: ${\mathrm{Mat}}_2(\mathbb{C})$ has the maximal two-sided ideal $0$, but has no scalar-valued character, since such a character would have zero kernel and give an impossible injective linear map into $\mathbb{C}$.

For use in the next question, the spectrum equals character values in a commutative Banach algebra. We have proved one inclusion. For the other, if $a-\lambda 1$ is noninvertible, its generated ideal is proper, hence lies in a maximal ideal. The corresponding character takes $a$ to $\lambda$. ThusGive ${\mathcal{M}}_A$ the Gelfand topology, namely the weak-star topology of pointwise convergence on $A$ through these characters. It is Hausdorff and compact: the characters lie in the dual unit ball and form a weak-star closed subset, described by the closed equations $\varphi (1)=1$ and $\varphi (ab)=\varphi (a)\varphi (b)$. Apply the Banach-Alaoglu theorem. The Gelfand transform $a\mapsto \hat{a}$, where $\hat{a}(\varphi )=\varphi (a)$, is consequently a contractive unital algebra homomorphism into $C({\mathcal{M}}_A)$; it separates distinct characters. This argument proves compactness and continuity, but does not assert injectivity or isometry for an arbitrary Banach algebra.