Must an abstract $C^*$-algebra that is an integral domain be a field? I'd like to understand a little bit better why the maximal (as opposed to prime) spectrum is the appropriate notion of spectrum for the theory of $C^*$-algebras.
The canonical answer is that $C^*$-algebras have no non-maximal prime ideals: every $C^*$-algebra integral domain is a division ring.  In the commutative case, this is discussed here and here, but both answers use the Gelfand transform in an essential way.  But I'm trying to use that fact to motivate the Gelfand transform in my notes, so this seems a little circular.
Is there an abstract, "algebraic" proof that every $C^*$ integral domain is a division ring?
Some notes:

*

*In general, the class of rings for which maximal spectra suffice to characterize the topology of the prime spectrum is known as Gelfand rings.  But I don't know how to prove that a Gelfand integral domain is a field, or even that a $C^*$-algebra is Gelfand (!!).   (I'm about to ask those questions on MSE too, and will link them when I do.)

*My guess is that one should try to mimic the proof that every finite integral domain is a field.  Note that $m_a=(x\mapsto ax)$ is a injective linear operator on $A$.  Showing that $m_a$ is surjective should be possible by some Fredholm theory, whence the claim.

 A: No; as written, your statement is false.  The correct statement is:

A prime ideal of a $C^*$-algebra is maximal iff it is closed.

In particular, call a $x\in A$ an approximate divisor of $0$ if there exists $\{y_n\}_n$ such that no $y_n=0$ and $\lim_{n\to\infty}{xy_n}=0$.  It is straightforward to see that a prime ideal $P\subseteq A$ is closed iff $A/P$ lacks approximate divisors of $0$.  But then a theorem of Kaplansky (proof below) states that any normed Banach algebra without approximate divisors of $0$ is a division ring.
Kaplansky's proof (in the $C^*$ case)
First, note that we can in fact prove more: the Gelfand-Mazur theorem states that the only normed division algebras over $\mathbb{R}$ are $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$.  $C^*$ algebras are already over $\mathbb{C}$, so we can rule out $\mathbb{R}$ and $\mathbb{H}$.  Thus it suffices to give a proof that $A\cong\mathbb{C}$.
Fix $x\in A$.  $\mathrm{Spec}{(x)}$ is compact and nonempty, so there exists $a+bi\in\partial\mathrm{Spec}{(x)}$.  Swapping $x-(a+bi)$ for $x$, we may assume $a+bi=0$ w/oLoG.
$x$ is not invertible, whilst a limit of invertible elements — say, $\{u_n\}_n$.  If $\{u_n^{-1}\}_n$ were bounded, then $xu_n^{-1}-1=(x-u_n)u_n^{-1}\to0$.  Passing to a weak-*-convergent subsequence of $\{u_n^{-1}\}_n$, we would find an (impossible) inverse of $x$.
Thus $\{u_n^{-1}\}$ is not bounded, and (by passing to a subsequence) increasing in norm.  But $\frac{u_n^{-1}}{\|u_n^{-1}\|}$ is still bounded, so $$0=\lim_{n\to\infty}{(x-u_n)\left(\frac{u_n^{-1}}{\|u_n^{-1}\|}\right)}=\lim_{n\to\infty}{\frac{xu_n^{-1}+1}{\|u_n^{-1}\|}}=\lim_{n\to\infty}{x\cdot\frac{u_n^{-1}}{\|u_n^{-1}\|}}$$  That is, $x$ is an approximate divisor of $0$.
But $A$ lacks approximate divisors of $0$.  So $x=0\in\mathbb{C}$, whence the claim.
