Where does Gelfand Theory fail for non-commutative algebras.

I'm trying to get my head around Gelfand theory, and I can't seem to find the subtleties between commutative and non-commutative algebras. Why is there not a one-to-one correspondence between maximal ideals of a non-commutative algebra and the character homomorphisms from the algebra to the complex plane? Doesn't the Gelfand Mazur theorem apply to these algebras to, so if $\mathfrak{a}$ is maximal, the map

$$A \to A/\mathfrak{a} \cong \mathbf{C}$$

is a homomorphism with kernel $\mathfrak{a}$. Conversely, if $\phi: A \to \mathbf{C}$ is a homomorphism, then $\phi$ is surjective, so if $\mathfrak{a} = \ker(\phi)$, $\tilde{\phi}: A/\mathfrak{a} \to \mathbf{C}$ is an isomorphism, hence $A/\mathfrak{a}$ is a field, so $\mathfrak{a}$ is maximal. What's going wrong here?

For a noncommutative algebra $A$ with maximal ideal $M$, $A/M$ need not be a field, so it is not always $\Bbb C$. It is just a simple ring, and those can get pretty unusual.
• What's wrong with this argument? Let $[N]$ be the image of $N \in A$ under the quotient of $A$ to $A/\mathfrak{a}$. If $[N] \neq 0$, then $N \not \in \mathfrak{a}$, so $(N) + \mathfrak{a}$ is an ideal properly containing $\mathfrak{a}$, so $(N) + \mathfrak{a} = A$. This means there is $K$ such that $KN + L = 1$, for $L \in \mathfrak{a}$, which implies $[KN] = 1$. Similarily, there is $R$ such that $[NR] = 1$. But then $[K] = [R] = [N]^{-1}$. – Jacob Denson Mar 7 '16 at 4:23
• @JacobDenson There does not necessarily exist $K,L$ such that $KN+L=1$. The only thing you can conclude is that there exists $r_i,s_i$ and $L$ such that $\sum r_iNs_i +L=1$, where that summation is finite. – rschwieb Mar 7 '16 at 4:29
• @JacobDenson besides, you should also be aware of counterexamples like $M_n(\Bbb C)$ for any integer $n>1$. – rschwieb Mar 7 '16 at 4:32