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.

  • $\begingroup$ 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 $[K] = [R] = [N]^{-1}$. $\endgroup$ – Jacob Denson Mar 7 '16 at 4:23
  • 1
    $\begingroup$ @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. $\endgroup$ – rschwieb Mar 7 '16 at 4:29
  • $\begingroup$ There we go! I knew focusing on commutative algebra rather than noncommutative would bite me in the ass one day... $\endgroup$ – Jacob Denson Mar 7 '16 at 4:31
  • $\begingroup$ @JacobDenson besides, you should also be aware of counterexamples like $M_n(\Bbb C)$ for any integer $n>1$. $\endgroup$ – rschwieb Mar 7 '16 at 4:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.