0
$\begingroup$

I tried to prove:

Let $A$ be a commutative unital complex Banach algebra. Then there is a bijection between the maximal ideals in $A$ and the set of non-zero homomorphisms $A \to \mathbb C$.

But I can't do the last step of the proof. Here is what I have so far:

Let $\chi : A \to \mathbb C$ be a character. Then $\mathrm{ker}(\chi)$ is a maximal ideal in $A$. Then $\mathfrak m$ is closed and hence $A/ \mathfrak m$ is a Banach algebra in which every non-zero element is invertible. Let $\varphi$ denote the map that takes $\chi$ to $\mathrm{ker}(\chi)$. Let $\psi$ be the map taking a maximal ideal $\mathfrak m \subseteq A$ to the homomorphism $A \to \mathbb C$ corresponding to the Gelfand-Mazur isomorphism $A / \mathfrak m \to \mathbb C$.

Then $\varphi$ is a bijection because $\psi$ is a left and a right inverse of $\varphi$:

$$ \varphi (\psi (\mathfrak m)) = \varphi (f_{\mathfrak m}: A \to \mathbb C ) = \mathfrak m$$

is clear hence $\psi $ is a right inverse of $\varphi$.

$$ \psi (\varphi (\chi )) = \psi (\mathrm{ker}(\chi)) = f_{\mathrm{ker}(\chi)}: A \to \mathbb C$$

where $f$ is the map mapping $a$ to the unique element in $\sigma(a)$ or $0$ if $a$ is in the kernel of $\chi$.

How to show that $f_{\mathrm{ker}(\chi)} = \chi$?

$\endgroup$
  • $\begingroup$ $\chi$ and $f_{\ker(\chi)}$ are two linear functionals vanishing on the $1$-codimensional subspace $\ker(\chi)$. So if there is any $x\notin \ker(\chi)$ with $\chi(x) = f_{\ker(\chi)}(x)$, the two linear functionals are the same. But $e \notin \ker(\chi)$ and $\chi(e) = 1 = f_{\ker(\chi)}(e)$. $\endgroup$ – Daniel Fischer Mar 31 '14 at 18:16
  • $\begingroup$ @DanielFischer Thank you for your comment. Now I realize that the reason I couldn't finish the proof was because I was missing an assumption. The homomorphisms have to be unital. $\endgroup$ – Student Apr 3 '14 at 9:56
  • $\begingroup$ That's something you should prove: a non-zero (algebra) homomorphism is unital. The proof isn't hard. $\endgroup$ – Daniel Fischer Apr 3 '14 at 9:59
  • $\begingroup$ @DanielFischer Is the statement I want to prove the following? ''If $R$ is a commutative unital ring, $S$ is a field and $f: R \to S$ is a ring homomorphism then $f$ is unital. '' $\endgroup$ – Student Apr 4 '14 at 6:46
  • $\begingroup$ Would I have to add the assumption that $f$ is not the zero map? $\endgroup$ – Student Apr 4 '14 at 6:51

Your Answer

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

Browse other questions tagged or ask your own question.