Bijection between $\Delta(A)$ and $\mathrm{Max}(A)$

Question

For a commutative Banach algebra, the mapping $$\varphi \to \ker\varphi$$ is a bijection between $\Delta(A)$ and $\mathrm{Max}(A),$ where $\Delta(A)$ is the set of all multiplicative non-zero linear functionals on $A$ and $\mathrm{Max}(A)$ is the set of all maximal modular ideals of $A.$

I've seen the proof of above given in A Course in Commutative Banach Algebras by Kaniuth. However, I was wondering if there is a way to prove injectivity without using the fact that $\ker \varphi$ has codimension $1$ in $A?$

Answer 1

Let $\ker \varphi_1 = \ker \varphi_2=I,$ say. Let $x \in A$. We need to show that $\varphi_1(x)=\varphi_2(x).$

Case $1: \varphi_1(x)=0$. Then we're done.

Case $2: \varphi_1(x) \neq 0$. Let $$y= \dfrac{x}{\varphi_1(x)}.$$ Then $\varphi_1(y)=1.$ Hence, $y$ is identity modulo $I$. Thus, $x-yx \in I=\ker \varphi_2.$ So, \begin{align*} \varphi_2(x-yx)&=0\\ \varphi_2(x)&=\varphi_2(y)\varphi_2(x)\\ \varphi_2(y)&=1. \end{align*} Thus, we have $$\varphi_1(y)=\varphi_2(y).$$ From this it follows that $\varphi_1(x)=\varphi_2(x).$

Answer 2

Suppose $\ker \varphi_1=\ker\varphi_2\neq 0$, then $\varphi_1(1)=\varphi_2(1)=1$, hence $x-\varphi_1(x)\in \ker\varphi_1=\ker\varphi_2.$ Therefore $\varphi_2(x-\varphi_1(x))=0 $, i.e. $\varphi_1(x)=\varphi_2(x)\forall x\in X$.