Kernel of Character on Unital Commutative Banach Algebra is Maximal Ideal/Proof 2

Proof
Let $I$ be an ideal of $A$ such that:
 * $\ker \phi \subsetneq I$

We need to show $I = A$.

That is, we need to show:
 * ${\mathbf 1}_A \in I$

Let:
 * $x \in I \setminus \ker \phi$

Then:
 * $\map \phi x \ne 0$

Thus we can define:
 * $\ds \tilde x := {\map \phi x}^{-1} x$

Then:

Thus:
 * ${\mathbf 1}_A = \underbrace{ {\mathbf 1}_A - \tilde x}_{\in \ker \phi} + \underbrace{\paren { {\map \phi x}^{-1} {\mathbf 1}_A} }_{\in A} \underbrace{x}_{\in I} \in I$