On a proof about Jacobson radical

Question

I have encountered the following proof that shows if an element $a$ of a ring $A$ is in the Jacobson radical $J(A)$, intersection of all maximal ideals, then $1-ab$ must be invertible $\forall b \in A$. Here is the proof (taken from "Commutative algebra" by Andrea Ferretti).

I don't quite get the first line of the proof:

"if $a \in J(A)$ then $1-ab$ is not contained in any maximal ideal, so $<1-ab> = A$".

What is the reasoning behind this?

Answer 1

Suppose for contradiction that $1-ab$ is contained in a maximal ideal $M$. Since $a$ is in the Jacobson radical, $a\in M$, so $ab\in M$, and thus $1 = (1-ab) + ab\in M$, contradicting the fact that $M$ is proper.

Now we know that $1-ab$ is not contained in any maximal ideal. Let $I = (1-ab)$ be the principal ideal generated by $1-ab$. Since every proper ideal is contained in a maximal ideal, $I$ must be improper, so $(1-ab) = A$.

Answer 2

Suppose $1-ab$ is contained in maximal ideal $I$.

Then since $a$ is in each maximal ideal (and so is the multiple $ab$), $(1-ab) + ab = 1$ would be in maximal ideal $I$.