TITLE: How to show $a$ is an element of every maximal ideal of ring $R$ iff $1-ab$ is a unit for all $b \in R$?
QUESTION [5 upvotes]: Let $R$ be a commutative unitary ring. The task is to prove the statement which says that for an element $a\in R$ stands: $a$ is an element of every maximal ideal of $R$ iff $1-ab$ is a unit for all $b \in R$.
My effort
Further I suppose ring is non-trivial, otherwise situation is clear.
For an arbitrary $b \in R$ we consider principal ideal $(b)$ and find a maximal ideal $\mathfrak{m}_b\supset (b)$ (according to Zorn's Lemma), so we have $a\in \mathfrak{m}_b, b \in \mathfrak{m}_b\Rightarrow ab \in \mathfrak{m}_b$. $\mathfrak{m}_b$ is maximal, hence particularly prime, so $1\notin \mathfrak{m}_b$ and $(1-ab)+\mathfrak{m}_b\neq0+\mathfrak{m}_b\in R/\mathfrak{m}_b$. $R/\mathfrak{m}_b$ is a field, so we find $p \in R : p(1-ab)+\mathfrak{m}_b=1+\mathfrak{m}_b$. But seemingly this is not enough to find inverse of $1-ab$ in $R$.
In other direction I can only show that: $1-ab$  is invertible $\Rightarrow ab\neq1\Rightarrow a$ is not a unit and hence is an element of a maximal ideal. But I can't figure out how to show that $a$ is in every maximal ideal.
I give up because I tried to solve this too long, the statement seems pretty natural for me, but I don't see something obvious so I would love it if someone gives me an advice. 
Thanks in advance!

REPLY [7 votes]: An idea:
Assume $$\;a\in \bigcap_{M\le R\;max.}M\;$$
but nevertheless for some $\;b\in R\; $ we have that
$$1-ab\;\;\text{not a unit}\implies \text{the ideal}\;\langle 1-ab\rangle\;\;\text{is contained in some maximal ideal}\;\;M\le R$$
But $\;a\in M\implies 1\in M\iff M=R\;$ , and this cannot be.
Now suppose that for some maximal ideal $\;M\le R\;$ we have that $\;a\notin M\;$, so for maximality
$$\langle a\,,\,M\rangle =R\implies\;\exists \;b\in R\;,\;m\in M\;\;s.t.\;\; ab+m=1\implies$$
$$1-ab=m\in M\implies\;\;1-ab\;\;\text{cannot be a unit}$$