$I$ a maximal ideal in ring $R$ if and only if for all $a \in R\setminus I, \exists r\in R, b \in I$ such that $b+ar=1$. Let $R$ be a commutative Ring with $1\neq 0$. Let $I$ be a proper ideal of $R$. Show that $I$ is maximal $\iff$ for all $a \in R\setminus I$ there are $r\in R$ and $ b \in I$ such that $b+ar=1$.
I'm not sure how to approach proving this.
 A: Hint:
To show $\Rightarrow$, let $I$ maximal and $a \in R \setminus I$.
Then $\langle I, a \rangle \supsetneq I$. Since $I$ is maximal, what must $\langle I, a \rangle$ be? Do you see why this shows $1 = b  +ar$?
To show $\Leftarrow$, we use a slick trick: recall $I$ is maximal if and only if $R/I$ is a field (see here for a proof).
Then for some $a \in R \setminus I$, saying $b+ar = 1$ is saying exactly that $ar = 1 \text{ (mod $I$) }$. Do you see how to show that $R/I$ must be a field, and thus $I$ must be maximal?

I hope this helps ^_^
A: It is helpful to view this as an ideal analog of $\,B\,$ prime, $\,B\nmid A\Rightarrow \gcd(A,B) = 1.\,$ Indeed, in PIDs  we have $\, B\supseteq A\iff B\mid A\,$ and $\,A+B = \gcd(A,B)\,$ (and similarly in Dedekind domains). Keep this analogy in mind when reading the generalization below.
Theorem $\ $ The following are equivalent for an ideal $\,B\,$ in a commutative ring $\,R$.
$(1)\ \ A\not\subseteq B \Rightarrow \ \, A\,+\,B = (1)$
$(2)\ \ \ a\not\in B \Rightarrow\ (a)+B=(1)$
$(3)\ \ \ a\not\in B \Rightarrow\ \  ar\,+\,b\, =\, 1,\,$ for some $\,b\in B.\ r\in R$
$(4)\ \ \,B\,$ is a maximal ideal
Proof $\ \  (1\Rightarrow 2\Rightarrow 3)\  $ clear.
$(3\Rightarrow 4)\ \,$ If $\,A\supsetneq B\,$ then there is $\,a\in A,\,a\not\in B\,$ so $\,1 = ar+b\in A\,$ so $\,A = (1)$
$(4\Rightarrow 1)\ \ $ The ideal $\,A+B\supsetneq A\,$ so $\,A+B = (1)$ by maximality of $B$.
A: It is well-known that if
$I \subset R \tag 1$
is a maximal ideal, then $R/I$ is a field; and thus if
$a \in R \setminus I, \tag 2$
we have
$a \ne 0 \mod I \tag 3$
which implies $a + I$ has an inverse in $R/I$; that is, there is some
$r \in R \tag 4$
with
$ar = 1 \mod I, \tag 5$
or
$ar - 1 \in I; \tag 6$
this of course implies that
$1 - ar = -(ar - 1) \in I, \tag 7$
and thus
$\exists b \in I \; 1 - ar = b, \tag 8$
from which we readily infer that
$ar + b = 1. \tag 9$
Now suppose that for every $a$ as in (2) there are
$r \in R, \; b \in I \tag{10}$
such that
$ar + b = 1 \tag{11}$
or
$ar - 1 = -b \in I, \tag{12}$
whence
$ar = 1 \mod I; \tag{13}$
this implies that
$r \in R \setminus I \tag{14}$
lest
$r = 0 \mod I \tag{15}$
in contradiction to (13), which now implies $r + I$ is the inverse to $a + I$ in $R/I$, which is thus seen to be a field, implying $I$ is maximal in $R$. $OE\Delta$.
