If $R$ is a ring and $A$ is a maximal ideal of $R$ then $R/A$ is a field I've to prove one way :
If $R$ is a ring and $A$ is a maximal ideal of $R$ then $R/A$ is a field.
Now suppose that $A$ is maximal and let $b \in R$ but $b \notin A$. It suffices to show that $b+A$ has a multiplicative inverse. (All other properties for a field follow trivially.)  
Consider $B=\{br+ a | r\in R, a \in A\}$. This is an ideal of $R$ that properly contains $A$ .
Since $A$ is maximal, we must have $B=R$. Thus, $1\in B$, say, $1=bc+a'$, where $a'\in A$.
Then $:$
$1 + A = bc + a'+A = bc + A = (b + A)(c + A).$
I can't understand why we chose $br+A$ in proof and not any other set .What does choosing this set ensures...
Also how do we prove that $B$ is an ideal of $R$ properly containing $A$.
Please help....
 A: Here we are given that $A$ is a maximal ideal of $R$ and are required to show that $R/A$ is a field.
We need to show three things for proving a field,that under multiplication operation:
$1.)$ it consists unity.
$2.)$ every element in it has a multiplicative inverse.
$3.)$ it is commutative.
$1.)$ it contains unity:
$(1+M)$ is unity of $R/A$.
$3.)$ it is commutative : as $R$ is commutative so is $R/A$ .
$2.)$ every element in it has multiplicative inverse :
as in comment We have to show that there's $r$ such that $(b+A)(r+A)=1+A.$ That's the same as showing $1-br$ is in $A.$

This task can be accomplished by showing that $bR+A$ is an ideal containing both $A$ and $bR$ then it has to be whole of $R$ ($\because$ it is given that $A$ is maximal ideal) .
Now to do this consider $bR=\{br|r \in R\}$
Clearly $b\in R$ and  $bR+A$ is an ideal  ($\because$ sum of 2 ideals is an ideal containing both ideals.)
since $A$ is maximal ideal of $R$ $\implies A+bR=R \implies 1\in R. \implies 1=a+br $
$\implies 1-a=br \implies 1-a+A=br+A$
$\implies 1+A=(b+A)(r+A)$.

