# 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 $A$.

• Our goal is to show that there's $r$ such that $(b+A)(r+A)=1+A$. That's the same as showing 1 is in $br+A$. So that's why we look at $B$. $B$ contains $A$ because you can take $r=0$ in $br+a$. The containment is proper because you can take $r=1$, $a=0$. Now can you show $B$ is an ideal? – Gerry Myerson Oct 8 '14 at 6:23
• Or go this way: Let $B$ be the smallest ideal containing both $A$ and $b$. Since $b\notin A$ it is strictly larger than $A$, hence equals $R$ and contains $1$. Only now try to find a better description of the elements of $B$ and find that it contains exactly the elements of the form $br+a$. – Hagen von Eitzen Oct 8 '14 at 6:46
• Thanks both of you....I got it.. – coool Oct 8 '14 at 8:25
• @coool could you write what you learned into a short answer? – rschwieb Oct 8 '14 at 10:21
• A very important property to keep in mind (which might be introduced in your class later, but it doesn't hurt to start thinking about it now) is that there is a bijection between the ideals of $R$ which contain $A$ and the ideals of $R/A$. (The result then follows trivially: if there is no proper ideal of $R$ which contains $A$, then $R/A$ has no proper non-trivial ideal, which is the definition of a field.) – fkraiem Oct 8 '14 at 12:14

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 $R$ $\implies A+bR=R \implies 1\in R. \implies 1=a+br$
$\implies 1-a=br \implies 1-a+A=br+A$
$\implies 1+A=(b+A)(r+A)$.

• Very nice job writing it up! I only have one nitpick: "That's the same as showing $1$ is in $br+A$." I think what you actually meant is "showing $1-br$ is in $A$", which is correct. That establishes that $1\equiv br$ mod $A$, that is, $b$ has a multiplicative inverse in $R/A$. – rschwieb Oct 8 '14 at 12:43
• @rschwieb thanks.Yes exactly what I meant was showing $1−br$ is in $A$ will imply the existence of multiplicative inverse as you stated.I'll edit it in answer... – coool Oct 8 '14 at 12:59