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....

  • 1
    $\begingroup$ 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? $\endgroup$ – Gerry Myerson Oct 8 '14 at 6:23
  • $\begingroup$ 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 $br+a$. $\endgroup$ – Hagen von Eitzen Oct 8 '14 at 6:46
  • $\begingroup$ Thanks both of you....I got it.. $\endgroup$ – coool Oct 8 '14 at 8:25
  • 1
    $\begingroup$ @coool could you write what you learned into a short answer? $\endgroup$ – rschwieb Oct 8 '14 at 10:21
  • 1
    $\begingroup$ 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.) $\endgroup$ – 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 $
$\implies 1-a=br \implies 1-a+A=br+A$
$\implies 1+A=(b+A)(r+A)$.

  • $\begingroup$ 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 $R/A$. $\endgroup$ – rschwieb Oct 8 '14 at 12:43
  • $\begingroup$ @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... $\endgroup$ – coool Oct 8 '14 at 12:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.