# If $I$ is a maximal ideal of $R$, why is $R/I$ a field?

If $I$ is a maximal ideal of $R$, why is $R/I$ a field?

I'm trying to use the fact that $I$ is maximal to show that $R/I$ only have ideals $\{0\}$ and $R/I$. Can anyone help me with this method. Many thanks.

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What if $R/I$ has an ideal $J$ which is neither $\{0\}$ nor $R/I$? What can you say about the preimage of $J$ in $R$? – t.b. Jan 10 '12 at 18:32
All you need to show is that every element of $R/I$ has a multiplicative inverse. Hint: if $[a] \neq 0$, then $a \notin I$, so $<a,I> = R$ by the maximality of $I$. – user18063 Jan 10 '12 at 18:34
@t.b.: So you're saying if $\bar{J}=\{x+I:x \in J\}$ is a proper ideal in $R/I$ then $J$ is an ideal in $R$. But then as $I$ is maximal $J\subset I \Rightarrow \bar{J}=0$? – Freeman Jan 10 '12 at 18:40
Dear LHS: Do you know the correspondence between ideals of $R/I$ and ideals of $R$ containing $I$? (This is what t.b. is referring to.) – Pierre-Yves Gaillard Jan 10 '12 at 19:10
@BenjaminLim: Yes I do, thanks so much to you all, sorry I didn't write a personal response before, I was planning on when I wasn't so snowed under, but I wanted to give you credit :) Thanks! – Freeman Jan 15 '12 at 18:40

I believe several people have already told you of some relation between ideals in the ring $R$ and the quotient $R/I$ of a maximal ideal $I$ in $R$. It is called the Lattice Theorem:

If $R$ is a commutative ring and $I$ an ideal of $R$, let $\phi : R \longrightarrow\!\!\!\!\!\!\!\!\to R/I$. Then there is a one-to-one order-preserving correspondence between ideals $\mathfrak{b}$ of $R$ that contain the $I$ and ideals $\mathfrak{\overline{b}}$ of $R/I$. The correspondence is given by $\mathfrak{b}$ being the inverse image of $\mathfrak{\overline{b}}$.

Now you already know that a commutative ring $R$ is a field iff it has no non-trivial ideals. So suppose that $R/I$ is not a field and $I$ a maximal ideal of $R$. Then since $R/I$ is not a field there is a proper ideal $\mathfrak{\overline{a}}$ of $R/I$. But then the inverse image of $\mathfrak{\overline{a}}$ in $R$, say $\mathfrak{a}$ must contain $I$ by the Lattice Theorem above. But then by maximality of $I$ either $\mathfrak{a} = I$ or $\mathfrak{a} = R$. This contradicts $\mathfrak{\bar{a}}$ being a proper ideal of $R/I$ so that $R/I$ is a field.

You should now attempt the following problems to strengthen your understanding of taking quotients of maximal ideals:

All rings are commutative with a unit.

1) Suppose that $R$ is a domain and $p$ a prime element of $R$. Prove that $p$ is also a prime element in $R[x]$. Hint: look at $R/(p)$.

2)Atiyah - Macdonald Problem 1.12 Prove that a local ring has no non-trivial idempotents.

Regards.

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+1 In problem (1), I think you mean "... Prove that $p[x]$ is also prime in $R[x]$ ..." ($p$ is never an ideal of $R[x]$ unless $p=0$ whereas $p[x]$, the set of polynomials with coefficients in $p$, is a prime ideal of $R[x]$). Exercise 2 is an interesting exercise but its solution does not use the language of quotient rings; in fact, it seems to have very little to do with quotient rings. Did you have some solution in mind with relevance to the question? – Amitesh Datta Jan 17 '12 at 1:01
(I think it is a good exercise even if it is not very relevant to the question but since you included it, I thought you had something in mind [regarding its relevance to quotient rings].) – Amitesh Datta Jan 17 '12 at 1:02
@AmiteshDatta For problem (2) you can pass to a quotient. I have to say though that was not the way I did it. – user38268 Jan 17 '12 at 1:06
@AmiteshDatta For (1) I mean to show that if $p$ is a prime in $R$, and $p |fg$ where $f,g \in R[x]$ then $p$ divides all the coefficients of $f$ or $g$. – user38268 Jan 17 '12 at 1:09
Thanks for the clarification! No need to apologize, it is just a minor notational point; in any case, the exercises are interesting and the answer is very informative (I upvoted the answer as I mentioned in my first comment). – Amitesh Datta Jan 17 '12 at 2:10