Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Could anyone explain to me why maximal ideals are prime?

I'm approaching it like this, let $R$ be a commutative ring with $1$ and $A$ be a maximal ideal. Let $a,b\in R:ab\in A$

I'm trying to construct a ideal $B$ such that $A\subset B \neq A$ As this would be a contradiction. An alternative idea I had was to prove that $R/A$ is an integral domain, but this reduces to the same problem.

EDIT: Ergh.. just realized that I've learnt a theorem that states is $A$ is a maximal ideal then $R/A$ is a field

share|cite|improve this question

5 Answers 5

up vote 7 down vote accepted

Let $A$ be a maximal ideal. Then $R/A$ contains no proper ideals, by the correspondence theorem. Indeed, $R/A$ is a field (assuming that $R$ contains an identity).

Thm: R/A is a field.

Proof: Let $i+A\in R/A$ such that $i+A\neq 0+A$. We want to prove that $i+A$ is a unit. Then set $B=A+Ra=\{a+ri: a\in A, r\in R\}$.

Now, you (yourself!) need to prove that $B$ is an ideal, and that $A\subset B$ properly. So, since $A$ is maximal this means that $B=R$.

As $B=R$ we have that $1\in B$, so $(1+ri)+A=ri+A=(r+A)(i+A)$, and so $i+A$ is a unit, as required.

share|cite|improve this answer
Thanks! That's a much neater proof than the one I have here.. will be making a note of your answer for writing up in my notes! :) – Freeman Sep 29 '11 at 11:06

For a completely different approach: An ideal is prime if and only if it is maximal with respect to the exclusion of a nonempty multiplicatively closed subset. (This theorem is extremely useful in commutative ring theory.) By definition, maximal ideals are maximal with respect to the exclusion of {1}.

For the proof of the nontrivial direction of that theorem, let $P$ be an ideal maximal with respect to the exclusion of a nonempty multiplicatively closed subset $S$. Then $P$ is proper. Pick $a,b \notin P$. Since $(P + (a))(P + (b)) \subseteq P + (ab)$ contains an element of $S$, we conclude that $ab \notin P$.

share|cite|improve this answer

Below I show that the proof is an ideal-theoretic form of a well-known proof of $ $ Euclid's Lemma $ $ for integers. To highlight the analogy I give four proofs: first the common proof for integers using Bezout's Identity for the gcd; second, a translation in terms of gcds, then ideal translations.

Euclid's Lemma in Bezout form, gcd form and ideal forms

$\smash[t]{\begin{align}\\ \\ Ax\!+\!ay=&\,\color{#c00}1,\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\,\ A\ \mid\ Ab,ab\, \Rightarrow\, A\,\mid Abx\!\!+\!aby\! =\, (\!\overbrace{Ax\!+\!ay}^{\large\color{#c00} 1}\!) b = b\\ (A,\ \ \ a)=&\,\color{#c00}1,\,\ A\ \mid\ ab\ \ \ \Rightarrow\, A\ \mid\ b.\ \ \ {\bf Proof}\!:\,\ A\ \mid\ Ab,ab\, \Rightarrow\, A\,\mid (Ab,\ \ ab) = (A,\ \ \ a)\ \ b =\, b\\ A\!+\!(a)=&\,\color{#c00}1,\,\ A\supseteq\! (ab)\, \Rightarrow\, A \supseteq\! (b).\: {\bf Proof}\!:\ A\, \supseteq Ab,ab \,\Rightarrow A\supseteq Ab\!+\!(ab)\! =(A\!+\!(a))b =\! (b)\\ A +{\cal A}\ =&\,\color{#c00}1,\,\ A\supseteq {\cal A B}\, \Rightarrow\, A \supseteq\, {\cal B}.\:\ {\bf Proof}\!:\ A\, \supseteq\! A{\cal B},\!{\cal AB}\!\Rightarrow A\supseteq A{\cal B}\!+\!\!{\cal AB} =(A+{\cal A}){\cal B} = {\cal B} \end{align}}$

The third ideal form is precisely the same proof as in Brian's answer. The fourth form shows that the proof works more generally for coprime (i.e. comaximal) ideals $\ A,\, {\cal A},\ $ i.e. $\ A+{\cal A}= 1. $ In the second proof for integers, we can read $\,(A,a)\,$ either as a gcd or an ideal. If read as a gcd then the proof uses the universal property of the gcd and the gcd distributive law. In the first proof the gcd arithmetic is traded off for integer arithmetic, so the use of the gcd distributive law then becomes the use of the distributive law in the ring of integers.

share|cite|improve this answer

Here’s a proof that doesn’t involve the quotient $R/A$.

Suppose that $A$ is not prime; then there are $a,b\in R\setminus A$ such that $ab\in A$. Let $B$ be the ideal generated by $A \cup \{a\}$; $B = \{x+ar: x\in A\text{ and }r\in R\}$. Clearly $A \subsetneq B$, so $B = R$, $1_R \in B$, and hence $1_R = x + ar$ for some $x\in A$ and $r\in R$. Then $$b = b1_R = b(x+ar) = bx + bar.$$ But $bx \in bA \subseteq RA = A$, and $bar \in Ar \subseteq AR = A$, so $b \in A$. This contradiction shows that $A$ is prime.

share|cite|improve this answer
Thanks for showing me this, good to learn some less standard proofs – Freeman Sep 29 '11 at 11:15
@Freeman The above proof is simply an ideal-theoretic translation of a well-known proof of Euclid's Lemma for integers - see my answer – Bill Dubuque Feb 25 '14 at 20:51

$A$ is an maximal ideal $\Rightarrow$ $R/A$ is a field $\Rightarrow$ $R/A$ is an integral domain $\Rightarrow$ $A$ is prime

share|cite|improve this answer
@JyrkiLahtonen Haha thanks.. silly me. But I accepted the other guy to be nice. – Freeman Sep 29 '11 at 11:06
I suggested that you accept your own answer, 'cause I didn't see right away that Swlabr had already answered (I should be used to the delays by now). I deleted that comment as I felt that it was a tad rude at that point. I'm upvoting all the correct answers. – Jyrki Lahtonen Sep 29 '11 at 11:57

protected by user26857 Sep 5 at 7:32

Thank you for your interest in this question. Because it has attracted low-quality answers, posting an answer now requires 10 reputation on this site.

Would you like to answer one of these unanswered questions instead?

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