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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

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4 Answers 4

up vote 5 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.

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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

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

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@JyrkiLahtonen Haha thanks.. silly me. But I accepted the other guy to be nice. –  Freeman Sep 29 '11 at 11:06
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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

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.

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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 at 20:51

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 three proofs: first the common proof for integers using Bezout's Identity for the gcd; second, the translation in terms of gcds; third, the ideal translation.

$ \begin{eqnarray} Ax\!+\!ay&=&1,\ \ A &\mid&\, ab&\Rightarrow&\ A&\mid&b. &&{\bf Proof}\!:\,\ A&\mid& Ab,ab\,&\Rightarrow&\ A\,\mid Abx\!+\!aby &=& (Ax\!+\!ay) b\! &=& b\\ (A,\ \ a)&=&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)&=&1,\ A&\!\!\supseteq\!& (ab)\, &\Rightarrow&\, A&\!\supseteq\!&(b). &&{\bf Proof}\!:\,\ A&\!\supseteq\!& Ab,ab \,&\Rightarrow&\, A\supseteq Ab\!+\!(ab) &=&(A\!+\!(a))b &=&\! (b) \end{eqnarray}$

The final ideal form is precisely the same proof as in Brian's answer. Notice that the proof works more generally for any comaximal ideals $\,A,\ (a),\,$ i.e. $\,A+(a)= 1.\,$ 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, e.g. the use of the gcd distributive law then becomes the use of the distributive law for the ring of integers.

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