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Start with a ring R. An ideal I of R is called maximal if there exists no ideal J of R which contains I and doesn't equal I except for R.

Let's give some examples (justified below) immediately. If the ring is Z, the ring of integers, then the maximal ideals are:

2Z, 3Z, 5Z, 7Z, ...
that is each ideal pZ, for a prime p. Recall from ideal what this notation means: pZ consists of all integer multiples of p. Note that the ideal {0} is not maximal in this ring. But is in maximal in, say, Q.

From the above you will already begin to suspect that at least for rings like Z maximal ideals are closely related to factorization. This is indeed the case.

Theorem Let R be a commutative integral domain and let p in R. Then pR is maximal implies that p is prime. If R is a principal ideal domain then conversely, p is prime implies that pR is maximal.

Proof: Suppose that pR is maximal and that p divides ab. We must show that p divides either a or b. Suppose WLOG that p does not divide a. It follows that a is not in pR. Thus we have R=aR + pR, since pR is maximal, by assumption. Therefore 1=ar+ps, for some r,s in R. Multiplying both sides by b we get b=(ab)r+p(sb). Now p divides both terms on the right and so it follows that p|b as needed.

Suppose that R is a PID and p is prime. Let x be an element of R that is not in pR. Then p and x have highest common factor 1. It follows that there exist r,s in R such that 1=pr+xs. Immediately then pR+xR=R. Thus pR is maximal.

We conclude with some more general remarks. First, the existence of maximal ideals in any ring.

Lemma If I is an ideal that doesn't equal R then there exists a maximal ideal of R containing I.

Proof: First of all by replacing R by the quotient ring R/I we can reduce to the case when I={0}. So we just have to show that there exists a maximal ideal. Let S be the set of ideals of R which don't equal R. I will show that each totally ordered subset has an upper bound. Then by Zorn's Lemma the result follows. So let T be such a totally ordered subset and let J be the union of all the ideals in T. I claim that J is itself an ideal. This follows because if I take any two elements a,b of J then there exists an ideal in T so that both these elements lie in the ideal. Evidently then a-b and ar,ra lie in J, for any r in R. Furthermore, J is proper because if it contains 1 then so shall one of the ideals in T which is impossible. Hence the result.

Finally, another useful characterization of maximal ideals.

Lemma An ideal I of a ring R is maximal if and only if R/I is a simple ring. If R is commutative, an ideal I is maximal if and only if R/I is a field.