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.