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:

2**Z**, 3**Z**, 5**Z**, 7**Z**, ...

that is each ideal

*p***Z**, for a

prime *p*.
Recall from

ideal what this notation means:

*p***Z**
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.