Prime ideals are a generalization
of

prime numbers to rings more general
than the

integers. They are useful in subjects like

number theory
and

algebraic geometry. For simplicity we will
restrict to the case of commutative rings.

Fix *R* a commutative ring.

**Definition** An ideal *I* of *R*
is called a prime ideal if it is not *R* and whenever
*ab* is in *I*, for some elements
*a,b* of *R*, the either
*a* is in *I* or *b* is in *I*.

__Example__ If the ring is **Z**, the ring
of integers, then for each prime number
*p* the ideal *p***Z** consisting
of all multiples of *p* is a prime ideal.
In fact these are all the prime ideals
of **Z**.

To prove this just observe that in an integral
domain then *y* is in *xR* if and only if
*x* divides *y*. In fact this shows that
in an integral domain *yR* is a prime ideal
if and only if *y* is a prime element of
*R*.

__Example__ The polynomial

*f=y*^{2} - x^{2} - x^{3}

is irreducible in the

polynomial ring

**C***[x,y]* by

Eisenstein's irreducibility criterion. Thus, since
in a

unique factorization domain irreducible elements are
prime we deduce that

*f***C***[x,y]* is a prime ideal.

We can get an equivalent formulation of prime ideals
in terms of quotient rings.
The proof is straight from the definitions.

**Lemma** An ideal *I* of *R* is prime if and only if
the quotient ring *R/I* is an integral domain.

As a corollary of this we can get some important examples.

**Proposition**
A maximal ideal of *R* is prime.

**Proof:** Let *I* be maximal. Then
*R/I* is simple and commutative, hence a field.
In particular, it is an integral domain, so *I* is
a prime ideal by the lemma.

This allows us to give examples of prime ideals that are not
cyclic. By Hilbert's Nullstellensatz we know that the
maximal ideals of **C***[x,y]*
are exactly *(x-a, y-b)*, for *(a,b)* in
**C**^{2}. Thus the proposition tells us that
these ideals are all prime ideals, but it is easy to see that
none of them are cyclic.

Let's talk a little bit more about prime ideals in the polynomial
ring. So fix *k* an algebraically closed field.
We know about the correspondence between closed sets
for the Zariski topology and radical ideals in the polynomial
ring. This begs the question under this correspondence what is the
geometric property of a closed set that makes its ideal prime?
Well actually we are getting a little ahead of ourselves. First:

**Lemma**
A prime ideal is radical.

**Proof:**
Let *I* be a prime ideal. Suppose that
*a*^{n} is in *I* for some *n>0*.
We must show that *a* is in *I*. We proceed by induction
on *n*. In the case *n=1* there is nothing to show.
But

*
a*^{n}=a.a^{n-1}

So by the definition of a prime ideal either

*a* is in

*I* in which case we are done
or the

*n-1*st power of

*a* is. By induction, we're
through.

Back to the polynomial ring

*R=k[x*_{1},..., x_{n}]

If

*I* is a prime ideal and

*I=I(X)*,
for a closed subset of

*k*^{n} in the

Zariski topology
what can we say about

*X*?

**Definition** If *X* is any toplogical space we say that
*X* is reducible if there exist two proper closed subsets
of *X* called *Y,Z* such that *X=Y U Z*.
Otherwise *X* is called **irreducible**.

**Theorem** A closed subset *X* of *k*^{n}
is irreducible if and only if *I(X)* is a prime ideal.

**Proof:** Suppose that *I=I(X)* is prime
but *X=Y U Z*. We have that *Y=Z(I(Y))* and
*Z=Z(I(Z))* and further that *X = Z(I(Y)I(Z))*.
Thus *I=rad(I(Y)I(Z))*. Since *I* is prime it follows
from the definition that either *I(Y)* or *I(Z)*
lies inside *I*. WLOG
let's take the first case. Then apply *Z(-)* to both sides
we get that *Y* contains *X* and hence is not
proper after all.

On the other hand, suppose that *X* is irreducible
and that *ab* lies in *I(X)*.
Let *Y=Z(a)^X* and let
*Z=Z(b)^X*. Then we have
*X=Y U Z*. Since *X* is irrreducible, WLOG
we have *X=Y* which says that *X* is contained in
*Z(a)* It follows that *a* is in *I(X)*,
so *I(X)* is prime.