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 pZ 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=y2 - x2 - x3
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 fC[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 C2. 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 an 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
So by the definition of a prime ideal either a is in I in which case we are done or the n-1st power of a is. By induction, we're through.

Back to the polynomial ring

R=k[x1,..., xn]
If I is a prime ideal and I=I(X), for a closed subset of kn 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 kn 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.

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