Irreducibility

Definition A topological space is called reducible if it can be written as the union of two proper nonempty closed subsets. A space that is not reducible is called irreducible.

Remark If X is an irreducible topological space, then every nonempty open subset is both irreducible and dense. Moreover, any two nonempty open subsets of an irreducible topological space have a nonempty intersection. The closure of an irreducible subset of a topological space is also irreducible.

Proposition An algebraic set X is irreducible if and only if the ideal I(X) is prime.

Proof: If I(X) is not prime, then there exist polynomials f, g∉ I(X) with product fg∈ I(X). So, we can write

X = (X∩ Z(f)) ∪ (X∩ Z(g))

as a nontrivial decomposition of X. Conversely, suppose X= X1∪ X2 is a decomposition of X as a union of nonempty proper closed subsets. Then each ideal I(Xi) contains I(X) properly. So, we can choose functions fi∈ I(Xi)\ I(X). Then the product f1f2∈ I(X), so the ideal I(X) is not prime.

Corollary There is a one-to-one correspondence between irreducible algebraic sets in An and prime ideals in k[x1, …, xn].

Theorem Let X be an algebraic set in An. Then there exists a unique collection of irreducible algebraic sets X1, …, Xm such that
  1. X= X1∪ X2∪…∪ Xm, and
  2. Xi⊂ Xj if and only if i=j.

Proof: (Existence) Suppose the theorem does not hold for an algebraic set X. In particular, X must be reducible. Write X=X1∪ Y1. The theorem must also fail for at least one of the pieces in this decomposition; say, X1. Now write X1=X2∪ Y2 and repeat. In this way, one constructs an infinite decreasing chain of algebraic sets

X ⊃ X1 ⊃ X2 ⊃ ….

However, the chain of associated ideals

I(X) ⊂ I(X1) ⊂ I(X2) ⊂ …

must terminate, by Hilbert's Basis Theorem. Therefore, X has a finite decomposition as a union of irreducibles. The condition prohibiting containments can be realized by throwing out any redundant items in the decomposition. (Uniqueness) Suppose

X=∪iXi = ∪jYj

are two irredundant decompositions of X as a union of irreducible algebraic sets. For each i, we can write

Xi = Xi∩ X = Xi∩(∪jYj) = ∪j(Xi∩ Yj).

Because Xi is irreducible, there exists some j with the property that Xi = Xi∩ Yj ⊂ Yj. By interchanging the roles of the decompositions, there exists some i' with

Xi ⊂ Yj ⊂ Xi'.

Because the decompositions are irredundant, this last chain of inclusions must actually consist of equalities, and the result follows.

Corollary Let f∈ k[x1,…,xn]. Write

f=f1n1… fsns

as a product of irreducible factors. Then

Z(f) = Z(f1) ∪…∪ Z(fs)

is the unique decomposition of its zero set into irreducible components. Furthermore,

I(Z(f)) = (f1f2… fs).

Proof: The function f1 must vanish on some component X1 of X. But then X1⊂ Z(f1), which forces X1 = Z(f1)∩ X.

Definition An algebraic set defined by one polynomial in An is called a hypersurface. The irreducible hypersurfaces are in one-to-one correspondence with the irreducible polynomials.

Theorem The product of two irreducible algebraic sets is irreducible.

Proof: Let X and Y be irreducible, and suppose that X× Y = Z1∪ Z2 is a decomposition of the product. For each point x∈ X, we have

Y = x× Y = ((x× Y)∩ Z1)∪ ((x× Y)∩ Z2).

So, at least one of the factors on the right contains this copy of the irreducible set Y. Now define Xi⊂ X to be the set of all x∈ X such that x× Y ⊂ Zi. We have already observed that X=X1∪ X2. However, X is also irreducible and the Xi are closed subsets. Therefore, we must have X=X1 and X× Y = Z1.

Definition An affine algebraic variety over an algebraically closed field k is an irreducible algebriac set in An with the induced topology and the (reduced) induced structure.

Examples

  1. An is a variety.
  2. Among the curves of the first section, only (v), (x), and (xi) are not varieties.
  3. A point is a variety; two points are not.

Comments on this web site should be addressed to the author:

Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210