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.


  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