Irreducible varieties

We can work either with affine varieties or with projective varieties. Thus, if we consider an affine variety X $\subset$ Aⁿ, we say that X is irreducible if it is not the union of two properly smaller nonempty affine varieties. Thus:

if X = Y $\cup$ Z, where Y and Z are nonempty affine varieties,
then either Y = X or Z = X. We have a completely parallel definition in the case of a projective variety X $\subset$ Pⁿ.

Some examples:

A set consisting of one point is obviously irreducible.
A¹ is irreducible Indeed, the proper closed subsets of A¹ are finite sets, while A¹ is infinite. (We are assuming that our field is algebraically closed, and thus infinite.)
By extension, any curve with a rational parametrization is irreducible.
Aⁿ and Pⁿ are irreducible.

PROOF OF 4). We will consider the case of Aⁿ, since the case of Pⁿ is similar.Suppose that Aⁿ = X $\cup$ Y, where X and Y are affine varieties, both nonempty and distinct from Aⁿ. Let f and g be nonzero elements of I(X) and I(Y) respectively, i.e. f and g are nonzero polynomials which vanish identically along X and Y respectively. It follows that

Aⁿ = V(f) $\cup$ V(g), so that Aⁿ = V(fg). Therefore, we can deduce the conclusion from the following:

LEMMA. Let K be an algebraically closed field, and let f(x₁,..., x_n) be a nonzero element of K[x₁,..., x_n]. Then V(f) is nonempty, and V(f) $\neq$ Aⁿ.

PROOF. We proceed by induction on n, the case n = 1 being immediate. So, in the inductive step, we can assume (by renumbering the variables) that x_n actually occurs nontrivially in f. Therefore, we can write: f(x₁,..., x_n) = g₀( x₁,..., x_n-₁) + g₁( x₁,..., x_n-₁) x_n + ... + g_d( x₁,..., x_n-₁) x_n^d, where g_i is homogeneous of degree = i, while d > 0 and g_d is nonzero. By the inductive hypothesis, there exists (a₁,..., a_n-₁) in A^n-1 such that g_d( a₁,..., a_n-₁) $\neq$ 0. It follows that $\varphi$ (x) := f (a₁,..., a_n-₁,x) is a nonzero polynomial of degree = d > 0. Hence, there exist finitely many values of a_n in K such that f(a₁,..., a_n) = $\varphi$ (a_n) = 0, and infinitely many values of the last coordinate such that the corresponding value of f is nonzero. This proves our claim.

We note that irreducibility is really a property of closed sets in the Zariski topology, and does not really involve any other aspects of varieties. Here are a couple of basic properties of irreducible varieties.

PROPOSITION 1. If X is an affine or projective variety, then X is irreducible if and only if every nonempty open subset U $\subset$ X is dense in X.

PROOF. Every open subset is dense if and only if every two nonempty open subsets have a nonempty intersection. By taking complements, this is equivalent to the statement that the union of two proper closed subsets is not equal to X.

PROPOSITION 2. Let X $\subset$ Aⁿ be an irreducible affine variety. If X $\subseteq$ Y $\cup$ Z, where Y and Z are affine varieties, then either X $\subseteq$ Y or X $\subseteq$ Z.

PROOF. If X $\subseteq$ Y $\cup$ Z, then X = (X $\cap$ Y) $\cup$ ( X $\cap$ Z). By irreducibility,

either X = X $\cap$ Y or X = X $\cap$ Z. This proves our claim.

Now we can state and prove our main result.

THEOREM. Let X be an affine variety (respectively a projective variety). Then X is the union of finitely many irreducible affine varieties (respectively finitely many irreducible projective varieties). Thus:

X = Y₁ $\cup$ ... $\cup$ Y_m, where Y₁ ,..., Y_m are irreducible. If we require that Y_i does not contain Y_j when i $\neq$ j, then Y₁ ,..., Y_m are uniquely determined. The corresponding statements are also true in the projective case.

DEFINITION: The irreducible closed subsets which occur in this uniquely determined decomposition are called the irreducible components of X.

PROOF OF THE THEOREM. In proving the existence of the decomposition for Zariski closed subsets of Aⁿ, we use the fact that the Zariski topology of Aⁿ is Noetherian. By definition, this means that every descending closed subsets stabilizes after finitely many steps. An equivalent characterization is that every nonempty collection of closed subsets has a minimal member, i.e., there is a member of the collection which is not contained in any other set in the collection.

So, assume that there exists a Zariski-closed subset of Aⁿ which is not the union of finitely many irreducible closed subsets, and let S be the collection of closed subsets which are not finite unions of irreducibles. Then S contains a minimal member, say X₀ . If X₀ were irreducible, this would immediately contradict the fact that no member of S is a union of finitely many irreducible closed subsets. So, we have X₀ = Y $\cup$ Z, where Y and Z are proper closed subsets of X₀ . Since X₀ is a minimal member of S, neither Y nor Z is a member of S; therefore, both Y and Z are unions of finitely many irreducible closed subsets. This implies that S is a union of finitely many irreducible closed subsets, thus contradicting the hypothesis that S is nonempty and thereby proving the first part of the theorem.

To prove uniqueness, suppose that some X $\subset$ Aⁿ has two different decompositions as a union of irreducible closed subsets, both satisfying the condition that there be no inclusion relation among different subsets in the collection. Specifically, suppose that:

X = Y₁ $\cup$ ... $\cup$ Y_m and also: X = Z₁ $\cup$ ... $\cup$ Z_p , where there is no inclusion relation among the various Y's and also no such relation among the Z's

Consider a particular subset Y_i from the first decomposition. Since Y_i $\subseteq$ Z₁ $\cup$ ... $\cup$ Z_p, we can use Proposition 2 to show that Y_i $\subseteq$ Z_j for some j. Similarly, we show that Z_j $\subseteq$ Y_k for some k. Thus, Y_i $\subseteq$ Z_j $\subseteq$ Y_k . Because there are no nontrivial inclusions among the Y's, it follows that i = k, and Y_i = Z_j . Hence, all of the Y_i occur among Z₁, ... ,Z_p . Now, we can show by similar methods that all of the Z_j occur among Z₁, ... , Z_p . Therefore, the same irreducible closed subsets occur in the two decompositions.

Another example.

We would now like to claim that if f is an irreducible polynomial in n variables (respectively, if F is an irreducible homogeneous polynomial in n + 1 variables), then the zeroset V(f) is an irreducible closed subset of Aⁿ (respectively that the zeroset V(F) is an irreducible closed subset of Pⁿ ). A little work is required for this, so that it will be saved for Exercises 7 and 8, which are linked below. It will follow from this claim that if f is an arbitrary polynomial in n variables, then the irreducible components of the hypersurface V(f) correspond bijectively to the distinct irreducible factors of f [and similarly in the projective case.]

Related Exercises

Last updated November 18, 1997.

Please send comments and/or corrections to:

Joel Roberts
351 Vincent Hall
625-1076
e-mail: roberts@math.umn.edu
http://www.math.umn.edu/~roberts