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
Proposition An algebraic set X is irreducible if and
only if the ideal I(X) is prime.
is not prime, then there exist
, g∉ I(X)
with product fg∈ I(X)
. So, we can write
X = (X∩ Z(f)) ∪ (X∩
as a nontrivial decomposition of X
Conversely, suppose X= X1∪ X2
decomposition of X
as a union of nonempty proper
closed subsets. Then each ideal I(Xi)
properly. So, we can choose functions
fi∈ I(Xi)\ I(X)
. Then the
product f1f2∈ I(X)
, so the ideal
is not prime.
Corollary There is a one-to-one correspondence between irreducible
algebraic sets in An and prime ideals in
k[x1, …, xn].
Let X be an algebraic set in
. Then there exists a unique collection of
irreducible algebraic sets X1, …, Xm
- X= X1∪ X2∪…∪
- Xi⊂ Xj if and only if
(Existence) Suppose the theorem does not hold for an algebraic
. In particular, X
reducible. Write X=X1∪ Y1
. The theorem must
also fail for at least one of the pieces in this decomposition; say,
. Now write X1=X2∪
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,
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.
X=∪iXi = ∪jYj
are two irredundant decompositions of X
as a union
of irreducible algebraic sets. For each i
, we can
Xi = Xi∩ X = Xi∩(∪jYj) =
is irreducible, there exists some
with the property that Xi = Xi∩
Yj ⊂ Yj
. By interchanging the roles of
the decompositions, there exists some i'
Xi ⊂ Yj ⊂ Xi'.
Because the decompositions are irredundant, this last chain of
inclusions must actually consist of equalities, and the result
Let f∈ k[x1,…,xn]
as a product of irreducible factors. Then
Z(f) = Z(f1) ∪…∪ Z(fs)
is the unique decomposition of its zero set into irreducible
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 =
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.
irreducible, and suppose that X× Y = Z1∪
is a decomposition of the product. For each point
, 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⊂
to be the set of all x∈ X
that x× Y ⊂ Zi
. We have already
observed that X=X1∪ X2
is also irreducible and the
are closed subsets. Therefore, we must have
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.
- An is a variety.
- Among the curves of the first section, only (v), (x), and
(xi) are not varieties.
- A point is a variety; two points are not.
Comments on this web site should be addressed to the
Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210