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 Biostatistics and Applied Mathematics
University of Texas M.D.Anderson Cancer Center
1515 Holcombe Blvd., Box 447
Houston, TX 77030