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.
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= X_{1}∪ X_{2} is a
decomposition of
X as a union of nonempty proper
closed subsets. Then each ideal
I(X_{i}) contains
I(X) properly. So, we can choose functions
f_{i}∈ I(X_{i})\ I(X). Then the
product
f_{1}f_{2}∈ I(X), so the ideal
I(X) is not prime.
Corollary There is a one-to-one correspondence between irreducible
algebraic sets in A^{n} and prime ideals in
k[x_{1}, …, x_{n}].
Theorem
Let X be an algebraic set in
A^{n}. Then there exists a unique collection of
irreducible algebraic sets
X_{1}, …, X_{m} such
that
- X= X_{1}∪ X_{2}∪…∪
X_{m}, and
- X_{i}⊂ X_{j} 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=X_{1}∪ Y_{1}. The theorem must
also fail for at least one of the pieces in this decomposition; say,
X_{1}. Now write
X_{1}=X_{2}∪
Y_{2} and repeat. In this way, one constructs an infinite
decreasing chain of algebraic sets
X ⊃ X_{1} ⊃ X_{2} ⊃
….
However, the chain of associated ideals
I(X) ⊂ I(X_{1}) ⊂ I(X_{2}) ⊂
…
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=∪_{i}X_{i} = ∪_{j}Y_{j}
are two irredundant decompositions of
X as a union
of irreducible algebraic sets. For each
i, we can
write
X_{i} = X_{i}∩ X = X_{i}∩(∪_{j}Y_{j}) =
∪_{j}(X_{i}∩ Y_{j}).
Because
X_{i} is irreducible, there exists some
j with the property that
X_{i} = X_{i}∩
Y_{j} ⊂ Y_{j}. By interchanging the roles of
the decompositions, there exists some
i' with
X_{i} ⊂ Y_{j} ⊂ X_{i'}.
Because the decompositions are irredundant, this last chain of
inclusions must actually consist of equalities, and the result
follows.
Corollary Let
f∈ k[x_{1},…,x_{n}]. Write
f=f_{1}^{n1}… f_{s}^{ns}
as a product of irreducible factors. Then
Z(f) = Z(f_{1}) ∪…∪ Z(f_{s})
is the unique decomposition of its zero set into irreducible
components. Furthermore,
I(Z(f)) = (f_{1}f_{2}… f_{s}).
Proof: The function f_{1} must vanish on some
component X_{1} of X. But then
X_{1}⊂ Z(f_{1}), which forces X_{1} =
Z(f_{1})∩ X.
Definition An algebraic set defined by one polynomial in
A^{n} 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 = Z_{1}∪
Z_{2} is a decomposition of the product. For each point
x∈ X, we have
Y = x× Y = ((x× Y)∩ Z_{1})∪
((x× Y)∩ Z_{2}).
So, at least one of the factors on the right contains this copy of the
irreducible set
Y. Now define
X_{i}⊂
X to be the set of all
x∈ X such
that
x× Y ⊂ Z_{i}. We have already
observed that
X=X_{1}∪ X_{2}. However,
X is also irreducible and the
X_{i} are closed subsets. Therefore, we must have
X=X_{1} and
X× Y = Z_{1}.
Definition An affine algebraic variety over an
algebraically closed field k is an irreducible
algebriac set in A^{n} with the induced topology
and the (reduced) induced structure.
Examples
- A^{n} 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
author:
Kevin R. Coombes
Department of Biomedical Informatics
The Ohio State University
Columbus, Ohio 43210