Proof: Write
S for the set on the right-hand
side. Clearly,
A(X)⊂ S. Conversely,
suppose
φ∈ S. Then for each
x∈ X, we can write
φ =
fx/gx with
gx(x)≠0. Now consider
the ideal
I = (I(X), gx : x∈ X)⊂
k[x1,…,xn].
Since
Z(I) = {x∈ X : φ is not
regular at x},
the Weak Nullstellensatz implies that
I=(1). Reducing modulo
I(X), one
has a relation
1 = ∑ ui gxi,
for some
ui∈ A(X). Multiplying by
φ yields
φ = ∑ ui fxi ∈ A(X).
Definition Any open subset of an affine variety is called a
quasi-affine variety. If
S⊂
An is any quasi-affine variety, then it has a ring of
functions defined by
A(S) = [
{φ∈ k(An):φ is regular at
all s∈ S}
/
{φ∈ k(An):φ(s)=0 at all
s∈ S}].
By the proposition, if
X is an affine variety, then
this definition agrees with the previous one.
Lemma The complement of a hypersurface in
An
is an affine variety. Moreover, if
U=An\
Z(f), then
A(U) = k[x1, …, xn, 1/f] =
A(An)f.
Proof: Let
V⊂An+1 be defined by
V=Z(xn+1f-1). There is a commutative diagram
U = | An\ Z(f) | ⊂ |
An |
| ↓φ ↑ψ | |
↑pr |
| V | ⊂ | An+1 |
where
φ(a1, …, an) = (a1, …, an,
1/f(a1, …, an)),
and
ψ(a1,…,an+1) =
(a1,…,an).
It is easy to check that
φ°ψ =
1, that
ψ°φ=1, and
that
φ and
ψ induce
isomorphisms on the rings of functions.
Corollary Let
f∈ A(X) be a function on an affine
algebraic variety
X. Then
U=X\
Z(f) is an affine variety and
A(U) = A(X)f = A(X)[1/f].
Lemma Let X be an algebraic set. Then there
exists a basis for the topology on X consisting of
affine open algebraic sets.
Proof: Let
U⊂ X be any open subset, and
let
P∈ U be a point. Since
P and
X\ U are
disjoint closed subsets in
X, there must exist a
function
f∈ A(X) with
f(P)=1 and
f∈ I(X\
U). Then
X\ U⊂
Z(f), so
P∈ X\ Z(f) ⊂ U
gives an affine open neighborhood of
P inside
U.
Lemma Let X be an algebraic variety. Let
U⊂ X be a nonempty open subset. If
φ, ψ∈ A(X) are
regular functions such that φ|U = ψ|U
agree on U, then
φ=ψ.
Proof: Let
ρ=(φ,ψ): X→
A1×A1 =A2. By hypothesis, the image
ρ(U) is contained in the diagonal,
Δ, which is an algebraic set defined by
x=y. Since
U is dense in
X (by irreducibility), and
Δ is closed, we have
ρ(X)⊂ ρ(U)-
⊂Δ
and, therefore,
φ=ψ.
Theorem Let
X be an algebraic variety. Then
k(X) =
[{(U,f) : U is a nonempty open
subset of X and f:U→A1}
/
(U,f)∼(V,g) iff f|U∩ V =
g|U∩ V]
Proof: Since the affine opens form a basis for the topology, we only
need to consider affine opens in the description. If
U and V are nonempty opens, then
so is their intersection (by irreducibility). Therefore,
∼ is an equivalence relation. By definition,
every element in the set on the right-hand side comes from a rational
function. But every rational function φ∈
k(X) has at least one representation
φ=f/g, yielding a pair
(X\ Z(g), f/g).
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