## Rational Functions

Definition Let X be an affine algebraic variety over an algebraically closed field k. Then I(X) is a prime ideal, so A(X) is an integral domain. Define k(X) to be the field of fractions of the integral domain A(X). Elements of k(X) are called rational functions on X.

Definition A rational function φ k(X) is called regular at a point x X if it can be represented in the form φ = f/g where f, g A(X) and g(x)0.

Proposition Let X be an algebraic variety. Then

A(X) = { φ k(X) : φ is regular at all points x X}.

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.

### Examples

1. Consider the quasi-affine variety U=A1\{0}. Since U is open in the irreducible variety A1, it is irreducible and dense. Therefore, A(U) is a subring of k(A1)= k(x). In order to determine which rational functions φ=f(x)/g(x) are regular on U, we only need to determine which polynomials g(x) are zero only at the point 0. Since these polynomials consist only of the powers of x, we see that A(U)= k[x, x-1]. We recognize this ring of functions as the coordinate ring of the affine plane curve V defined by xy=1. Moreover, projection on the x-coordinate is a topological homeomorphism V U that is an isomorphism on the ring of regular functions. In this way, we can identify the quasi-affine variety U with the affine variety V.
2. Let's try to repeat the previous example in a larger dimension. This time, take U=A2\ {(0,0)}. The same analysis shows that A(U) is a subring of k(X). Writing a rational function φ=f(x,y)/g(x,y) as a quotient of relatively prime polynomials, we see that φ is not regular on the zero set Z(g). If g is a nonconstant polynomial, then we have seen that this zero set is infinite. It follows that A(U) A(A1) = k[x,y], and that this isomorphism is induced by the inclusion map (which is not a homeomorphism). Since coordinate rings determine their varieties up to isomorphism, it follows that the quasi-affine variety U is not isomorphic to any affine variety.

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 VAn+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:UA1} / (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).