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Let Fun(S, C) be the set of all complex valued functions on S

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1. Coordinate Ring of an affine Variety

Let S be any nonempty set. Let Fun(S, C) be the set of all complex valued functions on S. For f, g ∈ Fun(S, C), and a ∈ C, we define f + g, af, by

(f + g)(x) = f (x) + g(x), (af )(x) = af (x),

for all x ∈ S. Then Fun(S, C) is a complex vector space. Moreover, we define (f g)(x) = f (x)g(x), x ∈ S.

Then Fun(S, C) becomes a complex algebra.

Let V be a nonempty affine variety in Cn. A function f ∈ Fun(V, C) is said to be a polynomial function on V if there exists a polynomial F ∈ C[X1, · · · , Xn] such that

f (a1, · · · , an) = F (a1, · · · , an),

for all (a1, · · · , an) ∈ V. Let Γ(V ) be the subset of Fun(V, C) consisting of polynomial functions on V. Then Γ(V ) is a complex subalgebra of Fun(V, C). Notice that two polyno- mials F, G ∈ C[X1, · · · , Xn] determine the same polynomial function on V if and only if F − G ∈ I(V ), where I(V ) is the ideal defining V. This observation leads to the following conclusion:

Proposition 1.1. Let V be an affine variety of Cn. Then we have the following isomorphism of complex algebras:

Γ(V ) ∼= A(V ).

Proof. Let ϕ : C[X1, · · · , Xn] → Γ(V, C) be the canonical map. (The restriction of F to V for F ∈ C[X1, · · · , Xn].) Then ϕ is an algebra epimorphism. The kernel of ϕ consisting of polynomials F so that F = 0 on V, in other words, ker ϕ = I(V ). The assertion follows from the first isomorphism of algebras.

 This proposition tells us that the coordinate ring of V is (can be identified with) the algebra of polynomial functions on V. Since V is a variety, Γ(V ) is an integral domain. We can consider its field of fraction C(V ) called the function field of V or the field of rational functions on V.

Definition 1.1. Let f ∈ C(V ) be a rational function. We say that f is defined at P ∈ V if there exists a, b ∈ Γ(V ) such that f = a/b with b(P ) 6= 0. The set of all rational functions defined at P is denoted by OP(V ) or simply OP when V is specified. The set of points P ∈ V where f is not defined is called the pole set of f.

Lemma 1.1. The set OP forms a subring of C(V ) containing Γ(V ).

Proof. Let f, g be defined at P. Choose a, b, c, d ∈ Γ(V ) so that f = a/b and c/d with b(P ), d(P ) 6= 0. Then f + g = (ad + bc)/bd. Since b(P ), d(P ) 6= 0, b(P )d(P ) 6= 0. We find f + g is defined at P. Similarly, f g = ac/bd is defined at P. The zero function is defined at P. We find OP is a subring of C(V ). Moreover, for each polynomial function f on V, f /1 is

define at P. Hence Γ(V ) is contained in OP. 

Proposition 1.2. Let V be an affine variety.

(1) Let f be a rational function on V. The pole set of f is an affine algebraic subset of V.

(2) Γ(V ) =T

P ∈V OP.

1

(2)

2

Proof. Let us denote the canonical map ϕ : C[X1, · · · , Xn] → Γ(V ) by ϕ as above. Let If be the set of G ∈ C[X1, · · · , Xn] so that ϕ(G)f ∈ Γ(V ). Then If is an ideal of C[X1, · · · , Xn] containing I(V ). The affine algebraic set V (If) is exactly the pole set of f and contained in V. This proves (1).

By lemma, Γ(V ) ⊂ T

P ∈V OP. If f ∈ T

P ∈V OP, then f is defined at all P ∈ V. Hence V (If) = ∅. By the Nullstellensatz, 1 ∈ If. Hence If = C[X1, · · · , Xn]. We see that 1 · f = f ∈ Γ(V ). Hence T

P ∈V OP ⊂ Γ(V ). We prove (2).

 Let f ∈ OP. Then f = a/b for some a, b ∈ Γ(V ) with b(P ) 6= 0. We set f (P ) = a(P )/b(P ) called the value of f at P. One can check that the value of f at P is independent of choice of a, b. The evaluation map P : OP → C defined by f 7→ f(P ) is a ring epimorphism. The kernel of P denoted by mP is a maximal ideal of OP. In fact, we can check (in the following proposition) mP is the unique maximal ideal in OP. Hence OP/mP ∼= C. We call mP the maximal ideal of V at P.

Proposition 1.3. The ring OP is a Noetherian local domain.

Proof. We check that mP is exactly the set of all non units in OP. Then all proper ideals of OP is contained in mP. Hence mP is the unique maximal ideal of OP; OP is a local ring.

To show that OP is a Noetherian ring, we need to show that all the ideal I in OP is finitely generated. Let J = I ∩ Γ(V ). Then J is an ideal of Γ(V ). Since Γ(V ) is Noetherian, J is finitely generated. Choose a generator {f1, · · · , fr} of J. Let f ∈ I. Choose a, b ∈ Γ(V ) such that f = a/b with b(P ) 6= 0. Then bf = a ∈ Γ(V ). Since I is an ideal of OP, bf ∈ J.

Then we can find a1, · · · , ar∈ Γ(V ) so that

bf = a1f1+ · · · + arfr=⇒ f = a1

b f1+ · · · +ar

br

fr.

Since b(P ) 6= 0, a1/b, · · · , ar/b ∈ OP. Since f1, · · · , fr ∈ J = I ∩ Γ(V ), f1, · · · , fr∈ I. Since I is an ideal of OP, (a1/b)f1+ · · · + (ar/b)fr ∈ I. We see that {f1, · · · , fr} also forms a set of generators of I over OP.

 Definition 1.2. We call OP the local ring of V at P.

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