The affine n-space over k denoted by An(k) is the set of n-tuples: {(a1

1. Affine Space and affine algebraic sets

Let A = k[X₁, · · · , X_n] be the ring of polynomials in variables X₁, · · · , X_n over an alge- braically closed field k. For any index set I = (i₁, · · · , i_n), we denote X^I = X₁ⁱ¹· · · X_nⁱⁿ. Elements of A are of the form f (X) = P

Ic_IX^I, where c_I ∈ k. The affine n-space over k denoted by Aⁿ(k) is the set of n-tuples: {(a1, · · · , an) : a1, · · · , an ∈ k}. For any P (a₁, · · · , a_n) ∈ Aⁿ(k), we set

f (P ) =X

I

c_Iaⁱ₁¹· · · aⁱ_nⁿ.

Then f : Aⁿ(k) → k defines a k-valued function on Aⁿ(k). The zero set V (f ) of f is the set V (f ) = {P ∈ Aⁿ(k) : f (P ) = 0}.

We also denote Aⁿ(k) \ V (f ) by D(f ). When f is nonconstant, V (f ) is called a hypersurface in Aⁿ(k). A hypersurface in A²(k) is called an affine plane curve. When deg f = 1, V (f ) is called a hyperplane in Aⁿ(k) and called a line if n = 2.

If S is a family of polynomials in A, the common zero set of S is defined to be V (S) = \

f ∈S

V (f ) = {P ∈ Aⁿ(k) : f (P ) = 0, ∀f ∈ S}.

A subset Y of Aⁿ(k) is called an affine algebraic set if Y = V (S) for some family of polynomials S in A. A subset U of Aⁿ(k) is said to be Zariski open if Aⁿ(k) \ U is an affine algebraic set.

Proposition 1.1. The Zariski open sets of Aⁿ(k) forms a topology on Aⁿ(k). The topology defined by the Zariski open sets is called the Zariski topology of Aⁿ(k).

Proof. It is equivalent to prove the following.

(1) ∅, Aⁿ(k) are affine algebraic.

(2) Any intersection of affine algebraic sets is affine algebraic.

(3) Any finite union of affine algebraic sets is affine algebraic.

(1) follows from the fact that V (0) = Aⁿ(k) and V (1) = ∅, i.e. Aⁿ(k) and ∅ are affine algebraic sets and their complement ∅ and Aⁿ(k) in Aⁿ(k) are Zariski open by definition.

(2) follows from the fact that T

iV (T_i) = V (S

iT_i). Here T_i are sets of polynomials for all i.

Let T₁, T₂ be two sets of polynomials and T be the set of polynomials {f g : f ∈ T₁, g ∈ T2}. Then V (T₁) ∪ V (T2) = V (T ). (3) can be proved by induction. 

From now on, we only consider the Zariski topology on Aⁿ(k).

A topological space X is reducible if it is a union of two proper closed subsets, i.e.

X = X₁∪ X₂ where X₁, X₂ are closed and X₁, X₂ 6= X. A topological space is irreducible if it is not reducible. A nonempty subset Y of a topological space is irreducible if Y with the subspace topology induced from X is an irreducible space.

Definition 1.1. An affine variety is an irreducible closed subset of Aⁿ(k). A quasi-affine variety is an open subset of an affine variety.

Lemma 1.1. If T₁ ⊂ T₂ are subsets of A, then V (T₁) ⊃ V (T₂).

Proof. If P ∈ V (T₂), then f (P ) = 0 for all f ∈ T₂. Since T₁ ⊂ T₂, we see that f (P ) = 0 for all f ∈ T₁. This implies that P ∈ V (T₁). Hence V (T₂) ⊂ V (T₁). 

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Corollary 1.1. Let a be the ideal of A generated by a set of polynomials T. Then V (a) = V (T ).

Proof. We know that a is the smallest ideal containing T. By the previous lemma, V (a) ⊂ V (T ). Let P ∈ V (T ). We find f (P ) = 0 for all f ∈ T. If h ∈ a, there exist a₁, · · · , a_r ∈ A and f1, · · · , fr ∈ T so that h = a₁f1+ · · · + arfr. Since fi ∈ T for 1 ≤ i ≤ r, f_i(P ) = 0 for 1 ≤ i ≤ r. Hence h(P ) = 0. We see that P ∈ V (a). We conclude that V (a) = V (T ).  Corollary 1.2. Every affine algebraic subset of A¹(k) is either finite or A¹(k) itself.

Proof. Let Y be an affine algebraic subset of A¹(k) so that Y 6= A¹(k). Then there is a proper ideal I of k[X] so that Y = V (I). Since every polynomial in k[X] only has finite number of zeros, V (f ) is a finite subset of A¹ for all f (X) ∈ k[X]. Since V (I) = ∩_{f ∈I}V (f ), V (I) is the intersection of finite subsets of A¹(k). This implies that V (I) is either the empty

set or a finite subset of A¹(k). 

Since k is an algebraically closed field, A¹(k) has infinite number of elements. Since every proper closed subsets of A¹(k) is finite, A¹(k) can not be a union of two proper closed subsets. Hence A¹(k) is irreducible.

2. The Ideal of a Set of Points Let P be any point of Aⁿ(k), we denote

I_P = {f ∈ A : f (P ) = 0}.

Then I_P is a maximal ideal of A. This is because I_P is the kernel of the ring epimorphism

_P : A → k defined by f 7→ f (P ) and k[X]/I_P ∼= k is a field. If P = (a1, · · · , a_n), then I_P is the ideal hX1− a₁, · · · , Xn− a_ni.

For any subset Y of Aⁿ(k), we define

I(Y ) = \

P ∈Y

I_P.

Since the intersection of ideals of a ring is again an ideal of that ring, I(Y ) is an ideal of A.

Moreover I(Y ) is the set of all f in A so that f (P ) = 0 for all P ∈ Y.

Lemma 2.1. If Y₁ ⊂ Y₂ are subsets of Aⁿ(k), then I(Y₁) ⊃ I(Y₂).

Proof. Let f ∈ I(Y2). Then f (P ) = 0 for all P ∈ Y2. Since Y1 is contained in Y2, f (P ) = 0 for all P ∈ Y1. Therefore f ∈ I(Y1). We find I(Y2) ⊂ I(Y1).  Lemma 2.2. For any two subsets Y1, Y2 of Aⁿ(k), I(Y1∪ Y₂) = I(Y1) ∩ I(Y2).

Proof. Let f ∈ I(Y₁ ∪ Y₂). Then f (P ) = 0 for all P ∈ Y₁∪ Y₂. Hence f (P ) = 0 for all P ∈ Y₁ and all P ∈ Y₂. Then f ∈ I(Y₁) and f ∈ I(Y₂). We see f ∈ I(Y₁) ∩ I(Y₂). Then I(Y1∪ Y₂) = I(Y1) ∩ I(Y2).

Conversely, let f ∈ I(Y₁) ∩ I(Y₂). Then f (P ) = 0 for all P ∈ Y₁ and all P ∈ Y₂. Hence f (P ) = 0 for all P ∈ Y1 ∪ Y₂. This implies that f ∈ I(Y1 ∪ Y₂). Hence I(Y1) ∩ I(Y2) ⊂ I(Y₁∪ Y₂).

 Let R be any commutative ring with identity and I be an ideal of R. The radical of I denoted by √

I is the set of all r so that there exists n > 0 so that rⁿ∈ I. We leave to the reader to verify that√

I is an ideal of R containing I. When I = (0) is the trivial ideal of R, p(0) is also denoted by Nil(R) called the nilradical of R.

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Theorem 2.1. (Hilbert Nullstellensatz) Let k be an algebraically closed field and I be an ideal of A. Let f ∈ A be a polynomial so that f (P ) = 0 for all P ∈ V (I). Then f^r∈ I for some r > 0.

Proposition 2.1. For any ideal a of A, I(V (a)) =√ a.

Proof. Suppose f ∈ I(V (a)). Then f (P ) = 0 for all P ∈ V (a). By the Hilbert-Nullstellensatz, f^r∈ a for some r ≥ 0. In other words, f ∈√

a. Conversely, if f ∈√

a. Then f^r ∈ a for some r ≥ 0. Hence for all P ∈ V (a), (f (P ))^r = 0. Since k is a field, f (P ) = 0. This shows that

f ∈ I(V (a)). We prove the assertion. 

Proposition 2.2. Let Y be any subset of Aⁿ(k). Then V (I(Y )) = Y .

Proof. For P ∈ Y, for all f ∈ I(Y ), we have f (P ) = 0 by definition. Hence P ∈ V (I(Y )).

In other words, Y ⊂ V (I(Y )). Since V (I(Y )) is closed containing Y and Y is the smallest closed subset of Aⁿ(k) containing Y, V (I(Y )) contains Y .

Since Y is closed in Aⁿ(k), Y = V (T ) for some set T of polynomials. Then we know that for all P ∈ Y, and all f ∈ T, f (P ) = 0. Hence T ⊂ I(Y ). By Lemma, V (T ) ⊃ V (I(Y )), i.e.

Y ⊃ V (I(Y )). This proves our assertion. 

Proposition 2.3. An algebraic set is irreducible if and only if its ideal is a prime ideal.

Proof. Let us assume Y is an affine variety. Then we want to show that I(Y ) is a prime ideal in A. Let f g ∈ I(Y ). Then V (f g) ⊃ Y. Since Y is closed, Y ∩ V (f g) is closed. Since V (f g) = V (f ) ∪ V (g),

Y ∩ V (f g) = (Y ∩ V (f )) ∪ (Y ∩ V (g)).

We know that Y ∩ V (f ) and Y ∩ V (g) are closed subsets of Y. By the irreducibility, either V (f ) = Y or V (g) = Y. Assume that V (f ) = Y. Then I(Y ) = I(V (f )) = p(f). Since f ∈p(f), f ∈ I(Y ). We prove that I(Y ) is a prime ideal of A.  Corollary 2.1. The affine n-space Aⁿ(k) is irreducible.

Proof. The ideal defining Aⁿ(k) is (0). Since A = k[X1, · · · , Xn] is an integral domain, the zero ideal (0) is a prime ideal of A. Proposition 2.3 implies that Aⁿ(k) is irreducible.  Corollary 2.2. There is a one-to-one inclusion reversing correspondence between algebraic sets in Aⁿ(k) and radical ideals in A.

Proof. The correspondence is given by Y 7→ I(Y ). 

Definition 2.1. The coordinate ring of an affine algebraic set Y is the quotient ring A(Y ) = A/I(Y ).

Proposition 2.4. Let B be an algebra over k which is also a domain. B is finitely generated if and only if it is isomorphic to the coordinate ring of an affine variety.

Proof. Since A is generated by {x1, · · · , xn} as an algebra over k, A is a finitely generated complex algebra. If B = A(Y ) for some affine variety in Aⁿ(k), then B is generated by {π(x₁), · · · , π(xn)}, where π : A → B is the quotient map. Since Y is an affine variety, I(Y ) is a prime ideal. Then A(Y ) = A/I(Y ) is an integral domain.

Conversely, assume that B is a finitely generated l-algebra which is also a domain. Let {b₁, · · · , b_n} be a set of generators of B as a C-algebra. Define a ring homomorphism

ϕ : k[X1, · · · , Xn] → B

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sending Xi to bi. Then ϕ is surjective. By the first isomorphism theorem, B ∼= k[X1, · · · , X_n]/I,

where I = ker ϕ. Since B is a domain, I is a prime ideal. Let Y = V (I). Then Y is an algebraic set in Aⁿ. Moreover I(Y ) = I(V (I)) = √

I. Let f ∈ I(Y ). Then f ∈ √

I. There is r > 0 so that f^r ∈ I. Since I is prime, f ∈ I. We find I(Y ) ⊂ I. Since I ⊂ √

I and

√

I ⊂ I(Y ), I ⊂ I(Y ). We conclude that I = I(Y ). Since I = I(Y ) is prime, Y is an affine

variety. This prove our assertion. 

Proposition 2.5. Aⁿ(k) with the Zariski topology is a Noetherian space.

Proof. Let {Y_i} be a descending chain of closed subsets in Aⁿ(k) : Y_i ⊃ Y_i+1· · · . Then we obtain an ascending chain of ideals (ai) : ai ⊂ a_i+1, where ai = I(Yi). Since A is a Noetherian ring, there exists r > 0 so that a_i= a_r for all i ≥ r. In this case, Y_i = Y_r for all

i ≥ r. We prove that Aⁿ(k) is a Noetherian space. 

Since the subspace of any Noetherian space is also Noetherian, an affine algebraic set is a Noetherian space with the subspace topology of Aⁿ(k). Using the property of Noetherian space, we can decompose any algebraic set in Aⁿ(k) uniquely into a finite union of its irreducible components. Hence we conclude that:

Corollary 2.3. Every algebraic set in Aⁿ(k) can be expressed uniquely as a finite union of varieties, no one containing another.