AG1. Algebraic Sets, Zariski Topology and Affine Varieties - Ryan Lok-Wing Pang

Algebraic Sets, Zariski Topology and Affine Varieties

Ryan, Lok-Wing Pang

Department of MathematicsHong Kong University of Science and Technology

Introduction to Algebraic Geometry, 2015

Ryan, Lok-Wing Pang (HKUST) Algebraic Sets, Zariski Topology and Affine VarietiesIntroduction to Algebraic Geometry, 2015 1

Outline

1 Algebraic Sets

2 Zariski Topology

3 Affine Varieties

Affine Spaces and Algebraic Sets

Let k be a field and k be a fixed algebraic closure of k . We define theaffine n-space over k to be An

k = {(a1, · · · , an)|ai ∈ kn}.

Defintion (Algebraic Set)

An algebraic set X ⊆ Ank is the set of common zeros of a collection of

polynomials f1, · · · , fm ∈ k[x1, · · · , xn]. We denote this set byV (f1, · · · , fm) = {(a1 · · · , an) ∈ An

k |fi (a1, · · · , an) = 0 ∀i = 1, · · · ,m}.

More generally, for an ideal I E k[x1, · · · , xn], we defineV (I ) = {(a1 · · · , an) ∈ An

k |fi (a1, · · · , an) = 0 ∀fi ∈ I}.It is easy to see that if I = (f1, · · · , fm), then V (I ) = V (f1, · · · , fm).Conversely, since k[x1, · · · , xn] is a Noetherian ring (by Hilbert BasisTheorem), every ideal I is finitely generated, and hence every V (I )can be written as V (f1, · · · , fm) for some finite collection of fi .

k = {(a1, · · · , an)|ai ∈ kn}.

k |fi (a1, · · · , an) = 0 ∀i = 1, · · · ,m}.

k = {(a1, · · · , an)|ai ∈ kn}.

k |fi (a1, · · · , an) = 0 ∀i = 1, · · · ,m}.

k |fi (a1, · · · , an) = 0 ∀fi ∈ I}.

It is easy to see that if I = (f1, · · · , fm), then V (I ) = V (f1, · · · , fm).Conversely, since k[x1, · · · , xn] is a Noetherian ring (by Hilbert BasisTheorem), every ideal I is finitely generated, and hence every V (I )can be written as V (f1, · · · , fm) for some finite collection of fi .

k = {(a1, · · · , an)|ai ∈ kn}.

k |fi (a1, · · · , an) = 0 ∀i = 1, · · · ,m}.

Examples

Algebraic sets can be extremely hard to be determined.

The algebraic set V : xn + yn = 1 is defined over Q. Andrew Wilesproved the Fermat’s Last Theorem in 1995 which stated that for alln ≥ 3,

V (Q) =

{{(1, 0), (0, 1)} if n is odd,{(±1, 0), (0,±1)} if n is even.

In fact, by Hilbert’s tenth problem (which is now solved), given afinite system of polynomial equations with coefficients in Q, theredoes NOT exist a finite algorithm to determine the system has asolution in Q or not.

Examples

V (Q) =

Examples

V (Q) =

Zariski Topology

We then define the Zariski Topology on Ank . First, we need a

lemma:

Let I , J, Iα be ideals in k[x1, · · · , xn], then we have(i) V (I ) ∪ V (J) = V (I ∩ J),(ii)

⋂α V (Iα) = V (

∑α Iα),

(iii) V (I ) ⊃ V (J) for√I ⊂√J, where

√I = {f ∈ k[x1, · · · , xn]|f m ∈ I}

for some m ∈ Z+ is called the radical of I .

Remark. If I =√I , then we call I an radical ideal.

Remark 2. Hence algebraic set behaves like closed set.

Zariski Topology

lemma:

⋂α V (Iα) = V (

∑α Iα),

√I = {f ∈ k[x1, · · · , xn]|f m ∈ I}

Zariski Topology

lemma:

⋂α V (Iα) = V (

∑α Iα),

√I = {f ∈ k[x1, · · · , xn]|f m ∈ I}

Zariski Topology

lemma:

⋂α V (Iα) = V (

∑α Iα),

√I = {f ∈ k[x1, · · · , xn]|f m ∈ I}

Proof of lemma

Proof.

First observe that V (I ) ⊃ V (J) for I ⊂ J, henceV (I ∩ J) ⊃ V (I ),V (J) and V (I ) ∪ V (J) ⊂ V (I ∩ J).

Conversely, let P ∈ V (I ∩ J). If P 6∈ V (I ), then there exists apolynomial f ∈ I such that f (P) 6= 0.

Now, for an arbitrary element g ∈ J, we have h = fg ∈ I ∩ J. Henceh(P) = f (P)g(P) = 0 and g(P) = 0, which implies P ∈ V (J).Therefore V (I ∩ J) ⊂ V (I ) ∪ V (J), completing the proof of (i).

For (ii), since Iα ⊂∑

α Iα, we have V (Iα) ⊃ V (∑

α Iα) and hence⋂α V (Iα) ⊃ V (

∑α Iα) .

For each α, we can write Iα in terms of generators by Noetherianproperty: Iα = (fα1 , · · · , fαkα

). Then for P ∈⋂

α V (Iα), we havefαj (P) = 0 for all j = 1, · · · , kα. On the other hand, the set {fαk

}generates the ideal

∑α Iα and hence P ∈ V (

∑α Iα).

(iii) is left as an exercise.

Proof of lemma

Proof.

∑α Iα) .

∑α Iα).

Proof of lemma

Proof.

∑α Iα) .

∑α Iα).

Proof of lemma

Proof.

∑α Iα) .

∑α Iα).

Proof of lemma

Proof.

∑α Iα) .

∑α Iα).

Proof of lemma

Proof.

∑α Iα) .

∑α Iα).

Zariski Topology

Defintion (Zariski Topology)

We define the Zariski topology on Ank by taking the open subsets to be the

complements of the algebraic sets. By the previous lemma and the factthat ∅ = V (1),An

k = V (0), this is indeed a topology.

Let us consider the Zariski Topology on the Affine line A1k for an

algebraically closed field k. Since k[x ] is a PID, every algebraic set isthe set of zeros of a single polynomial. Since k is algebraically closed,f (x) = an(x − α1) · · · (x − αn) for f (x) ∈ k[x ], an, αi ∈ k . HenceV (f ) = {α1, · · · , αn}. Hence all algebraic sets not equal to A1

k arefinite.

Notice that this topology is not Hausdorff.

Zariski Topology

k arefinite.

Zariski Topology

k arefinite.

Affine Varieties

An algebraic set V in Ank is said to be irreducible if V = V1 ∪ V2

implies V1 = ∅ or V2 = ∅, where V1 and V2 are algebraic sets in Ank .

Defintion (Affine Varieties and Quasi-Affine Varieties)

An affine varieties is an irreducible closed subset of Ank with the induced

topology (i.e. an affine variety is an irreducible algebraic set V in Ank). An

open subset of an affine variety is called a quasi-affine variety.

Next, we show that there exists a bijective correspondence betweenradical ideals and affine varieties. Recall that we have constructed avariety (a set) V (I ) from an ideal I E k[x1, · · · , xn]. Conversely, wecan construct an ideal from a set.

Defintion (Ideal of X )

Let X ⊂ Ank be any subset, then the ideal of X is