# Algebraic Geometry Concepts

We have decided to introduce, at this early point, some basics of algebraic geometry in order to motivate the later concepts.

*In summary, algebraic geometry studies solutions to polynomial equations over a field*.

First we consider a case which has applications to number theory.

## Example

Let us compute all integer solutions to . For *c* = 0, we only have the trivial solution; for , we get so we need to solve in the field of *rational numbers*.

By observation we obtain our first solution *P* = (1, 1). For any other point with coordinates in ℚ, let us take the gradient of the line *PQ*, given by which is a rational number. Thus each *Q* gives us a rational number *m*.

[ Image edited from Geogebra plotter.]

Conversely, suppose we are given . The line through *P* = (1, 1) with gradient *m* intersects the ellipse in another point *Q* = (*x*, *y*). Now *x* and *y* are rational, because *x* is the root of a quadratic equation with rational coefficients; thus the sum of the two roots is rational; since one of the root is 1, the other root is also rational.

E.g. take . The line through *P* of gradient *m* is then , or . Substituting this into the equation of the ellipse then gives

which has two roots with sum . Since one of the roots is 1, the other is . And from we get . This gives the solution to the original equation .

Hence, we have shown that all solutions to our equation can be obtained in this way. This solution generalizes to all quadratic forms in three variables, e.g. . The case for cubic forms is much more complicated and involves *elliptic curves*, which is a huge topic.

**Exercise A**

Write down parametric solutions for and .

# The Affine Space

Let *k* be a fixed field.

Definition.The

affine n-spaceover the field k is the set of all n-tuples with each .If the base field is implicit, we often just write .

Set-theoretically, is just . However, we have chosen a different notation because one often mentally associates with an *n*-dimensional vector space over *k*.

Let be the ring of polynomials in *n *variables and coefficients in *k*. Each can be interpreted as a function on . For example,

.

Definition.Given a collection of polynomials, let

A subset of of this form is called a

closed subset.

In words, we say that *V*(*S*) is the subset of carved out by the polynomials in *S*. For example if *n* = 3 and comprises of and , the resulting *V*(*S*) is a unit circle on the *XY*-plane.

Proposition 1.The collection of all closed subsets in forms a topology for . Specifically, we have the following.

- and .
- For any collection of subsets , we have .
- Given we have , where .

**Proof**

The first property is obvious.

For the second, is equivalent to: for each *i*, , which is equivalent to: for each *i* and , we have , which is finally equivalent to: for each we have .

Finally clearly : if *P* satisfies for each , then certainly for all . Similarly so we have proven ⊆ in the claim.

For the reverse inclusion, suppose *P* lies outside *V*(*S*) or *V*(*T*); thus there exists such that , . But this means satisfies .♦

Definition.The above topology is called the

Zariski topologyon the affine n-space.

**Exercise B**

Prove that the Zariski topology on is the cofinite topology (i.e. a subset is closed if and only if it is finite, or the whole space).

Let . Prove that is not a closed subset of .

# Ideals of Polynomial Ring

In the previous section, a subset *S* of the polynomial ring *A* gives us a subset of the affine space. We now define the reverse.

Definition.Given any subset , let be the set of all such that for all .

Immediately we have:

Lemma 1.For any subset V of , is a radical ideal of the ring of polynomials A.

**Proof**

Clearly . Next if then for any we have . Similarly if and , then for any we have so we have .

Finally, if satisfies then for any we have and so . This gives . ♦

Hence we have the following diagram.

# Properties of the Correspondence

Clearly the above does not provide a bijection, but it will when we restrict the sets on both sides.

Proposition 2.Take any and .

- .
- .
- .
- .
- .
- . [Apologies for the awkward notation!]

**Note**

The fifth claim says that if *V* is a closed subset of , then *VI* takes *V* back to itself.

**Proof**

We will prove only half of the claims, since the remaining are similar.

First claim: if , then for any , any also lies in *V’* so and we have .

Third claim: if , we take any ; by the definition of *V*(*S*) we have . Hence *f* lies in *I*(*V*(*S*)).

Fifth claim: we have by the fourth claim. On the other hand since by the third claim we have by the second claim. ♦

Now we are very close to achieving a bijective correspondence, but we need a final additional condition.

Theorem (Nullstellensatz).Suppose k is an algebraically closed field (e.g. ). The above correspondence gives a bijection between:

- closed subsets ,
- radical ideals of .

**Exercise C**

Find a counter-example for the real field ℝ.

Unfortunately, we have to defer the proof till much later. For now, we will have to contend with exploring some examples.