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.
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.
Write down parametric solutions for and .
The Affine Space
Let k be a fixed field.
The affine n-space over 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,
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.
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 .
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 .♦
The above topology is called the Zariski topology on the affine n-space.
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.
Given any subset , let be the set of all such that for all .
Immediately we have:
For any subset V of , is a radical ideal of the ring of polynomials A.
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.
Take any and .
- . [Apologies for the awkward notation!]
The fifth claim says that if V is a closed subset of , then VI takes V back to itself.
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.
Suppose k is an algebraically closed field (e.g. ). The above correspondence gives a bijection between:
- closed subsets ,
- radical ideals of .
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.