# 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 $a^2 + 2b^2 = 3c^2$. For c = 0, we only have the trivial solution; for $c\ne 0$, we get $(\frac a c)^2 + 2(\frac b c)^2 = 3$ so we need to solve $x^2 + 2y^2 = 3$ in the field of rational numbers.

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

[ Image edited from Geogebra plotter.]

Conversely, suppose we are given $m\in \mathbb Q$. 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 $m = \frac 3 2$. The line through P of gradient m is then $(y-1) = \frac 3 2 (x-1)$, or $y = \frac 3 2 x - \frac 1 2$. Substituting this into the equation of the ellipse then gives

$x^2 + 2\left( \frac 3 2 x - \frac 1 2\right)^2 = 3 \implies \frac{11}2 x^2 - 3x - \frac 5 2 = 0$

which has two roots with sum $\frac 6 {11}$. Since one of the roots is 1, the other is $x=\frac 6{11} - 1 = -\frac 5 {11}$. And from $y = \frac 3 2 x - \frac 1 2$ we get $y = -\frac{13}{11}$. This gives the solution $a = -5, b=-13, c = 11$ to the original equation $a^2 + 2b^2 = 3c^2$.

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. $a^2 + ab + b^2 = 3c^2$. 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 $a^2 + 2b^2 = 3c^2$ and $a^2 + ab + b^2 = 3c^2$.

# The Affine Space

Let k be a fixed field.

Definition.

The affine n-space $\mathbb A_k^n$ over the field k is the set of all n-tuples $(x_1, \ldots, x_n)$ with each $x_i \in k$.

If the base field is implicit, we often just write $\mathbb A^n$.

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

Let $A = k[X_1, \ldots, X_n]$ be the ring of polynomials in variables and coefficients in k. Each $f\in A$ can be interpreted as a function on $\mathbb A_k^n$. For example,

\left.\begin{aligned} f = X^2 - Y^3 + 2Z \in \mathbb C[X, Y, Z]\\P = (2, 1, 0) \in \mathbb A_{\mathbb C}^3 \end{aligned}\right\} \implies f(P) = 2^2 - 1^3 + 2\cdot 0 = 3.

Definition.

Given a collection $S \subseteq A$ of polynomials, let

$V(S) = \{ P \in \mathbb A_k^n : f(P) = 0 \text{ for all } f\in S\}.$

A subset of $\mathbb A_k^n$ of this form is called a closed subset.

In words, we say that V(S) is the subset of $\mathbb A^n_k$ carved out by the polynomials in S. For example if n = 3 and $S \subset \mathbb R[X, Y, Z]$ comprises of $f = X^2 + Y^2 - 1$ and $g = Z$, the resulting V(S) is a unit circle on the XY-plane.

Proposition 1.

The collection of all closed subsets in $\mathbb A_k^n$ forms a topology for $\mathbb A_k^n$. Specifically, we have the following.

• $V(\{0\}) = \mathbb A_k^n$ and $V(\{1\}) = \emptyset$.
• For any collection of subsets $S_i \subseteq A$, we have $\cap_i V(S_i) = V(\cup_i S_i)$.
• Given $S, T\subseteq A$ we have $V(S) \cup V(T) = V(ST)$, where $ST = \{fg : f\in S, g\in T\}$.

Proof

The first property is obvious.

For the second, $P \in \cap_i V(S_i)$ is equivalent to: for each i, $P\in V(S_i)$, which is equivalent to: for each i and $f \in S_i$, we have $f(P) = 0$, which is finally equivalent to: for each $f\in \cup_i S_i$ we have $f(P) = 0$.

Finally clearly $V(S) \subseteq V(ST)$: if P satisfies $f(P) = 0$ for each $f\in S$, then certainly $(fg)(P) = f(P)g(P) = 0$ for all $fg\in ST$. Similarly $V(T) \subseteq V(ST)$ so we have proven ⊆ in the claim.

For the reverse inclusion, suppose P lies outside V(S) or V(T); thus there exists $f\in S, g\in T$ such that $f(P) \ne 0$, $g(P) \ne 0$. But this means $fg\in ST$ satisfies $(fg)(P) \ne 0$.♦

Definition.

The above topology is called the Zariski topology on the affine n-space.

Exercise B

Prove that the Zariski topology on $\mathbb A^1_k$ is the cofinite topology (i.e. a subset is closed if and only if it is finite, or the whole space).

Let $k = \mathbb R$. Prove that $\{(x, e^x) : x\in \mathbb R\}$ is not a closed subset of $\mathbb A^2_{\mathbb R}$.

# 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 $V\subseteq \mathbb A^n_k$, let $I(V)$ be the set of all $f\in A$ such that $f(P) = 0$ for all $P\in V$.

Immediately we have:

Lemma 1.

For any subset V of $\mathbb A^n_k$, $I(V)$ is a radical ideal of the ring of polynomials A.

Proof

Clearly $0\in I(V)$. Next if $f, g\in I(V)$ then for any $P\in V$ we have $(f-g)(P) = f(P) - g(P) = 0$. Similarly if $f\in I(V)$ and $g\in A$, then for any $P\in V$ we have $(fg)(P) = f(P)g(P) = 0$ so we have $fg\in I(V)$.

Finally, if $f \in A$ satisfies $f^n \in I(V)$ then for any $P\in V$ we have $f(P)^n = 0$ and so $f(P) = 0$. This gives $f\in I(V)$. ♦

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 $S, S' \subseteq A$ and $V, V' \subseteq \mathbb A^n_k$.

• $V \subseteq V' \implies I(V) \supseteq I(V')$.
• $S \subseteq S' \implies V(S) \supseteq V(S')$.
• $S \subseteq IV(S)$.
• $V\subseteq VI(V)$.
• $VIV(S) = V(S)$.
• $IVI(V) = I(V)$. [Apologies for the awkward notation!]

Note

The fifth claim says that if V is a closed subset of $\mathbb A^n_k$, then VI takes V back to itself.

Proof

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

First claim: if $V\subseteq V'$, then for any $f \in I(V')$, any $P\in V$ also lies in V’ so $f(P) = 0$ and we have $f \in I(V)$.

Third claim: if $f\in S$, we take any $P\in V(S)$; by the definition of V(S) we have $f(P) = 0$. Hence f lies in I(V(S)).

Fifth claim: we have $VI(V(S)) \supseteq V(S)$ by the fourth claim. On the other hand since $IV(S) \supseteq S$ by the third claim we have $VIV(S) \subseteq V(S)$ 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. $k = \mathbb C$). The above correspondence gives a bijection between:

• closed subsets $V\subseteq \mathbb A^n_k$,
• radical ideals of $A = k[X_1, \ldots, X_n]$.

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.

This entry was posted in Advanced Algebra and tagged , , , , , , , . Bookmark the permalink.