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Elliptic curves over the field with four elements (not finished)

algebraic geometry

Luca Leon Happel


March 1, 2022

Table of contents


The Field \(\mathbb{F}_4\) with four elements

The affine plane

We define the affine space \(\mathbb{A}^n_k\) over a ground field \(k\) to be the set of all tuples \((x_1, \dots, x_n)\) with elements in \(k\). In the case of \(n=2\) we call \(\mathbb{A}^n_k\) affine plane. A variety \(V(A)\) for a subset \(A\subset k[X_1, \dots, X_n]\) of polynomials is the set points in \(\mathbb{A}^n_k\) which are zeros for each polynomial in \(A\). We call \(V(A)\) a hypersurface, if \(A=\{f\}\) contains only one element.

We can define a topology on \(\mathbb{A}^n_k\) by defining the open sets to be the sets \(V(f)\) for each \(f\in k[X_1,\dots, X_n]\). I will not go into detail as to why this results in a topology. As an example, we can always construct \(V(f_1) \cup\dots\cup V(f_r) = V(f_1\cdot\dots\cdot f_r)\)

This image shows how solution sets can be united by multiplying the underlying functions. We use the affine real line \(\mathbb{A}^1_\mathbb{R}\) for this illustration. The green function is the product of the blue and red one and as can be seen, the green graph has exactly the same poles as the red and blue graph combined. Furthermore, we define a hypersurface \(V(f)\) to be irreducible if the underlying polynomial \(f\) is irreducible. A hypersurface is reducible if it is not irreducible.

Example of a reducible hypersurface

Consider \(f=(x-y)(x+y)\) which is reducible in \(\mathbb{R}[x,y]\) as it can be factored into the non-units: \((x-y)\) and \((x+y)\).

This can be seen in the graph of \(V(f)\), because \(V(f)\) can be split into two hypersurfaces \(V(x-y)\) and \(V(x+y)\).

Example of an irreducible hypersurface

Consider \(f=x^2+y^2-1\). This defines the circle \(V(f) = S^1\).

Example of an irreducible hypersurface that is an elliptic curve

Another irreducible hypersurface is \(V(y^2-x^3+x)\) in \(\mathbb{A}^2_\mathbb{R}\).

This also happens to define an elliptic curve. Also, note that an irreducible hypersurface need not be a connected space in the sense of topology.

The projective space