MATH 470 Lecture 22

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Algebraic Curves

Out of 52 fields medals , 25 have gone to work in or around algebraic geometry

Newton Polygons

If $K$ is a field (e.g. $\mathbb {Q}$, $\mathbb {R}$, $\mathbb {C}$, $\mathbb {Z} /19\mathbb {Z}$)

$K$ adjoint $x$ and $y$ (written $K[x,y]$ is the set of polynomials in the two variables $x$ and $y$ with coefficients in $K$.

For any polynomial $f\in K[x,y]$, written $f(x,y)=\sum _{(a_{1},a_{2})\in \mathbb {Z} ^{2}}c_{a}\,x^{a_{1}}\,y^{a_{2}}$

We define the support of $f$ to be $\mathrm {supp} (f)=\{(a_{1},a_{2})\mid c_{a}\neq 0\}$

For example, $x+y-3=x^{1}\,y^{0}+x^{0}\,y^{1}-3x^{0}\,y^{0}$ has support $\{(1,0),(0,1),(0,0)\}$

If we plot these as points on a graph, this forms the Newton polygon, a convex hull of $\mathrm {supp} (f)$

Algebraic Curves

An algebraic curve over $K$ ($/K$) is just the zero set in $K^{2}$ of a non-constant polynomial $f\in K[x,y]$.

For example, a circle represents the rational roots of $f(x,y)=x^{2}+y^{2}-1$.

what about $y^{2}-(x+1)$ over $\mathbb {Z} /5\mathbb {Z}$? The squares mod 5 are {0, 1, 4}:

$x$	$x+1$	$y$?
0	1	±1
1	2	no
2	3	no
3	4	±2
4	0	0

Singularities

$\zeta =\left(\zeta _{1},\zeta _{2}\right)$ is a singularity (or degeneracy, singular point, or degenerate point) for the curve defined by $f$ iff

$f(\zeta _{1},\zeta _{2})=\left.{\frac {\partial f}{\partial x}}\right|_{\left(\zeta _{1},\zeta _{2}\right)}=\left.{\frac {\partial f}{\partial y}}\right|_{\left(\zeta _{1},\zeta _{2}\right)}=0$

For example, the curve $f(x,y)=x^{2}+y^{2}-1$ represents a circle of radius 1 centered at the origin.

Then ${\frac {\partial f}{\partial x}}=2x$ and ${\frac {\partial f}{\partial y}}=2y$, so the only singularity would be at $(0,0)$, which is not on the curve. Therefore, the curve is smooth, (i.e. no singularities)

Another example: $y^{2}=x^{3}$ (rewritten as $y^{2}-x^{3}$) has ${\frac {\partial f}{\partial x}}=-3x^{2}$ and ${\frac {\partial f}{\partial y}}=2y$, which are both 0 at $(0,0)$, and that is a point on the curve!

Elliptic Curve

We call the zero set $C$ of $f\in K[x,y]$ an elliptic curve iff

1. the Newton polygon of $f$ has exactly one lattice point in its interior.
2. $C$ is smooth.

For example, $y^{2}+a_{1}\,x\,y-(a_{2}\,x^{3}+a_{3}\,x^{2}+a_{4}\,x+a_{6})$ (Weierstrass Normal Form) has newton polygon

Now for most choices of $a_{i}$, you get an elliptic curve!

For example, ${\vec {a}}=(0,1,-6,11,\mathrm {NULL} ,-6)$ gives a smooth curve:

Recall that $y=a\,x^{2}+b\,x+c$ has 0, 1, or 2 real roots, and the discriminant ($b^{2}-4a\,c$) tells you this.

For cubics, we also have a determinant. A simplified version for $y=x^{3}+b\,x+c$ is $4b^{3}+27c^{2}$ (note this is not an elliptic curve since its Newton polygon has no interior points)

FACT: $y^{2}=x^{3}+b\,x+c$ defines a smooth curve (i.e. an elliptic curve) iff the descriminant is nonzero. (this is what "most" meant)

Alternate Notation: Edwards Normal Form

$x^{2}+y^{2}-a^{2}+b^{2}\,x^{2}\,y^{2}$

Newton polygon is a square { (0,0), (2,0), (2,2), (0,2) }, and has exactly one lattice point on its interior, namely (1,1)

FACTs:

1. Any elliptical curve, after a change of variables, can be put in Weierstrass Normal Form (or Edwards Normal Form)
2. Elliptic curve addition is much faster for curves in Edwards Normal Form.