I'm currently studying elliptic curves and have the following problem:

Let E be an elliptic curve over $\mathbb{Q}$ defined by $y^2 +y = x^3 - x$. Let P = (0,0) be a point on the curve.

i) show that $x$ is a uniformizer for $E$ at $P$.

ii) compute $ord_P(y)$

iii) compute the divisor of $y+x^2$

Now I've got the following questions:

i) In general, how do I show that something is a uniformizer? I guess showing that $u(P) = 0$ and every function $f(x,y)$ can be written in the form $ f = u^r*g$ with $r \in \mathbb{Z}$ and $g(P) \neq 0 $ would satisfy the definition. Also showing that $x$ generates the maximal ideal of the local ring $\mathcal{O}_{E,P} = \{ \frac{g}h\in K(E) \ \vert \ h(P) \neq 0 \} \cup \{0\}$ would be another way to do it, but I don't know how to approach this. (Would it be enough, just to show that the line $x$ is not tangent to the curve?)

iii) Here I have to compute the zeroes and poles of $g := y+x^2$ ant then write $div(g)$ according to our definiton $div(g) = \sum v_P(g)P $. But in this part, I really struggle to get the $v_P(g)$-part. Is there a straightforward way to compute this for each $P$?


2 Answers 2

  1. It is enough to show that $x(P) = 0$ and that $x$ is not tangent to the curve. Try to prove that it is enough to do so.
  2. I find it much easier to work with short Weierstrass equations. If you transform $Y = y + \frac{1}{2}$ then the curve equation becomes $E': Y^2 = x^3 - x + \frac{1}{4}$. The point $P$ on $E$ has $y=0$, so on $E'$ it has $Y=\frac{1}{2}$ and its coordinates are $(0,\frac{1}{2})$. We also have $$\operatorname{ord}_{(0,0)}(y) = \operatorname{ord}_{(0,\frac{1}{2})}\left(Y-\frac{1}{2}\right) = 1.$$ In fact $$\operatorname{div}\left(Y-\frac{1}{2}\right) = 1\cdot\left(-1,\frac{1}{2}\right) + 1\cdot\left(0,\frac{1}{2}\right) + 1\cdot\left(1,\frac{1}{2}\right) - 3\cdot(\infty)$$ on $E'$.
  3. Again, we work on $E'$ since it is much easier. We have $$y+x^2 = \left(Y-\frac{1}{2}\right)+x^2 = x^2-\frac{1}{2}+Y.$$ To find $\operatorname{div}(x^2-\frac{1}{2}+Y)$, we take advantage of symmetry: \begin{align*} \operatorname{div}\left(x^2-\frac{1}{2}+Y\right) + \operatorname{div}\left(x^2 - \frac{1}{2} - Y\right) &= \operatorname{div}\left(x^2-\frac{1}{2}+Y\right)\left(x^2 - \frac{1}{2} - Y\right) \\ &= \operatorname{div}\left(x^4 - x^2 + \frac{1}{4} - Y^2\right) \\ &= \operatorname{div}\left(x^4 - x^3 - x^2 + x\right) \\ &= \operatorname{div}(x(x+1)(x-1)^2) \\ &= \operatorname{div}(x) + \operatorname{div}(x+1) + 2\cdot\operatorname{div}(x-1) \end{align*} Explanation: The first equation follows from $\operatorname{div}(fg) = \operatorname{div}(f)+\operatorname{div}(g)$. The second equation is just algebraic manipulation. The third equation comes from substituting $Y^2 = x^3 - x + 1/4$ which is the curve equation for $E'$. The fourth equation is algebraic manipulation. The fifth equation is again $\operatorname{div}(fg) = \operatorname{div}(f)+\operatorname{div}(g)$. From direct calculation, we have \begin{align*} \operatorname{div}(x) &= 1\cdot \left(0,\frac{1}{2}\right) + 1\cdot \left(0,-\frac{1}{2}\right) - 2\cdot(\infty)\\ \operatorname{div}(x-1) &= 1\cdot \left(1,\frac{1}{2}\right) + 1\cdot \left(1,-\frac{1}{2}\right) - 2\cdot(\infty)\\ \operatorname{div}(x+1) &= 1\cdot \left(-1,\frac{1}{2}\right) + 1\cdot \left(-1,-\frac{1}{2}\right) - 2\cdot(\infty) \end{align*} Hence $$ \operatorname{div}\left(x^2-\frac{1}{2}+Y\right) + \operatorname{div}\left(x^2 - \frac{1}{2} - Y\right) = \left(0,\frac{1}{2}\right) + \left(0,-\frac{1}{2}\right) + \left(1,\frac{1}{2}\right) + \left(1,-\frac{1}{2}\right) + \left(1,\frac{1}{2}\right) + \left(1,-\frac{1}{2}\right) + \left(-1,\frac{1}{2}\right) + \left(-1,-\frac{1}{2}\right) - 8\cdot(\infty) $$ To figure out which of these points belongs to $\operatorname{div}(x^2-\frac{1}{2}+Y)$, we just evaluate $x^2-\frac{1}{2}+Y$ at each point. We find that $(0,\frac{1}{2})$, $(1,-\frac{1}{2})$, and $(-1,-\frac{1}{2})$ lie on $x^2-\frac{1}{2}+Y=0$, and $(0,-\frac{1}{2})$, $(1,\frac{1}{2})$, and $(-1,\frac{1}{2})$ do not lie on $x^2-\frac{1}{2}+Y=0$. Hence $$ \operatorname{div}\left(x^2-\frac{1}{2}+Y\right) = \left(0,\frac{1}{2}\right) + 2\cdot\left(1,-\frac{1}{2}\right) + \left(-1,-\frac{1}{2}\right) - 4\cdot (\infty). $$ Translating this answer from $E'$ back to $E$ is left as an exercise to the reader.

Let $A = Q^2 = \operatorname{Spec} Q[x,y]$ and $E = V(y^2 +y - x^3 + x) \subset A$. Then $O_{E,P} = Q[x,y]_{(x,y)}/(y^2 +y - x^3 + x)$. To show that $x$ is the uniformizor, it suffices to show that $y \in xO_{E,P}$ (or the image of $y$ to be precise).

Note that $y^2 +y - x^3 + x = y (y+1) - x (x^2+1)$. Furthermore, the images of $y+1,x^2+1$ are units in $O_{E,P}$. Thus, $xO_{E,P} = yO_{E,P} = (x,y)O_{E,P}$ showing $(i)$, and $(ii)$ follows from this.

I am not sure about (iii). But think you wanted to consider it as a divisor on $E$. Notice that $$(y^2 +y - x^3 + x, y+x^2) = (x^4 - x^2 - x^3 + x, y + x^2) \\=(x(1-x-x^2+x^3), y + x^2) = (x(1-x)^2(1+x), y+x^2)$$ as an ideal in $Q[x,y]$ since $1-x-x^2+x^3 = (1-x)-x^2(1-x) = (1-x)(1-x^2) = (1-x)^2(1+x)$. So $div (y+x^2)$ on $E$ is supported on points corresponding to the prime ideals of the form $$ P = (x,y+x^2 ) = (x,y), Q_1 = (1+x, y+x^2) = (1+x,y+1), Q_2 = (1-x,y + x^2) = (1-x,y+1). $$ I believe you can determine the divisor of $y+x^2$ on $E$.


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