# computing uniformizer and principal divisor of elliptic curve

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 $$P$$?

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 $$E$$.