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I have an elliptic curve $E$ over $\mathbb{F}_{11}$ defined by $y^2=x^3+4x$ with the point at infinity $\mathcal{O}$

I have a divisor of $E$, defined by $$D=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right]$$

As $\text{deg}(D)=0$ and $\text{sum}(D)=\infty$ then $D$ is the divisor of a function - we want to find this function

I am given the following: \begin{align}\text{div}(y-2x)&=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right] \end{align}

I have already found (in Elliptic Curve and Divisor Example help (Step 1)) that $$D = \left[(2,-4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]$$

I am then told that we can write

$$[(4,5)]+[(6,3)]=[(2,4)]+[\mathcal{O}]+\text{div}\left(\frac{y+x+2}{x-2}\right)$$

And thus write $$D=[(2,-4)]+\text{div}\left(\frac{y-2x}{x-2}\right)+[(2,4)]+\text{div}\left(\frac{y+x+2}{x-2}\right)-2[\mathcal{O}]$$

However I am struggling to see how we can find the value of $$\text{div}\left(\frac{y+x+2}{x-2}\right)$$ from what we already know

I can find that \begin{align}\left[(4,5)\right]+\left[(6,3)\right] &=\left[(2,4)\right]+\left[\mathcal{O}\right]+\text{div}\left(\frac{y+x+2}{x-2}\right)\\ &=\left[(2,4)\right]+\left[\mathcal{O}\right]+\text{div}(y+x+2)-\text{div}(x-2)\\ &=\left[(2,4)\right]+\left[\mathcal{O}\right]+\text{div}(y+x+2)-\left(\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right]\right)\\ &=\text{div}(y+x+2)-\left[(2,-4)\right]+3\left[\mathcal{O}\right]\\ &\Downarrow\\ \text{div}(y+x+2)&=\left[(4,5)\right]+\left[(6,3)\right]-\left[(2,-4)\right]-3\left[\mathcal{O}\right] \end{align}

but how can I justify this from only knowing \begin{align}\text{div}(y-2x)&=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right] \end{align}


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I think you’ve overthought this. Let’s look closely, if you’ll excuse the sloppiness of my notation: \begin{align} D&=(0,0)+(2,4)+(4,5)+(6,3)-4\Bbb O\\ D - (y-2x)&=-(2,4)+(4,5)+(6,3)-\Bbb O\\ D-(y-2x)+(x-2)&=(2,-4)+(4,5)+(6,3)-3\Bbb O\\ D-(y-2x)+(x-2)&=(y+x+2)\,, \end{align} the last equation being justified because the line $Y=-X-2$ intersects $Y^2=X^3+4X$ in the three points $(2,-4)$, $(4,5)$, and $(6,3)$. Thus $$ D=\left(\frac{(y+x+2)(y-2x)}{x-2}\right)\,. $$

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  • $\begingroup$ Thanks, this makes sense. I think the paper I'm reading is just going about it a slightly convoluted way $\endgroup$
    – lioness99a
    Commented Mar 6, 2017 at 9:14

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