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I am reading this paper, specifically Example 2.3 on page 9, and am having a few problems understanding a part of it

We construct an elliptic curve $E$ on $\mathbb{F}_{11}$ defined by $y^2=x^3+4x$ with a point at infinity $\mathcal{O}$

We then find a divisor of $E$ $$D=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right]$$

We can then say that: $$\begin{align}\text{div}(y-2x)&=\left[(0,0)\right]+2\left[(2,4)\right]-3\left[\mathcal{O}\right]\\ \text{div}(x-2)&=\left[(2,4)\right]+\left[(2,-4)\right]-2\left[\mathcal{O}\right]\end{align}$$

We can then replace various elements in the definition of $D$ with these divisors:

$$\begin{align}D&=\left[(0,0)\right]+\left[(2,4)\right]+\left[(4,5)\right]+\left[(6,3)\right]-4\left[\mathcal{O}\right]\\ &=\left[(2,4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(4,5)\right]+\left[(6,3)\right]-3\left[\mathcal{O}\right]\\ &=\left[(2,-4)\right]+\text{div}\left(\frac{y-2x}{x-2}\right)+\left[(2,4)\right]+\text{div}\left(\frac{y+x+2}{x-2}\right)-2\left[\mathcal{O}\right]\\ &=\text{div}\left(x-2\right)+\text{div}\left(\frac{y-2x}{x-2}\right)+\text{div}\left(\frac{y+x+2}{x-2}\right)\end{align}$$

However, what I don't understand are the following steps:

$$\begin{align} D&=\text{div}\left(x-2\right)+\text{div}\left(\frac{y-2x}{x-2}\right)+\text{div}\left(\frac{y+x+2}{x-2}\right)\\ &=\text{div}\left(\frac{(y-2x)(y+x+2)}{x-2}\right)\\ &=\text{div}\left(x^{2}-y\right)\end{align}$$

Can anyone help explain to me how they have come to this conclusion please


This question comes in 3 parts:

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marked as duplicate by Brandon Carter, C. Falcon, Juniven, Leucippus, астон вілла олоф мэллбэрг Mar 3 '17 at 3:03

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  • $\begingroup$ Is the part you're missing that $\text{div}(f\cdot g) = \text{div}(f) + \text{div}(g)$? $\endgroup$ – Callus Mar 1 '17 at 13:11
  • $\begingroup$ That covers the first equality, thanks, but I'm still confused by the second one $\endgroup$ – lioness99a Mar 1 '17 at 13:14
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    $\begingroup$ Multiply it out, and then replace $y^2$ with $x^3+4x$ $\endgroup$ – Callus Mar 1 '17 at 13:16
  • $\begingroup$ Brilliant - thanks. I'd multiplied it out but not realised I could substitute in a value for $y^2$. Makes a lot more sense now! $\endgroup$ – lioness99a Mar 1 '17 at 13:21
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Conclusion:

$$\begin{align} D&=\text{div}\left(x-2\right)+\text{div}\left(\frac{y-2x}{x-2}\right)+\text{div}\left(\frac{y+x+2}{x-2}\right)\\ &=\text{div}\left((x-2)\left(\frac{y-2x}{x-2}\right)\left(\frac{y+x+2}{x-2}\right)\right)\tag{div$(f\cdot g)=$ div$f+$ div$g$}\\ &= \text{div}\left(\frac{(x-2)(y-2x)(y+x+2)}{(x-2)^2}\right)\tag{multiply out}\\ &= \text{div}\left(\frac{(y-2x)(y+x+2)}{x-2}\right)\tag{cancel $(x-2)$}\\ &= \text{div}\left(\frac{-2x^2-xy-4x+y^2+2y}{x-2}\right)\tag{multiply out}\\ &= \text{div}\left(\frac{-2x^2-xy-4x+\left(x^3+4x\right)+2y}{x-2}\right)\tag{substitute $y^2=x^3+4x$}\\ &= \text{div}\left(\frac{-2x^2-xy+x^3+2y}{x-2}\right)\tag{simplify numerator}\\ &=\text{div}\left(x^2-y\right)\tag{simplify fraction} \end{align}$$

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