Confusion about holomorphic differential on elliptic curve Let $(a:b)\in\mathbb{C}P^1$ and look at the elliptic curve $C$ given by $y^2=x^3+a^4x+b^6$. It is well known that on this elliptic curve we have the holomorphic differential $dx/y$. I have two questions about this:


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*I actually don't understand very well why $dx/y$ is a holomorphic differential. What happens when $y=0$? If you have a reference where this is explained well, that would be also nice.

*I also wanted to know how does this differential actually change when we use another representative of $(a:b)$, say $(\lambda a:\lambda b)$? We can outside $y=0$ write $\frac{dx}{\sqrt{x^3+a^4x+b^6}}$ and this changes to $\frac{dx}{\sqrt{x^3+\lambda^4a^4x+\lambda^6b^6}}$. Or does $dx$ itself also change?
 A: If you look at a neighbourhood of a point $(t,0)$, and you project it on the $x$-axis, you see that the projection is $2$-to-$1$.
This means that $(x-t)$ is not a uniformizer for the local ring at that point. The projection map $x : E \to \Bbb P^1(k)$ locally looks just like the map $z \mapsto z^2$ around $0$ (if you work over $\Bbb C$ there are actual charts that give this). So, just like $d(z^2) = 2zdz$ has a single zero, $dx$ has order $1$ at that point.
Meanwhile, if you project that neighbourhood on the $y$-axis, it is then $1$-$1$, and so $y$ is a uniformizer for the local ring, and $dy$ has neither a zero nor a pole. This also means that to get the order of a differential $\omega$ you can just write $\omega = f dy$ and look at the order of the rational function $f$.
Indeed, $dx/y = -2/(3x^2+a^4).dy$, and $-2/(3x^2+a^4)$ has order $0$ at $(t,0)$, so $dx/y$ has order $0$.
Next you need to check that $dx/y$ is holomorphic at the point at infinity.
There is only one point at infinity, where $x$ has a pole of order $2$ (so the projection on $\Bbb P^1(k)$ is $2$-to-$1$ around that point and looks just like $1/z^2$ around $0$, and so $dx$ has a pole of order $3$) and $y$ has a pole of order $3$ (and so $dy$ has a pole of order $4$)
So $dx/y$ has order $-3-(-3) = 0$, and so $dx/y$ is holomorphic everywhere.

When you change the representative, you also have to rescale $x$ and $y$ by some powers of $\lambda$. (you should give different variable names to $x$ and $y$ and write down the actual isomorphism between the two curves or else you will get confused). Then $dx/y$ will also get rescaled by a power of $\lambda$, so nothing much happens.
