This is a step in a proof of a longer theorem (Hartshorne IV.4.6). In this situation we have an elliptic curve $X$ and a fixed point $P \in X$ so that the linear system $|3P|$ is very ample and of dimension $2$, embedding $X$ into $\mathbb{P}^2$.
In the proof, we take a basis $1, x, y \in \Gamma(X, \mathcal{O}(3P))$ which satisfy $$y^2 = x(x - 1)(x - \lambda)$$ in the homogeneous coordinate ring of the embedding $\bigoplus_{n \in \mathbb{Z}} \Gamma(X, \mathcal{O}(3nP))$, which expresses the image of $X$ as the zero locus of $zy^2 = x(x - z)(x - \lambda z)$.
It is then claimed that $P \mapsto [0 : 1 : 0]$ under this embedding, since $1, x, y$ are chosen so that they have poles at $P$, which I don't understand. How can we calculate the image of $P$ and verify this?
Thanks!
I summarize how I tried to understand it below.
Explicitly, this map will be $Q \mapsto [1(Q): x(Q): y(Q)]$ so we just need to show $1(P) = y(P) = 0$. On the other hand, in the proof we chose $1, x$ to be so that they form a basis of $\Gamma(X, \mathcal{O}(2P)) \subset \Gamma(X, \mathcal{O}(3P))$. Moreover $Q \mapsto [1(Q) :x(Q)]$ is a map to $\mathbb{P}^1$. This is also the linear projection of $X$ onto the $x$-axis.
I'm also not sure why under this projection, $P \mapsto [0: 1]$ either since we could apply an automorphism of $\mathbb{P}^1$ to use the same linear system $|2P|$, to send $P$ to anything, right? I assume the conditions (the equation in particular) on the sections we chose would impose some kind of obstruction to this, but I don't see it. Perhaps one could argue that the the morphism to $\mathbb{P}^1$ is induced by the field extension $$k(x) \subset k(x)[y]/(y^2 - x(x - 1)(x - \lambda))$$ for example, but I'm not sure.
The map $X\to\Bbb P^2$ sends $P_0\mapsto [x(P_0):y(P_0):1]$. $x$ is chosen to have a pole of order 2 at $P$, and $y$ is chosen to have a pole of order $3$ at $P$. In order to make $[x(P):y(P):1]$ be represented by regular functions near $P$ which aren't all zero at $P$, we should multiply by a function $f$ that has a zero of order 3 at $P$. This gives that $P$ is mapped to $[f\cdot x(P):f\cdot y(P): f]$, and since $f\cdot x$ and $f$ have a zero of positive order at $P$ while $f\cdot y$ has neither a zero nor a pole, we have that $P\mapsto [0:1:0]$.
The reason this works is that the local ring at $P$ is a discrete valuation ring, so we can find a function $f$ of the specified valuation, lift it to an open neighborhood of $P$, and then use this to extend our map over $P$ by noting that on the intersection of $D(f)$ and our original domain of definition, $[x:y:1]=[fx:fy:f]$. This is a (very computational!) instance of the curve-to-projective extension theorem, which you can find a fair amoutn of material about both here on MSE and in Vakil's notes (or by googling).