Equivalence between two definitions of compactifications of an elliptic curve Suppose we have the curve $y^2=h(x)=x(x-1)(x-2)$ in $\mathbb{C}^2$. Then this is clearly not compact and I saw two definitions for compactifying this and I want to prove that they are equivalent.
Method $1$: Consider the homogenization of the curve given by $y^2z=x(x-z)(x-2z)$. Now just take the zero set in $\mathbb{C}P^2$. Now this is compact since it is a closed subset of a compact set.
Method 2: This is the method I saw in the book "Algebraic curves and Riemann surfaces" by Rick Miranda. The definition used here is two take two patches and glue them by an isomorphism. 
So the curve we consider is $w^2=z^4h(\frac{1}{z})=z(1-z)(1-2z)$. Now we have an open set $V=\{(z,w): z \neq 0\}$ and an open set $U=\{(x,y): x \neq 0\}$ and they are isomorphic via a map $\phi(z,w)=(\frac{1}{x},\frac{y}{x^2})$. Now wee can glue these curves using this isomorphism.
Now I am really not sure if these two definitions are equivalent. Any help is appreciated. Thanks.
 A: Yes, both constructions are equivalent, in the sense that they give isomorphic projective curves containing your affine curve. Maybe you can show directly that the explicit map from one curve to the other, defined as the identity on the original affine curve and sending one point at infinity to the other, is an isomorphism of Riemann surfaces. I will follow another route.
Key idea: it is not surprising that both constructions are equivalent, because compactifications of algebraic curves are unique. This is a simple consequence of the following theorem.
Theorem. Let $k$ be a field, $X$ be a smooth curve over $k$ and $U\subset X$ be an open subset. If $Y$ is a proper variety over $k$ (not necessarily of dimension 1), then any $k$-morphism $f:U \to Y$ extends uniquely to $X$.
The existence follows from the so-called valuative criterion of properness, and the uniqueness form separatedness. See Proposition 4.1.16 in Q. Liu, Algebraic Geometry and Arithmetic Curves, for a more general statement. For your question, the following statement is sufficient (and easy to prove).
Theorem. Let $X$ be a Riemann surface and $U\subset X$ be an open subset such that $X\setminus U$ is finite. If $Y$ is a compact Riemann surface, then any morphism $f:U \to Y$ extends uniquely to $X$.
Now, let $X_1\subset \mathbb{C} P^2$ be the elliptic curve obtained from the method 1 in your question, $X_2$ be the curve obtained from method 2, and $U$ be the original affine curve given by the equation $y^2=x(x-1)(x-2)$.
The first thing you must do is to prove that $X_2$ is smooth; that is, that it really defines a Riemann surface (this is easy, you must only check one point). Then, it follows from the above theorem that the inclusion $U\to X_1$ extends to a unique morphism $f: X_2 \to X_1$.
Now, notice that, if you prove that $X_2$ is compact (i.e. that it indeed defines a compactification of $U$), then by applying the same theorem, you would obtain a unique morphism $g: X_1 \to X_2$ extending the inclusion $U \to X_2$. By uniqueness, $f$ and $g$ are inverses of each other, and you are done.
Now, how to prove that $X_2$ is compact (or proper, in the language of algebraic geometry)? The idea is to define an explicit closed embedding of $X_2$ in some projective space.
Let $i: X_2 \to \mathbb{C}P^3$ be the morphism defined on $U$ by
$$
(x,y)\mapsto (1:x:x^2:y)
$$
and on $V$ by
$$
(z,w)\mapsto (z^2:z:1:w).
$$
Note that these two morphism indeed glue via the transition map $\phi$. It is easy to prove that $i$ is a closed immersion, and thus it identifies $X_2$ with it's image in $\mathbb{C}P^3$, which is by definition a projective curve (thus compact). This finishes the proof.
Let me remark that I didn't guess the morphism $i$. In algebraic geometry jargon, what is happening here is that we are proving that the line bundle $\mathcal{O}(4\cdot [\infty])$ (of meromorphic functions which are holomorphic on $U$ and have a pole of order at most 4 at the point $\infty \in X_2\setminus U$) is very ample. Of course, if you want to prove that $X_2$ is isomorphic to $X_1$, then this must be true because it is for the curve $X_1$, as the usual theory of elliptic curves show (cf. comments below).
One final remark on why this question is interesting. The method 2 is actually how you should compactify hyperelliptic curves in general, i.e. curves of the form $y^2 = f(x)$. The point is that, if $\deg f\ge 4$, then the curve obtained by method 1 will be singular at infinity. It turns out that for $\deg f = 3$ both methods give smooth curves, so in text books on elliptic curves one usually prefers method 1 as it is simpler: it gives a plane curve, instead of a spacial curve.
