Proving algebraicity of compact Riemann surfaces without Chow's theorem I am trying to write a report for a complex analysis class where I prove Riemann-Roch and apply it to prove algebraicity of compact Riemann surfaces. While writing this, I found that Riemann-Roch quite nicely yields a very ample line bundle and hence a holomorphic embedding into projective space. However, it was not so obvious to me why the image of a holomorphic embedding $f: X \longrightarrow \mathbb{CP}^N$ (for $X$ a compact Riemann surface) is actually an algebraic variety.
The resources I have consulted tend to resolve this either by not mentioning algebraicity (as Forster does) or appealing to Chow's theorem (as Griffiths and Harris does). Also, Miranda does a strange thing where an "algebraic curve" is not the zero set of polynomials, but a Riemann surface whose meromorphic functions separate points and tangents. I find the algebraicity of compact Riemann surfaces to be a very striking and beautiful fact, but Chow's theorem is quite advanced for what I am intending this report to be, and I don't believe I can include a proof of it. Furthermore, Chow's theorem applies very generally to any analytic subvariety of $\mathbb{CP}^N$. As I am only interested in the case of $\dim X = 1$, I was wondering if a more elementary argument for algebraicity is out there.
In attempting to find such an argument, I tried to compute the Zariski closure of $im(f)$ and use that as a candidate, but I was not able to show equality. Specifically, I wrote $f$ as $x \mapsto [f_0(x) : \dots : f_N(x)]$. Given that the transcendence degree of the field of meromorphic functions $\mathcal M(X)$ over $\mathbb C$ is $1$, we can ensure that the map $\mathbb C[x_0, \dots, x_N] \longrightarrow \mathcal{M}(X)$ sending $x_i \mapsto f_i$ has a nontrival kernel $\mathfrak p$ of height $n$. As such, letting $Z(\mathfrak p)$ be the variety in $\mathbb{CP}^N$ defined by $\mathfrak p$, we can show that $im(f) \subseteq Z(\mathfrak p)$. By dimension considerations, I would expect to have equality $im(f) = Z(\mathfrak p)$, but it's not at all obvious to me why the reverse inclusion must hold. $Z(\mathfrak p)$ will be the Zariski closure of $im(f)$, so what I'm asking is of course tantamount to showing that $im(f)$ is Zariski closed.
I was also recommended to try using Riemann-Roch to get a finite covering $X \longrightarrow \mathbb{CP}^1$ and use this to prove algebraicity. I suspect that I am to analyze the associated map on sheaves of holomorphic functions $f^*: \mathcal O_{\mathbb{CP}^1} \longrightarrow \mathcal O_X$ and exploit finiteness as well as the fact that $\mathcal O_{\mathbb{CP}^1}$ is the analytification of an algebraic sheaf. But this too is not clear to me.
Is there any known elementary argument I can use?
 A: Given an embedding $X\to \Bbb P^n$, it's possible to construct polynomial equations for the image.

Theorem (Narasimhan 9.6.2): Let $X$ be a compact Riemann surface and $f$, a nonconstant meromorphic function on $X$. Then, there exists an integer $p>0$ with the following property:
For any $F$ meromorphic on $X$, there exist rational functions $a_1,\dots,a_p$ (in one variable) such that $$(F(x))^p+a_1(f(x))(F(x))^{p-1}+\cdots+a_p(f(x))=0 \quad \text{on } X.$$

The idea of the proof is that the map $f:X\to\Bbb P^1$ is a covering map with fibers of constant size $p$ over some set $U\subset \Bbb P^1$ with finite complement in $\Bbb P^1$. Let $a_k(z)$ be the $k^{th}$ elementary symmetric polynomial in the values of $F$ along the $p$ points in $f^{-1}(z)$. Then the $a_k(z)$ are meromorphic functions on $\Bbb P^1$ and hence rational, and the relation above holds everywhere by continuity.
Letting $f$ and $F$ vary among the meromorphic functions you used to embed $X$ in $\Bbb P^n$, after clearing denominators you get polynomial equations for the image. Alternatively, a little extra work with the above result shows that the field of meromorphic functions on $X$ is a finitely generated field of transcendence degree one over $\Bbb C$, which is exactly an algebraic function field. By the standard theory of such objects, it's equivalent to a smooth projective curve over $\Bbb C$ where the (closed) points correspond to (nontrivial) discrete valuations trivial on $\Bbb C$, and this correspondence between points and valuations also holds on the Riemann surface side.

Narasimhan, Raghavan; Nievergelt, Yves, Complex analysis in one variable., Boston, MA: Birkhäuser. xiv, 381 p. (2001). ZBL1009.30001.
