Algebraicity of compact Riemann surfaces

Question

I am taking a course in Riemann surfaces, in which the classical result about algebraicity of compact riemann Surfaces has been proven. However, I think there are some dubious points in the proof.

Here is the outline: let X be our compact connected Riemann surface, let g be its genus. If D is a divisor whose degree d satisfies $d\geq2g+1$, then the associated map $\varphi_{|D|}:X\rightarrow \mathbb{P}^N(\mathbb{C})$, with $N=d-g$, is a holomorphic embedding, so $Y=\varphi_{|D|}(X)$ is a closed submanifold of dimension $1$ in $\mathbb{P}^N$.

Now consider the graded $\mathbb{C}$-algebra $R(D):=\bigoplus_{n\geq0} H^0(X,\mathcal{O}_X(n\cdot D))$, with multiplication given by the product of global sections. Using what has been presented as the Castelnuovo-Mumford theorem, under these conditions there exists, for every $n\geq0$, a natural surjection $$ H^0(X,\mathcal{O}_X(D))^{\otimes n}\rightarrow H^0(X,\mathcal{O}_X(n\cdot D))$$ which sends $s_1\otimes\dots\otimes s_n$ to $s_1\dots s_n$. Indeed, since the product of sections is commutative, we obtain a surjection $$ \mathrm{Sym}^n(H^0(X,\mathcal{O}_X(D)))\rightarrow H^0(X,\mathcal{O}_X(n\cdot D)) $$ whence a surjective map $$ \mathbb{C}[z_0,\dots,z_N]_n\rightarrow H^0(X,\mathcal{O}_X(n\cdot D))$$ where $\mathbb{C}[z_0,\dots,z_N]_n\cong\mathrm{Sym}^n(H^0(X,\mathcal{O}_X(D)))$ is the space of homogeneous polynomials of degree $n$. Assembling these maps together, we obtain a surjective homomorphism of algebras $$ \alpha:\mathbb{C}[z_0,\dots, z_N]\rightarrow R(D)$$ whose kernel $I$ is a homogeneous ideal by construction of $\alpha$; therefore we can define the projective variety $Z=V(I)\subseteq\mathbb{P}^N$. Furthermore, since $R(D)$ is a domain, $I$ is prime and $Z$ is irreducible. Using the isomorphism $\mathbb{C}[z_o,\dots,z_N]/I\cong R(D)$ and applying the Riemann-Roch theorem, we can see that the Hilbert polynomial of $Z$ is $p_Z(t)=\mathrm{deg}(D)t+1-g$, so by the Hilbert-Serre theorem $Z$ has dimension $1$ (here by dimension I mean the supremum of the lengths of chains of irreducible subvarieties). Now a straightforward calculation shows $Y\subseteq Z$. Furthermore, $Y$ is irreducible in the Zariski topology and has dimension $1$ being the image of $X$ via an embedding (here I mean, a priori, the dimension as a complex manifold). This implies $Y=Z$.

Here are my doubts:

• Why is $Y$ irreducible? My guess: suppose there exist homogeneous polynomials $F_1,\dots,F_r, G_1,\dots, G_s$ such that $Y=(Y\cap V(F_1,\dots,F_r))\cup (Y\cap V(G_1,\dots, G_s))$. If $Y\cap V(F_1,\dots,F_r)$ is a proper non empty subset of $Y$, its complement in $Y$ is open and contained in every $Y\cap V(G_j)$. Since the restrictions of the $G_j$ on $Y$, which is a Riemann surface, are holomorphic, they must be identically zero on $Y$, hence $Y\subset V(G_1,\dots,G_s)$.
• $Y$ is certainly a $1$-dimensional complex manifold, but why is it $1$-dimensional as a topological space with the Zariski topology? My guess: let $C$ be a Zariski closed subset of $\mathbb{P}^N$: then $Y\cap C$ is either $Y$ or a finite set of points, by holomorphicity of the functions defining $C$ and compactness and "one-dimensionality" of $Y$. If $Y\cap C$ is a proper Zariski irreducible subset of $Y$, then it must be a point.
• Assuming the two previous points, why does the equality of the dimensions imply $Y=Z$? It would certainly be true if $Y$ were Zariski closed in $Z$, but then $Y$ would be Zariski closed in $\mathbb{P}^N$, which is exactly what we want. In general, a projective variety has the same dimension of any of its proper open subsets.

Thank you kindly to anyone who will have the patience to read and help!