You might like Griffiths's proof of Riemann-Hurwitz (Introduction to Algebraic Curves page 91, theorem (8.5) ) for a morphism $f:X\to Y$ .
It is very natural: he starts from a meromorphic differential form $\omega \in \Omega(Y)$ (whose existence he provisionally admits ) , lifts it to $f^*\omega \in \Omega(X)$, then computes and compares the divisors $div(f^*\omega), f^*(div(\omega))$ and $R$ locally in coordinates.
Riemann-Hurwitz is not difficult but somewhat explains the slightly confusing result that taking divisors and lifting do not commute for differential forms.
Edit
Stichtenoth gives a proof for fields of any characteristic in Corollary 3.4.13 in the language of function fields, but only under a separability assumption.
The most general formula I'm aware of (but I'm not in the least a specialist) and which does not require separability assumptions is Tate's Genus Formula.
You can find it as Corollary 9.5.20 in Villa's Topics in the Theory of Algebraic Function Fields.
$\def\ZZ{\mathbb{Z}}$I now understand the issue that user10676 is concerned about: For $X$ a reasonable topological space, why is $H^k(X, \ZZ_{\ell})$ isomorphic to $\lim_{\infty \leftarrow n} H^k(X, \ZZ/\ell^n)$? I thought that this would be most easily done from the universal coefficient theorem but, now that I try to write out a proof, I find it easiest to do directly.
Every complex algebraic variety is homeomorphic to a finite simplicial complex. (If I recall correctly, you can find a proof in "Algorithms in Real Algebraic Geometry".) So I'll use simplicial rather that singular cohomology. We work with a fixed triangulation for the rest of this proof.
Let $C^k$ be the group of $k$-co-chains with $\ZZ$ coefficients. Let $Z^k$ be the co-cycles and $B^k$ the co-boundaries. So $H^k(X, \ZZ) = Z^k/B^k$. Let the finitely generated abelian group $H^k(X, \ZZ)$ be $\bigoplus_{1 \leq i \leq r} \ZZ/d_i \oplus \ZZ^{\oplus s}$ where $d_1$, $d_2$, ..., $d_r$ is a sequence of positive integers where $d_1$ divides $d_2$ divides ... divides $d_r$. The following lemma is useful in many computations about cohomology:
Lemma Let $C^{k-1} \to C^{k} \to C^{k+1}$ be a complex of free finitely generated abelian groups. Then we can write this complex as a direct sum of complexes of the following forms:
$$\ZZ \longrightarrow 0 \longrightarrow 0 \quad (1)$$
$$0 \longrightarrow \ZZ \longrightarrow 0 \quad (2)$$
$$0 \longrightarrow 0 \longrightarrow \ZZ \quad (3)$$
$$\ZZ \stackrel{d}{\longrightarrow} \ZZ \longrightarrow 0 \quad (4)$$
$$0 \longrightarrow \ZZ \stackrel{d}{\longrightarrow} \ZZ \quad (5)$$
where $d$ is a positive integer. (The case $d=1$ is permitted.)
Proof First, choose arbitrary bases for $C^{k-1}$ and $Z^k$. The map $C^{k-1} \to Z^k$ is given by an integer matrix. Now change bases so that this matrix is in Smith normal form. In these bases, $C^{k-1} \to Z^k \to 0$ is a sum of complexes of types (1), (2) and (4).
Now, $B^{k+1}$ is a subgroup of the free $\ZZ$-module $C^{k+1}$, so it is free. Thus, we can choose a splitting of $C^k \to C^k/Z^k \cong B^{k+1}$. Let $A$ be the image of this splitting, so $C^k \cong Z^k \oplus A$. Put the map $A \to C^{k+1}$ into Smith normal form as before. Then $0 \to A \to C^{k+1}$ is a direct sum of complexes of the form (3) and (5). (Since $A \to C^{k+1}$ is injective, there are no zeroes on the diagonal of the Smith normal form, and (2) does not occur.) Our original complex is the direct sum of $C^{k-1} \to Z^k \to 0$ and $0 \to A \to C^{k+1}$, so it breaks up as a direct sum of complexes of the five types listed above. $\square$
If we now change coefficients to a new abelian group $G$, we get the same complexes with $\ZZ$ replaced by $G$. All our computations distribute over direct sum, so we just need to work out what each of these five complexes do for the groups $G = \ZZ_{\ell}$ and $G = \ZZ/\ell^n$.
For complexes (1) and (3), we get $0$ in both cases.
For complex (2), we get $\lim_{\infty \leftarrow n} \ZZ/\ell^n$ on one side and $\ZZ_{\ell}$ on the other (with the obvious surjections in the inverse limit).
For complex (4), let $d = \ell^i m$ with $GCD(\ell, m) =1$. For $n \geq i$, we get $\lim_{\infty \leftarrow n} \ZZ/\ell^i$ on one side and $\ZZ/\ell^i$ on the other, where the maps in the inverse limit are the identity for $n > i$.
Complex (5) is the hard one. On the $\ZZ_{\ell}$ side, we get $0$. Write $d = \ell^i m$ as above. For $n \geq i$, we again get $\lim_{\infty \leftarrow n} \ZZ/\ell^i$. However, this time the map in the inverse limit is multiplication by $\ell$ (for $n>i$). This inverse limit is $0$, so we win.
Remark It is probably worth stating the general version of this result: If $R$ is a PID and $C^{\bullet}$ is a complex of free $R$-modules, then $C^{\bullet}$ can be written as a direct sum of complexes of the forms $\cdots \to 0 \to 0 \to R \to 0 \to 0 \to \cdots$
and $\cdots \to 0 \to 0 \to R \stackrel{d}{\longrightarrow} R \to 0 \to \cdots$ for various $d \in r$, and where the nonzero terms can occur in any position. The proof is basically the same: submodules of a free module over a PID are free; surjections to free modules split; Smith normal form is valid over a PID. A lot of modern ring theory can be thought of as classifying the types of complexes which exist over different rings.
Best Answer
They were probably referring to a certain equivalence of categories. Let me (try to) formulate it for $k = \mathbb{C}$ if I manage to recall correctly:
The following categories are equivalent:
i) The category of algebraic curves with dominant rational maps
ii) The category of smooth projective curves with dominant morphisms
iii) The category of riemann surfaces with non-constant holomorphic maps
vi) The category of finitely generated (as $\mathbb{C}$-algebra) field extensions $\mathbb{C} \subset L$ of $\text{trdeg}_{\mathbb{C}}(L) = 1$ with homomorphisms of $\mathbb{C}$-algebras
We can also state vi) as
vi') The category of function fields in one variable with homomorphisms of $\mathbb{C}$-algebras
I think you can find that statement (or a similar one) in Introduction to Compact Riemann Surfaces and Dessins d'Enfants by Girondo and González-Diez.