[proofplan] The two statements are proved separately. For (1): the fiber $f^{-1}(q)$ is a closed subvariety of $C$, and it is proper because $f$ is nonconstant — the entire image of $C$ under $f$ is not collapsed to $q$. By the theorem that [proper subvarieties of a curve are finite](/theorems/2166), $f^{-1}(q)$ is a finite set. For (2): the pullback map $f^*: \mathcal{O}_D(\eta) \to \mathcal{O}_C(\eta)$ is well-defined and injective (because $f$ is dominant, which follows from being nonconstant on an irreducible curve), so we identify $\mathcal{O}_D(\eta)$ with a subfield of $\mathcal{O}_C(\eta)$. Pick any nonconstant $t \in \mathcal{O}_D(\eta)$; both function fields have transcendence degree $1$ over $k$ ([the function field of an irreducible variety has transcendence degree equal to its dimension](/theorems/???)), so the tower $k \subset k(t) \subset \mathcal{O}_D(\eta) \subset \mathcal{O}_C(\eta)$ has $\mathcal{O}_C(\eta)/k(t)$ algebraic — and since $\mathcal{O}_C(\eta)$ is finitely generated over $k$ (hence over $k(t)$), an algebraic finitely generated extension is finite. Hence $\mathcal{O}_C(\eta)/\mathcal{O}_D(\eta)$ is finite. [/proofplan] [step:Show $f^{-1}(q)$ is a closed proper subvariety of $C$ for each $q \in D$] Fix $q \in D$. The fiber $f^{-1}(q) = \{p \in C : f(p) = q\}$ is the preimage of the singleton $\{q\}$ under the morphism $f: C \to D$. The singleton $\{q\}$ is closed in $D$ (every point of a quasi-projective variety is closed in the Zariski topology, since $D$ is a Hausdorff-style $T_1$ space in the Zariski sense — equivalently, the maximal ideal corresponding to $q$ defines $\{q\}$ as a closed subscheme). Since morphisms of varieties are continuous in the Zariski topology, $f^{-1}(q)$ is a closed subset of $C$, equipped with its induced reduced subvariety structure. We claim $f^{-1}(q) \subsetneq C$. If instead $f^{-1}(q) = C$, then $f(p) = q$ for every $p \in C$, so $f$ is the constant morphism with value $q$. This contradicts the hypothesis that $f$ is nonconstant. Hence $f^{-1}(q)$ is a proper closed subvariety of $C$. [/step] [step:Conclude $f^{-1}(q)$ is finite via the proper-subvarieties theorem] By [Proper Subvarieties of a Curve Are Finite](/theorems/2166), every proper closed subvariety of an irreducible curve is a finite union of points. We verify the hypotheses: \begin{itemize} \item $C$ is a curve (irreducible variety of dimension $1$), by assumption. \item $f^{-1}(q) \subsetneq C$ is a closed proper subvariety, by Step 1. \end{itemize} The conclusion is that $f^{-1}(q)$ has only finitely many points, i.e. $f^{-1}(q)$ is a finite set, proving (1). [/step] [step:Set up the pullback $f^*: \mathcal{O}_D(\eta) \to \mathcal{O}_C(\eta)$ and verify it is an injective field homomorphism] For a morphism of irreducible varieties $f: C \to D$, the pullback on rational functions is defined as follows. A rational function $\psi \in \mathcal{O}_D(\eta)$ is represented by a regular function on some Zariski-open dense $V \subseteq D$. The pullback $f^*\psi$ is, formally, the rational function on $C$ represented by $\psi \circ f$ on the Zariski-open subset $f^{-1}(V) \subseteq C$ — provided $f^{-1}(V)$ is nonempty. We need $f^{-1}(V) \neq \varnothing$ for every nonempty open $V \subseteq D$, i.e. $f$ is **dominant** (its image is Zariski-dense in $D$). The image $f(C) \subseteq D$ is irreducible (the image of an irreducible set under a continuous map is irreducible) and contains more than one point (else $f$ would be constant). Since $D$ is itself an irreducible curve (dimension $1$, irreducible), the only irreducible closed subsets of $D$ are points and $D$ itself. The Zariski closure $\overline{f(C)}$ is an irreducible closed subset of $D$ containing more than one point (image of $C$ under a nonconstant map is not a single point; so the image, hence its closure, is not a single point), so $\overline{f(C)} = D$. Thus $f$ is dominant, and the pullback $f^* : \mathcal{O}_D(\eta) \to \mathcal{O}_C(\eta)$ is well-defined. The map $f^*$ is a field homomorphism: it preserves addition and multiplication because composition with $f$ is functorial, and $f^*(1) = 1$. Any nonzero ring homomorphism between fields is injective (its kernel is a proper ideal of a field, hence the zero ideal). So $f^*$ is injective, and we may regard $\mathcal{O}_D(\eta)$ as a subfield of $\mathcal{O}_C(\eta)$ via $f^*$. [/step] [step:Build a tower $k \subset k(t) \subset \mathcal{O}_D(\eta) \subset \mathcal{O}_C(\eta)$ and locate $\mathcal{O}_C(\eta)$ over $k(t)$] Both $\mathcal{O}_C(\eta)$ and $\mathcal{O}_D(\eta)$ are function fields of irreducible curves over $k$. By [Function Field is Finitely Generated](/theorems/2142), each is a finitely generated field extension of $k$. The transcendence degree of the function field of an irreducible variety equals the dimension of the variety, so \begin{align*} \operatorname{tr.deg}_k \mathcal{O}_C(\eta) = \dim C = 1, \qquad \operatorname{tr.deg}_k \mathcal{O}_D(\eta) = \dim D = 1. \end{align*} Pick any element $t \in \mathcal{O}_D(\eta) \setminus k$. We claim such $t$ exists: since $\mathcal{O}_D(\eta)$ has transcendence degree $1 > 0$ over $k$, it is not algebraic over $k$, so there exists $t \in \mathcal{O}_D(\eta)$ that is transcendental over $k$ (and in particular $t \notin k$). Fix such $t$. The chain of subfields is now \begin{align*} k \subset k(t) \subset \mathcal{O}_D(\eta) \subset \mathcal{O}_C(\eta). \end{align*} The middle inclusion $k(t) \subset \mathcal{O}_D(\eta)$ uses $t \in \mathcal{O}_D(\eta)$ and the universal property of $k(t)$ as the smallest subfield containing $k$ and $t$. The third inclusion is via the injection $f^*$ of Step 3. We compute degrees: \begin{itemize} \item $\operatorname{tr.deg}_k k(t) = 1$ (because $t$ is transcendental over $k$, so $k(t)/k$ is purely transcendental of transcendence degree $1$). \item $\operatorname{tr.deg}_k \mathcal{O}_D(\eta) = 1$, so $\mathcal{O}_D(\eta) / k(t)$ is algebraic (transcendence degree is additive in towers, and the difference is $1 - 1 = 0$). \item $\operatorname{tr.deg}_k \mathcal{O}_C(\eta) = 1$, so $\mathcal{O}_C(\eta) / k(t)$ is algebraic similarly. \end{itemize} [guided] The tower-of-fields argument is the standard technique for showing extensions of function fields of irreducible varieties of the same dimension are finite. Why does it work? The key invariant is the **transcendence degree**: for a finitely generated field extension $K/k$, $\operatorname{tr.deg}_k K$ counts the size of any maximal algebraically independent subset of $K$ over $k$ (Krull's transcendence basis theorem says this is well-defined). For function fields of irreducible varieties, this number equals the geometric dimension. A morphism between varieties of the same dimension cannot collapse the dimension — and it does not, on the level of function fields, because the pullback of rational functions is injective (the function field has more information than the morphism alone could destroy). Concretely: $C$ and $D$ are both curves, so both have $\operatorname{tr.deg}_k = 1$. Pick a transcendental element $t \in \mathcal{O}_D(\eta) \setminus k$ (such $t$ exists because the function field is not algebraic over $k$). Then $k(t)$ is a copy of the rational function field, of transcendence degree $1$ over $k$. Both $\mathcal{O}_D(\eta)$ and $\mathcal{O}_C(\eta)$ contain $k(t)$ (the latter via $f^*$), and both have transcendence degree $1$ over $k$. So both are algebraic extensions of $k(t)$. This is the hard part of the argument done. The remaining work is to upgrade "algebraic" to "finite". This requires "finitely generated", which is where the assumption of [Function Field is Finitely Generated](/theorems/2142) comes in. An algebraic finitely generated extension is automatically finite (in fact, generated by finitely many algebraic elements, each contributing a finite multiplicative factor to the degree). [/guided] [/step] [step:Promote algebraic to finite using finite generation] We show $\mathcal{O}_C(\eta) / \mathcal{O}_D(\eta)$ is finite. By [Function Field is Finitely Generated](/theorems/2142), $\mathcal{O}_C(\eta)$ is a finitely generated field extension of $k$. Concretely, $\mathcal{O}_C(\eta) = k(\alpha_1, \ldots, \alpha_r)$ for some elements $\alpha_1, \ldots, \alpha_r \in \mathcal{O}_C(\eta)$. Since $k(t) \supset k$, this also gives \begin{align*} \mathcal{O}_C(\eta) = k(t)(\alpha_1, \ldots, \alpha_r), \end{align*} i.e. $\mathcal{O}_C(\eta)$ is finitely generated as a field extension of $k(t)$. By Step 4, $\mathcal{O}_C(\eta) / k(t)$ is algebraic. Each generator $\alpha_i$ is therefore algebraic over $k(t)$, with some finite degree $d_i := [k(t)(\alpha_i) : k(t)]$. Building up the tower \begin{align*} k(t) \subset k(t)(\alpha_1) \subset k(t)(\alpha_1, \alpha_2) \subset \cdots \subset k(t)(\alpha_1, \ldots, \alpha_r) = \mathcal{O}_C(\eta), \end{align*} each successive extension is generated by a single algebraic element over the previous field, hence finite of degree at most $d_i$ (the degree may drop if $\alpha_i$ already lies in the previous field, but cannot exceed $d_i$). By the multiplicativity of degrees in towers, \begin{align*} [\mathcal{O}_C(\eta) : k(t)] \leq d_1 d_2 \cdots d_r < \infty. \end{align*} Now use the tower $k(t) \subset \mathcal{O}_D(\eta) \subset \mathcal{O}_C(\eta)$ from Step 4. Multiplicativity of degrees, \begin{align*} [\mathcal{O}_C(\eta) : k(t)] = [\mathcal{O}_C(\eta) : \mathcal{O}_D(\eta)] \cdot [\mathcal{O}_D(\eta) : k(t)], \end{align*} together with $[\mathcal{O}_C(\eta) : k(t)] < \infty$, forces both $[\mathcal{O}_C(\eta) : \mathcal{O}_D(\eta)]$ and $[\mathcal{O}_D(\eta) : k(t)]$ to be finite — both are positive divisors of a finite product. In particular, \begin{align*} [\mathcal{O}_C(\eta) : \mathcal{O}_D(\eta)] < \infty. \end{align*} A finite extension of fields is automatically algebraic (every element of a finite extension is algebraic, because $1, \alpha, \alpha^2, \ldots$ cannot all be linearly independent over a finite-dimensional vector space). This proves (2). [/step]