Lemma 4.2 Hartshorne IV The question I have is with regards to the proof given in Hartshorne IV lemma 4.2. Let $X$ be an elliptic curve and $P,Q\in X$ be closed points. One can show that the linear system $|P+Q|$ has dimension 1 and is base-point free, and thus induces a morphism $g:X \rightarrow \mathbb{P}^1$ of degree 2.
My confusion arises from what Hartshorne claimed after: he seem to imply that the each fibre of $g$ is of cardinality two (including ramification points). My understanding of the ‘degree of morphism’ is that if $\deg g =2$, then the dimension of the field extension $[K(X): K(\mathbb{P}^1)]=2$. For me it seems that Hartshorne concluded that each fibre of $g$ must then have a fibre of cardinality 2.
How did Hartshorne used this to conclude it? I have very little idea how to begin and any clues/help given would be greatly appreciated!
 A: $\newcommand{\Frac}{\mathrm{Frac}}$$\newcommand{\Spec}{\mathrm{Spec}}$Here's a down-to-earth way of understanding what's going on.
We have the following simple observation:

Observation: Suppose that we have an inclusion of integral domains
$A\hookrightarrow B$ such that $B$ is a finite free $A$-module. Then,
$$\mathrm{rank}_A(B)=[\Frac(B):\Frac(A)]$$

Proof: Note that we have a natural isomorphism of $\mathrm{Frac}(A)$-algebras
$$B\otimes_A \text{Frac}(A)\cong \mathrm{Frac}(B)$$
Indeed, we have the natural map $B\otimes_A\mathrm{Frac}(A)\to \Frac(B)$coming from the inclusions of $A$-algebras $B\hookrightarrow \Frac(B)$ and $\Frac(A)\hookrightarrow \Frac(B)$. This map is an inclusion since we have that
$$0\to \mathrm{Frac}(A)\to \mathrm{Frac}(B)$$
is an exact sequence of $A$-modules and thus, since $B$ is $A$-flat, this induces an inclusion
$$0\to B\otimes_A\Frac(A)\to B\otimes_A \Frac(B)$$
But, evidently $B\otimes_A\Frac(B)=\Frac(B)$.
Thus, we see that $B\otimes_A\Frac(A)$ is a domain containing $B$ in $\Frac(B)$. It's then a field since it's an integral domain which is finite as a $\Frac(A)$-vector space, and then use the usual argument (e.g. see the bottom of [1]). But, then it's a subfield of $\Frac(B)$ containing $B$, and so equal to $\Frac(B)$.
Since $B$ is a finite free module we see that
$$\mathrm{rank}_A (B)=\dim_{\Frac(A)}(B\otimes_A\Frac(A))=\dim_{\Frac(A)} \Frac(B)=[\Frac(B):\Frac(A)]$$
as desired $\blacksquare$
Why does this help us? Well, note that if $g:C\to D$ is any non-constant map of smooth projective geometrically integral curves over $F$ ($F$ is any field) then $g$ is finite flat. Both of these can be checked over $\overline{F}$, so we assume this. The finiteness maybe requires some small amount of work (e.g. here's an overkill proof: it's proper since $C$ and $D$ are, and quasi-finite since $C$ has the cofinite topology and $g$ is not constant--it follows then from Zariski's main theorem). The flatness is easy since $g$ is surjective (since $g(C)$ is an irreducible closed subset which is not a point) and a surjection of Dedekind schemes is flat (e.g. see [2, Proposition 3.9]).
Thus, we see that if $\Spec(B)$ is an affine open subset of $D$ then $g^{-1}(\Spec(B))=\Spec(A)$ for some affine open subset $\Spec(A)$ of $C$. But, by our assumptions we know that $A$ and $B$ are both integral domains and the map $A\to B$ is injective (since $\Spec(A)\to\Spec(B)$ is dominant). Moreover, by further shrinking we may assume that $B$ is a free $A$-module (e.g. since $B$ is finite flat it's locally free over $A$--for example see [3, Tag02KB]) Thus, by our lemma we have that
$$\mathrm{rank}_A(B)=[\Frac(B):\Frac(A)]$$
But, note that $\Frac(B)=K(D)$ and $\Frac(A)=K(C)$. So, we see that
$$\mathrm{rank}_A(B)=[K(D):K(C)]$$
But, if $p$ is any point of $\Spec(B)$, corresponding to a prime $\mathfrak{p}$ of $B$, then we know that
$$g^{-1}(p)=\Spec(B\otimes_A A_\mathfrak{p}/\mathfrak{p})$$
So then, it's easy to see that
$$\# g^{-1}(p)\leqslant \dim_{A_\mathfrak{p}/\mathfrak{p}}(B\otimes_A A_\mathfrak{p}/\mathfrak{p})=\mathrm{rank}_A(B)=[K(D):K(C)]$$
So, in summary, the above shows that if you have a non-constant map of curves $g:C\to D$ then the size of the fiber (let's say over a closed point) is bounded by $[K(D):K(C)]$ and, in fact, if you define 'size' to mean dimension of global sections over $F$ (where we assume that $F$ is algebraically closed for simplicity) is precisely $[K(D):K(C)]$--in other words, if you cound fiber size 'with multiplicity' account for nilpotents (i.e. ramification of $g$) then the fiber size is precisely $[K(D):K(C)]$.
[1] Homomorphism of $k$-algebras induce homomorphism of maximal spectrum
[2] 刘擎 (Qing Liu), 2002. Algebraic geometry and arithmetic curves (Vol. 6). Oxford University Press on Demand.
[3] Various authors, 2020. Stacks project. https://stacks.math.columbia.edu/
