Definition of trace of a rational differential on an algebraic curve

Question

In Serre's Algebraic Curves and Class Fields, Ch. II, no. 12 we find defined the notion of trace of a rational differential form (a.k.a. meromorphic differential form) on an algebraic curve (in the quoted text here one finds Serre's definition of these forms). I will reproduce here the relevant excerpt (p. 22):

Now let $X$ be any curve. We choose a function $\varphi$ on $X$ which is not constant. If $X^{\prime}$ denotes the projective line $\mathbf{P}_1(k)$, we can consider $\varphi$ as a map $X \rightarrow X^{\prime}$ which is evidently surjective; it makes $X$ a "covering" of $X^{\prime}$, possibly ramified. Putting $E=k\left(X^{\prime}\right)$ and $F=k(X)$, the map $\varphi$ defines an embedding of $E$ in $F$; the field $E$ is thus identified with the field $k(\varphi)$ generated by $\varphi$. Since $X$ has dimension $1,\left[F: F^p\right]=p$; if $F^{\prime}$ denotes the largest separable extension of $E$ contained in $F$, there thus exists an integer $n \geq 0$ such that $F^{\prime}=F^{p^n}$. The extension $F / E$ is separable if and only if $n=0$, in other words if $\varphi \notin F^p$; we assume this from now on.

If $f$ is an element of $F$, its trace in $F / E$ is well defined; it is an element of $E$ which we will write $\operatorname{Tr}_{F / E}(f)$. The operation of trace can be extended to differentials in the following way:

The injection $E \rightarrow F$ defines a homomorphism from $D_k(E)$ to $D_k(F)$; as $d \varphi$ is an $E$-basis of $D_k(E)$ and $\varphi \notin F^p$, this homomorphism is injective and extends to an isomorphism of $D_k(E) \otimes_E F$ with $D_k(F)$. On the other hand, $\operatorname{Tr}_{F / E}: F \rightarrow E$ is $E$-linear; applying this homomorphism to the second term of $D_k(E) \otimes_E F$, we finally deduce an $E$-linear map $$ \operatorname{Tr}_{F / E}: D_k(F) \rightarrow D_k(E) . $$ We can make this more explicit as follows: if $\omega$ is a differential on $X$, we write $\omega=f d \varphi$ and then $$ \operatorname{Tr}_{F / E}(\omega)=\left(\operatorname{Tr}_{F / E}(f)\right) d \varphi. $$ Thus, to every differential $\omega$ on $X$ we have associated a differential $\operatorname{Tr}(\omega)$ on $X'=\mathbf{P}_1(k)$.

My questions are:

1. In the second sentence, what is "a function $\varphi$ on $X$ which is not constant"? Is it just an element of $F\setminus k$?

2. If it is really $\varphi\in F\setminus k$, how does $\varphi$ induce a map $X\to X'$? Why is it "evidently surjective"?

3. Why "$X$ has dimension $1$" implies $[F:F^p]=p$? I have to say: I don't know anything about the Frobenius, and a "frobenius degree" search on Google/MSE doesn't give anything.

4. In third paragraph, first sentence, how does one see $\varphi\not\in F^p$ implies $d\varphi\neq 0$?

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I should say: Serre on his book always assumes $\overline{k}=k$ and with “variety” (in particular, “curve”) he means “variety in the sense of FAC” (locally ringed spaces over $k$ that are locally the maximal spectrum of a finitely generated reduced $k$-algebra, and that are separated).

Answer 1

1. This should be an element of $F$ which isn't algebraic over $k$ - if $k$ is algebraically closed, then it reduces to an element of $F\setminus k$. The point is that the morphism of curves $X\to X'$ should be nonconstant.

2. One may define a rational map $X\to\Bbb P^1$ by $[1:\varphi]$ (or maybe the reverse depending on your preferences) where $\varphi$ is defined, and then extend this to a morphism $X\to \Bbb P^1$ by repeated applications of the curve-to-projective extension theorem. To see surjectivity, irreducible algebraic curves have the topology where the closed sets are finite unions of points and the whole space (this follows quickly from noetherian + irreducible + dimension one). Therefore as the image of an irreducible set under a continuous map is again irreducible and the image of a proper variety is closed, the only options for the image of $X$ in $\Bbb P^1$ are a point or the whole space. Since the map is nonconstant, it can't be a point, so it's the whole space, and the map is surjective.

3. See here. The trick is to consider the following commutative diagram of fields: $$\require{AMScd} \begin{CD} K^p @>>> K\\ @AAA @AAA \\ k(t^p) @>>> k(t) \end{CD}$$ Since $[K:k(t)]$ is finite and $[k(t^p):k(t)]=p$ by direct calculation (assuming $k$ perfect), we must have that $k(t^p)\to K^p$ and $K^p\to K$ are finite as well. But $k(t^p)\to K^p$ is isomorphic to $k(t)\to K$, so $[K^p:K]=p$ as well by multiplicativity of extension degree.

4. Since $F^p\subset F$ is a degree-$p$ extension, it has a basis $\{1,t,t^2,\cdots,t^{p-1}\}$. Now write an arbitrary element $\psi$ in $F$ as $\sum_{i=0}^{p-1} q_it^i$ for $q_i\in F^p$. Then $d\psi = \sum_{i=0}^{p-1} (t^idq_i + q_idt^i)$, and as $dq_i=0$, we have $d\psi = \sum_{i=0}^{p-1} it^{i-1}qdt$. If $dt=0$, then $d\psi=0$ for all $\psi$, or $\Omega_{F/k}=0$, which contradicts the fact that $\Omega_{F/k}\neq 0$ since $k\subset F$ has positive transcendence degree. (ref)