How can function fields have different degrees over the projective line

Question

I'm confused.

Let $X$ be a curve over a field $k$. Let $K= K(X)$ be its function field. Then, $K(X)$ is a field. Each non-constant morphism $f:X\to \mathbf{P}^1_k$ gives a field extension $K(\mathbf{P}^1_k) = k(t) \subset K(X)$ of degree $\deg f$.

How is it possible that $\deg f$ can take infinitely many values? Aren't both fields fixed?

Maybe I'm not understanding the situation fully. A rational function $f$ on $X$ gives a field extension $k(t) \subset k(t,f)$ of degree $\deg f$. Maybe $k(t,f) \not= K(X)$? No...

Can somebody explain?

Answer 1

Each choice of $f$ produces a different embedding of $k(t) = K(\mathbb{P}^1_k)$ into $K(X)$, so it's no surprise that $\deg f$ depends on $f$. In other words, writing $[K(X) : K(\mathbb{P}^1_k)]$ is an abuse of notation because $K(\mathbb{P}^1_k)$ is not literally a subfield of $K(X)$.

Here's a simple example: I can embed $k(t)$ into $k(t)$ by the identity morphism, giving a field extension of degree $1$, or I can embed $k(t)$ into $k(t)$ by mapping any rational function $f(t)$ to the rational function $f(t^2)$, thus producing a field extension of degree $2$.