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I think this is going to be a silly question. I'm happy with the following fact:

If $\alpha : X \to Y$ is a non-constant morphism of irreducible curves, then $\alpha$ induces an embedding of field $k(Y) \hookrightarrow k(X)$ such that $[ k(X) : k(Y) ] = \mathrm{deg}(\alpha)$ is finite.

I have the following question:

Let $\phi = (1:f) : \mathbb P^1 \to \mathbb P^1 $ be a morphism given by a non-constant polynomial $f \in k[t] \subset k(\mathbb P^1)$. Prove that $\mathrm{deg}(\phi) = \mathrm{deg}\ f$.

I can't see why this is true. $\phi$ is a non-constant morphism of smooth irreducible curves, right? Why isn't $\mathrm{deg}(\phi) = 1$? (both domain and codomain have the same function field...)

Thanks

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    I'll try to write out something later, but in terms of just fields I think the point is that we can properly embed something like $k(x)$ in itself as, say, $k(x^2)$.2012-05-06

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It is absolutely true that the domain and codomain have abstractly isomorphic function fields, but the map $\phi$ provides a way of embedding the one function field as a subfield of the other. Let us take a simple example, where $f(t) = t^2$. In this case, the function fields of the domain and codomain are both abstractly isomorphic to $k(x)$, but the embedding of function fields induced by $f$ is $k(t^2)\subset k(t)$. For general polynomial $f$, the embedding of functions fields will be $k(f(t))\subset k(t)$. You should be able to derive what you want from this.

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    It should still be the same that the embedding of fields is $k(f(t))\subset k(t)$. I guess the degree of this extension is less clear, but I think it should be the degree of $f$, which is defined as follows: if $f(t) = P(t)/Q(t)$ with $P$ and $Q$ polynomials with no common root, then $\deg f = \max\{\deg P, \deg Q\}$.2012-05-06