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...)


  • 1
    $\begingroup$ The function field $k(X)$ can have many subfields isomorphic to it, e.g, $k(X^2)$ or $k(X^n)$ for any $n \geq 2$. In fact, for any nonconstant rational function $f \in k(X)$, the subfield $k(f)$ of $k(X)$ has codimension $\max(\deg a, \deg b)$, where $f(X) = a(X)/b(X)$ as a ratio of polynomials in reduced form. $\endgroup$
    – KCd
    May 6, 2012 at 17:50
  • $\begingroup$ 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, $k(x^2)$. $\endgroup$ May 6, 2012 at 17:50

1 Answer 1


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.

  • $\begingroup$ I can't quite see why the embedding is $k(f(t)) \subset k(t)$. Why is this true? $\endgroup$
    – Jonathan
    May 6, 2012 at 18:00
  • $\begingroup$ This boils down to the definition of the embedding. Generally, in the situation $\phi\colon X\to Y$, the embedding $k(Y)\to k(X)$ is given by taking a function $a\in k(Y)$ and pulling it back to $a\circ \phi\in k(X)$. To keep things very explicit, let us say that $k(y)$ is the function field of the codomain and $k(x)$ is the function field of the domain. The embedding $k(y)\to k(x)$ is obtained by sending $y\mapsto y\circ \varphi$. In the case $f(t) = t^2$, one computes $y\circ \varphi = x^2$. Thus the image of the embedding $k(y)\to k(x)$ is $k(x^2)$. $\endgroup$
    – froggie
    May 6, 2012 at 18:23
  • $\begingroup$ Ok, thanks. What happens if $f$ is instead a rational function? $\endgroup$
    – Jonathan
    May 6, 2012 at 21:57
  • $\begingroup$ 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 $\deg f = \max\{\deg P, \deg Q\}$. $\endgroup$
    – froggie
    May 6, 2012 at 22:26

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