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I am trying to understand the solution to the following exercise from Milne's notes on Galois theory (available here):


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At the end of the notes, the author presents the following solution to the problem:


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Although I understand the argument, it is not immediately clear to me that this contradiction implies $F(X^2)\cap F(X^2-X)=F$. What immediately follows is that there are no polynomials of degree $m>0$ in this intersection. However, $E=F(X^2)\cap F(X^2-X)$ is a field of rational functions, so, in principle, it is not clear why there couldn't be non-constant rational functions $p(X)/q(X)$ in this field, even if there are no polynomials. More precisely, I can formulate my question as follows:

  • How to show that, if there exists $p(X)/q(X)\in E$, with $p, q$ coprime, $q(X)\neq0$, and $\deg(q)>0$, then there exists a polynomial of degree $m>0$ in $E$?

This way, we would demonstrate that $E$ indeed contains no non-constant rational functions and, therefore, $E=F$.

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$let f(x)=p(x)/q(x)$.So $f(x)q(x)-p(x)=0$ is the polynomial that x satisfies on F(f(x)).Then we have that $[F(x):F(f(x))] \leq max \{deg(q(x)),deg(p(x))\}$.So the extension is finite.