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Let $k \subseteq F \subseteq K$ be fields, and let $z \in K$. Prove that if $k(z) \colon k$ is finite, then $[F(z):F] \leq [k(z):k]$. In particular, $[F(z):F]$ is finite.

If $k(z) \colon k$ is finite, then $[k(z):k]=\dim_{k}k(z)=n$ for some $n \in \mathbb{N}$. I was trying to prove that $F(z) \colon F$ is a subspace of $k(z) \colon k$. But doubt this happens.

The Hint over Rotman´s Advanced Algebra book says I should obtain an irreducible $p(x) \in k[x]$, and this polynomial should factor in $K[x]$. i dont really understand this hint.

Also, no problem seeing $[F(z):F]$ is finite once it is proved that $[F(z):F] \leq [k(z):k]$.

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  • $\begingroup$ Do you understand that there is an irreducible poly (non-zero) $p(x)\in k[x]$ such that $p(z) = 0$? $\endgroup$ – peter a g Apr 2 at 3:05
  • $\begingroup$ How I can justify that? $\endgroup$ – Cos Apr 2 at 3:22
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As $z$ is finite over $k$, then it has a minimum polynomial $p(X)$ over $k$. This means that $p(X)$ is monic, its coefficients are in $k$, $p(z)=0$ and $p(X)$ is a factor of all polynomials $f(X)$ with coefficients in $k$ such that $f(z)=0$.

But $k\subseteq F$, so that $z$ satisfies a polynomial equation over $k$ (namely $p(X)=0$). Therefore $z$ is finite over $F$, and so has a minimum polynomial $q(X)$ over $F$. This means that $q(X)$ is monic, its coefficients are in $F$, $q(z)=0$ and $q(X)$ is a factor of all polynomials $f(X)$ with coefficients in $F$ such that $f(z)=0$. But one of these polynomials is $p(X)$. Therefore $q(X)$ is a factor of $p(X)$. As a consequence, $\deg q(X)\le\deg p(X)$.

But the degree of the extension $|k(z):k|$ equals $\deg p(X)$. Likewise $|F(z):F|=\deg q(z)$. Therefore $|F(z):F|\le|k(z):k|$.

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  • $\begingroup$ So clearly explained! Thanks $\endgroup$ – Cos Apr 2 at 18:27

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