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

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 $$[F(z):F] \leq [k(z):k]$$.

• Do you understand that there is an irreducible poly (non-zero) $p(x)\in k[x]$ such that $p(z) = 0$? – peter a g Apr 2 at 3:05
• How I can justify that? – Cos Apr 2 at 3:22

## 1 Answer

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 $$|F(z):F|\le|k(z):k|$$.

• So clearly explained! Thanks – Cos Apr 2 at 18:27