# If $[K:F]$ is finite and $u$ is algebraic over $K$, prove that $[K(u):K]\leq [F(u):F]$.

I need to show:

If $$[K:F]$$ is finite and $$u$$ is algebraic over $$K$$, prove that $$[K(u):K]\leq [F(u):F]$$.

First if $$u$$ is algebraic over $$K$$, it is clear that it is also algebraic over $$F$$ because we can express the coefficients of polynomials over $$K$$ as linear combinations of elements of $$F$$ since $$[K:F]$$ is finite. What we would like to claim is that a polynomial over $$K$$ that is irreducible is also irreducible over $$F$$, because then we can use the minimal polynomial of $$[K(u):K]$$ and say it must have the lowest degree hence $$[K(u):K]\leq [F(u):F]$$. The problem though, is that a polynomial over $$K$$ might not exist as a polynomial over $$F$$ because $$K$$ contains elements that are not in $$F$$ by definition. So what I would like to do is show that the minimal polynomial $$[K(u):K]$$ must have a lower degree than the minimal polynomial $$[F(u):F]$$, which is where I am stuck on.

Let $$[K:F] =m$$ and $$[K(u) : K] =n$$ then by tower law we have $$[K(u) : F] =mn$$ so every element of $$K(u)$$ is algebraic over $$F$$ and in particular $$u$$ is algebraic over $$F$$.

Next let $$p(x) \in F[x]$$ be the minimal polynomial for $$u$$ and let degree of $$p(x)$$ be $$r$$ so that $$[F(u):F] =r$$. Then $$p(x)$$ is also a polynomial in $$K[x]$$ as $$F\subseteq K$$. But $$p(x)$$ may or may not be irreducible in $$K$$. If $$p(x)$$ is irreducible over $$K$$ then it is also the minimal polynomial for $$u$$ over $$K$$ and $$n=[K(u) :K] =r=[F(u) :F]$$ If $$p(x)$$ is reducible over $$K$$ then it is divisible by the minimal polynomial of $$u$$ over $$K$$ (which is of degree $$n$$) and hence degree of $$p(x)$$ ie $$r$$ is greater than $$n$$ so that $$[F(u) :F] > [K(u) : K]$$

You should observe that the argument holds even if $$[K:F]$$ is infinite. Thus if $$F, K$$ are fields with $$F\subseteq K$$ and $$u$$ is algebraic over both $$F$$ as well as $$K$$ then $$[F(u) :F] \geq [K(u) :K]$$. But in this case we need to assume that $$u$$ is algebraic over both the fields. If the extension $$K/F$$ is finite we only need to assume $$u$$ to be algebraic over $$K$$.

• your explanations are very clear! thank you
– Bill
Sep 3, 2021 at 1:15

Proving that $$u$$ is algebraic over $$F$$ is easier (your argument is good, but not clearly developed). You certainly know that an element $$b$$ in some extension field of $$E$$ is algebraic over $$E$$ if and only if $$[E(b):E]$$ is finite.

So you want to prove that $$[F(u):F]$$ is finite. Since $$F(u)$$ is a subfield of $$K(u)$$, it's sufficient to prove that $$[K(u):F]$$ is finite, but this is a consequence of $$[K(u):F]=[K(u):K][K:F]$$ Therefore $$u$$ is algebraic over $$F$$.

You also have $$[K(u):F]=[K(u):F(u)][F(u):F]$$ and therefore $$\frac{[K(u):K]}{[F(u):F]}=\frac{[K(u):F(u)]}{[K:F]}$$ Can you finish?

• It is a lemma that the right side is less than 1 so we can conclude from that.
– Bill
Sep 2, 2021 at 9:41
• @William That's it! Sep 2, 2021 at 9:49
1. $$[F(\alpha,\beta):F(\beta)]\leq [F(\alpha):F]$$ for $$\alpha,\beta$$ algebraic over $$F$$.
2. If $$[K:F]$$ is finite, $$K = F(\alpha_1,...,\alpha_n)$$ for some $$\alpha_1,...,\alpha_n\in K$$.