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

Question

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.

Answer 1

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 $$[F(u) :F] > [K(u) : K] $$

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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 $K$.

Answer 2

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?

Answer 3

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 $\alpha_1,...,\alpha_n\in K$.