If $b$ is algebraic over a finite extension $K$ of $F$ then $[K(b):K]\mid [F(b):F]$

Question

In A Book of Abstract Algebra by Charles C. Pinter, part 3 of exercise F page 299 is the following: ($a\mid b$ means $a$ divides $b$)

Let $F$ be a field, and $K$ a finite extension of $F$. Prove each of the following :

3. If $b$ is algebraic over $K$, then $[K(b):K]\mid [F(b):F]$. (Hint: The minimal polynomial of $b$ over $F$ may factor in $K[x]$, and $b$ will then be a root of one of its irreducible factors.)

I couldn't solve this exercise. If $p(x)$ and $q(x)$ are the minimal polynomials of $b$ over, respectively, $F$ and $K$, then as $p(b)=0$ and $p(x)\in K[x]$, we have $q(x)\mid p(x)$, which shows that $[K(b):K]\le [F(b):F]$. But I wasn't able to show that it divides it. I tried using the different formulas like $[K(b):F]=[K(b):F(b)][F(b):F]$ but I wasn't able to reach the desired conclusion. I also tried using the fact that $\{1,b,\dots,b^{n-1}\}$ is a basis of the vector space $K(b)$ over $K$, and see how this can relate with $F(b)$ over $F$; that linearly independent family doesn't necessarily span $F(b)$ over $F$. I tried using the fact that $K=F(a_1,\dots,a_m)$ since it is a finite extension of $F$, but I didn't manage to reach the conclusion.

Would you help me please? Thank you in advance!

Answer 1

This claim seems to be false.

Let $\sqrt[3]{2}$ be the real cube root of $2$, and let $\zeta$ be a primitive cube root of unity, so that $\sqrt[3]{2}$, $\zeta \sqrt[3]{2}$ and $\zeta^2 \sqrt[3]{2}$ are the three roots of $X^3-2$ over $\mathbb{Q}$.

Then $[\mathbb{Q}(\zeta\sqrt[3]{2}):\mathbb{Q}] = 3$, but $[\mathbb{Q}(\sqrt[3]{2}, \zeta\sqrt[3]{2}):\mathbb{Q}(\sqrt[3]{2})]=2.$

Taking $F=\mathbb{Q}$, $K=\mathbb{Q}(\sqrt[3]{2})$ and $b=\zeta\sqrt[3]{2}$, we seem to have a counter-example.

Answer 2

A counterexample of OP's statement is already given by @Pierre-Guy Plamondon.

With normality assumption for $K$ over $F$

Suppose $K$ is normal over $F$. Let $f$ be a minimal polynomial for $b$ over $F$. Then we will prove that $f$ factors into irreducible polynomials over $K$ all of the same degree.

Note that OP pointed out the degree of the minimal polynomial for $b$ over $K$ is the degree of field extension $[K(b):K]$. Let $b'$ be any root of $f$ in $\overline{K}$. Let $\sigma : K(b) \rightarrow \overline{K}$ be an embedding fixing $F$ with $\sigma(b)=b'$. We claim that $[K(b'):K]=[K(b):K]$.

To do this, note that $\sigma:K(b)\rightarrow \overline{K}$ is an injective $F$-linear mapping. Since $K$ is normal over $F$, the restriction $\sigma|_K$ is an automorphism of $K$ fixing $F$. Thus, $\sigma(K(b))= K(b')$. This shows $K(b)$ and $K(b')$ are isomorphic as $F$-vector spaces via an $F$-linear isomorphism $\sigma$. This shows that $[K(b):F]=[K(b'):F]$. Hence, we have $$ [K(b):K][K:F]=[K(b):F]=[K(b'):F]=[K(b'):K][K:F]. $$ It follows that $[K(b):K]=[K(b'):K]$. Thus, the irreducible factors of the minimal polynomial for $b$ over $F$ must factor into the irreducible polynomials over $K$ all of the same degree.

Therefore, we have $[K(b):K] | [F(b):F]$.