Dimension Field True/False.

Question

I'm having trouble approaching how to determine truthfulness and falsehood of the following type of problems.

$F$ and $K$ are fields.

1) Suppose that $F\subseteq K$ and $r\in K$. If $[F(r):F]=4$ then $F(r)=F(r^2)$.

I was previously asked to prove this same question except the conclusion was "then $F(r)=F(r^3)$". The question turned out to be true. I'm having a hard time seeing why they both aren't true for very similar reasons.

2) Suppose that $F\subseteq K$ and $r_1,r_2\in K$. If $[F(r_1):F]=6$ and $[F(r_2):F]=2$ with $r_2\notin F(r_1)$, then $[F(r_1,r_2):F]=12$

False? Can't think of a counterexample though. This is the one I'm mostly tripped up on. The part with $r_2\notin F(r_1)$ must somehow have an impact on the definition used below in 3). Otherwise, this would be true just by definition, right?

3) Suppose that $F\subseteq K$ and $r_1,r_2\in K$. If $[F(r_1):F]=3$ and $[F(r_2):F]=6$, then $[F(r_1,r_2):F]=18$.

True. This one seems straightforward from the definition from multiplicativity of field extension degree. That is, if I am not mistaking something between the definition and the way this problem is written.

4) If $r^5 \in F$ but $r\notin F$ then $[F(r):F]=5$.

For this, I was thinking maybe it was false and I could come up with a counterexample.

If we let $r=\sqrt{5}^{1/5}$ and $r^5=5$. If we let our field be something like $a+b\sqrt{5}$ then $r$ and $r^5$ could both be in there and then we just need to make it such that our dimension is $4$ and we'd have a counterexample, no?

Answer 1

1) You must see your previus post, I fixed it, and the question turned out to be false, sorry for the mistake. In this case, the question is again false(F(r)=F(r^2)), and the same example works again, the same $\omega$, but instead to take $w^3$ you must take $w^2=e^{\frac{2\pi i}{6}}$ and then $[Q(\omega^2):Q]=\phi(6)=2$.

Hint: 2,3,4,5 take $F=Q$.

4 must be direct, take $K$ the spliting field of $x^5-1$ over $Q$, and $\omega\in K$ a primitive root of $X^5-1$ ($\omega=e^{\frac{2\pi i}{5}}$), then $K=Q(\omega)$, so $[Q(\omega):Q]=\phi(5)=4$, and $\omega\not\in Q$, and $\omega^5=1\in Q$.

Answer 2

Regarding 2) and 3): The multiplicativity formula for field extension degrees gives $$[F(r_1,r_2):F] = [F(r_1,r_2):F(r_1)][F(r_1):F].$$

The minimal polynomial for $r_2$ over $F(r_1)$ divides the minimal polynomial for $r_2$ over $F$ so $$[F(r_1,r_2):F(r_1)] \leq [F(r_2):F].$$

2) We know that $[F(r_1,r_2):F(r_1)] \leq [F(r_2):F] =2$. We also know that $r_2 \notin F(r_1)$ which implies that $[F(r_1,r_2):F(r_1)] \neq 1$. Hence $[F(r_1,r_2):F(r_1)] = 2$ and $[F(r_1,r_2):F] = 2 \cdot 6 = 12$.

3) Let $F = \mathbb{Q}$, $r_1 = 2^{1/3}$ and $r_2 = 2^{1/6}$. Then $[F(r_1):F] = 3$ and $[F(r_2):F] = 6$. Also $F(r_1,r_2) = F(r_2)$ since $r_2^2 = r_1$, so $[F(r_1,r_2):F] = 6$.