$F$ and $K$ are fields.
Proof or counterexample: If $F\subseteq K$ and $r\in K$. If $[F(r):F]=4$ then $F(r)=F(r^3)$.
I think I need to find a polynomial in $F(r^3)[x]$ that has $r$ as a root. I can't seem to figure that out though and how to prove it true if that were the case.
Similarly, I'm having issues with the same idea else where.
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?
Any help would be great. I am having trouble understanding these types of problems.
It is false.
Take $F=\mathbb Q$ the rationals, and $\omega=e^{\frac{2\pi i}{12}}$. $\omega$ is a primitive $12$-root of unity, and we know that in this case $[\mathbb Q(\omega):\mathbb Q]=\phi(12)=4$. (If you do not know what this means, search for Euler totient.). Now see $\omega^3=e^{\frac{3(2\pi i)}{12}}=e^{\frac{2\pi i}{4}}$, so $[\mathbb Q(\omega^3):\mathbb Q]=\phi(4)=2$, and with this we conclude that the proposition is false.