I have been going through the exercises from Contemporary Abstract Algebra, and there are a few questions that I just couldn't solve.
Let $F=Q(\pi^3)$. Find a basis for $F(\pi)$ over $F$. I know the structure of $Q(\pi)$ (kinda like the quotient field on $Q$), but I don't know how to proceed from there.
Let $F$ be a field with characteristic $p\neq 0$. Show that the polynomial $x^{p^n}-x$ has distinct zeros. I know that a polynomial $f$ has a multiple zero if and only if $f$ and $f'$ have a common factor of positive degree. However, I'm kinda stuck on verifying that $x^{p^n}-x$ and $p^nx^{p^n-1}-1$ have no common factor.
If $F$ is a field and the multiplicative group of nonzero elements of $F$ is cyclic, show that $F$ is finite. I honestly don't even know how to start thinking about this question. A hint would be much appreciated.
Show that $\sqrt{2} \notin Q(\pi)$. Here is my train of thought: assume otherwise. Then \begin{equation} \sqrt{2} = \frac{a_m\pi^m+\cdots+a_1\pi+a_0}{b_n\pi^n+\cdots+b_1\pi+b_0} \end{equation} so that \begin{equation} 2 = \frac{(a_m\pi^m+\cdots+a_1\pi+a_0)^2}{(b_n\pi^n+\cdots+b_1\pi+b_0)^2} \text{.} \end{equation} But I'm not sure how to continue from here.
Suppose $a$ is algebraic over a field $F$. Show that $a$ and $1+a^{-1}$ have the same degree over $F$. The only observation that I can make is that $F(a)=F(1+a^{-1})$, which may or may not help me solve this question. I feel like there must be a simple trick that I'm simply overlooking.