What are elements a,b such that each has prime power degree but their product does not?

Question

I am studying from Artin and this is a practice problem I've come across.

For prime $p$, suppose there is an element $\alpha$ with degree $p^n$ over $\mathbb Q$ and an element $\beta$ with degree $p^m$ over $\mathbb Q$. Prove or find a counterexample for the following claim: the degree of $\alpha\beta$ must be a power of $p$.

I have not done Galois theory, and I have tried $\sqrt[3]{2}$ and $\omega$ which is the primitive root of unity of $x^3-1$. I think this is close, but I'm not sure.

Answer 1

Let $\alpha = \sqrt[3]{2}$ and $\beta = \omega/\sqrt[3]{2}$. Both have degree $3$ over $\mathbf Q$ while their product $\omega$ has degree $2$ over $\mathbf Q$.

More generally, let $\alpha = \sqrt[n]{2}$ and $\beta = e^{2\pi i/n}/\sqrt[n]{2}$, so $\alpha$ and $\beta$ have degree $n$ over $\mathbf Q$ while $\alpha\beta = e^{2\pi i/n}$, which has degree $\varphi(n)$ over $\mathbf Q$. When $n$ is a power of a prime $p$, $\varphi(n)$ is divisible by $p-1$, so when $n$ is an odd prime power, $\varphi(n)$ is not a power of $p$.

Answer 2

Try $2^{1/3}(1+\omega)$.

Let $\alpha=2^{1/3}(1+\omega)$. $\newcommand{\rat}{\mathbb{Q}}$ Let $\beta=2^{1/3}$. Let $K=\rat(\beta,\omega)$. We know that $[K:\rat]=6$. By a simple calculation, we see that $K=\rat(\omega,\alpha)$. In particular, we have $[\rat(\omega,\alpha):\rat(\omega)]=3$, so $\alpha$ has degree $3$ over $\rat(\omega)$. Now, $f:=x^3-2(1+\omega)$ is a polynomial over $\rat(\omega)$ that vanishes $\alpha$, but $\alpha$ has degree $3$, so $f$ is irreducible over $\rat(\omega)$. Then it divides the minimal polynomial $g$ of $\alpha$ over $\rat$, so $g$ must have a degree above $3$ because $f\not\in\rat[x]$. It can only be the case that $\deg(g)=6$ because $[K:\rat]=6$.