I'm wondering when, if ever, we can adjoin the four powers of $\sqrt[4]{x}$ to $\mathbb{Z} / p \mathbb{Z}$, for $p$ prime and $x$ a natural, to create a field with $p^4$ elements. Is it true that we can do this whenever those powers are not already elements within the field?
I assume that if we can't do this, we can adjoin the roots of the quartic equation without problems, as long as they are not elements within the field. Is this correct?
This is possible if and only if $p\equiv1\pmod4$.
A) If $p\equiv-1\pmod 4$ then it is not possible. Assume that $x$ has multiplicative order $m$. We know that $m\mid p-1$. On the other hand $4\mid p+1$, so $4m\mid p^2-1$. Given that the multiplicative group of $\Bbb{F}_{p^2}$ is cyclic of order $p^2-1$ it follows that $T^4-x$ has zeros in $\Bbb{F}_{p^2}$, and we cannot get a degree four extension.
B) If $p\equiv1\pmod 4$ then let $2^\ell$ be the highest power of two that is a factor of $p-1$. Let $x$ be an element of order $2^\ell$ in $\Bbb{F}_p^*$. Any root $\alpha$ of $T^4-x$ in some extension field of $\Bbb{F}_p$ will then have order $2^{\ell+2}$. Because $2^{\ell+2}$ is not a factor of either $p-1$ or $p^2-1$, it follows that the minimal polynomial of $\alpha$ has degree four. Therefore $\Bbb{F}_p(\alpha)=\Bbb{F}_{p^4}$.