Why in a field of characteristic $p$, $\zeta_p \sqrt[p]t$ is not a root of $X^p-t\in \mathbb F_p(t)[X]$

Question

I know that in a field such that $Car(K)=p$ is prime, the $X^p-t\in \mathbb F_p(t)[X]$ has a unique root (I know how to prove it and thus, it's not the question). But in the usual logic, $X^p=t\iff X=\zeta _p^k\sqrt[p]t$ where $\zeta_p=e^{\frac{2i\pi}{p}}$ and $k=0,...,p-1$. But why I can't do the in a field of characteristic $p$ prime ? Does $\zeta _p=1$ always ?

Answer 1

As you have guessed, it is the case that always $\zeta_p=1$. Let me be a little more specific: In a field of characteristic $p$ we have $x^p-1=(x-1)^p$. This implies that in no field extension of $\Bbb F_p(t)$, we will have a $p$-th root of unity which is distinct from $1$, because if $\zeta^p=1$, then $0=\zeta^p-1=(\zeta-1)^p$, so $\zeta-1=0,\zeta=1$.

While writing the answer, let me note that having roots $\zeta_p^k\sqrt[p]{t}$ with $\zeta_p=e^\frac{2i\pi}{p}$ is not the matter of using "usual logic", but rather the matter of working in the field of complex numbers, where expression $e^\frac{2i\pi}{p}$ makes sense and isn't equal to $1$.

Answer 2

Simply because for any $r$, $\sqrt[r]t$ doesn't exist in $\mathbf F_p(t)$ — I suppose $t$ in your context is an indeterminate. It is impossible for degree reasons: if there were polynomials $P(t), Q(t)$ such that $$\biggl(\frac{P(t)}{Q(t)}\biggr)^r=t\iff P(t)^r=t\bigl(Q(t)\bigr)^r.$$ The degree of the left-hand side is congruent to $0$ modulo $r$ while the degree of the right-hand side is congruent to $1$.