Is $X^p - T$ is irreducible over the separable closure of $\Bbb F_p(T)$?

Question

Is it true that the polynomial $X^p - T$ is irreducible over the separable closure $K = \Bbb F_p(T)^{sep}$ of $\Bbb F_p(T)$ ?

I know it is irreducible over $\Bbb F_p(T)$, by applying Eisenstein criterion (or see this). One can see that any root $a \in K^{alg} = \Bbb F_p(T)^{alg}$ does not belong to $K$. But this does not imply that it is an irreducible polynomial.

could you include a definition of separable closure...? is it like this? — cactus314, May 03 '18 at 11:50
Also https://math.stackexchange.com/questions/347162/xn-t-is-irreducible-over-kt — Alphonse, May 03 '18 at 18:13

score 4 · Accepted Answer · answered May 03 '18 at 11:51

4

Hint: since $X^p-T=(X-a)^p$, if $X^p-T$ factors, then the two factors are $(X-a)^k,(X-a)^l\in\mathbb F(T)^{sep}(X)$ for some $k+l=p$, hence $k,l$ are relatively prime. Deduce $X-a\in\mathbb F(T)^{sep}(X)$.

answered May 03 '18 at 11:51

Wojowu

26,600

score 0 · Answer 2 · answered May 03 '18 at 11:49

If $a\in \mathbb{F}_p(T)^{alg}$ is a root of $X^p-T$, then $X^p-T=(X-a)^p$. Therefore, $X^p-T$ is reducible if and only if it has a root in the field. But $a=T^{1/p}$ is not separable over $\mathbb{F}_p(T)$ because its minimal polynomial is $X^p-T$, which is not separable as we saw at the beginning. Therefore, $a\notin\mathbb{F}_p(T)^{sep}$.

Is $X^p - T$ is irreducible over the separable closure of $\Bbb F_p(T)$?

2 Answers2