let $f (x) = x^p - a \in F[x]$. Show that $f (x)$ is irreducible over $F$ or $f (x)$ splits in $F$.

Question

Let $F $ be a field of characteristic $p$ and let $f (x) = x^p - a \in F[x]$. Show that $f (x)$ is irreducible over $F$ or $f (x)$ splits in $F$.

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I am completely stuck on it.can someone help me please .thanks for your help

Answer 1

Suppose that $f(x)$ has a root $y$ in $F$. Then $y^{p} =a$. If $p$ is odd,

$$f(x) = x^p - a = x^p - y^p = x^{p} + (-y)^{p} = (x-y)^p$$

So $f(x)$ splits. If $p$ is even, then it is $2$, and if you have one root of a quadratic, you have the other.

Answer 2

Consider the splitting field $E$ for $f(x)$ over $F$.
There are three possible factorizations of $f(x)$ in $E[x]$.
(i) $f(x)=(x-r_1)(x-r_2)\cdots (x-r_p)$.
(ii) $f(x)=(x-r_1)^s(x-r_2)^s\cdots (x-r_t)^s$, where $s\geq 2$, $t\geq 2$.
(iii) $f(x)=(x-r_1)^p$.
Since $\gcd{(f(x),f'(x))}\neq 1$, the case (i) is impossible.
Since $p$ is a prime (the characteristic of a field), the case (ii) is impossible.
The remain possible is $f(x)=(x-r_1)^p\in E[x]$. Let $r=r_1$.

If $r\in F$, then $f(x)$ is splits in $F[x]$.

Suppose that $r\notin F$. Since $F[x]$ is a U.F.D.,
write $f(x)$ as a product of some irreducible polynomials $c_1(x), c_2(x), ..., c_n(x)$.

For each $i=1,2,...,n$, $f(x)$ has a root in $K_i=F[x]/\langle c_i(x)\rangle$.
But the only one root of $f(x)$ is $r$. Hence, $f(x)=(x-r)^p\in K_i[x]$.

Since $c_i(x)\mid f(x)$, we have $c_i(x)=(x-r)^{q_i}\in K_i[x]$

Note that $K_i=E$ for each $i=1,2,...,n$ by the uniqueness of the splitting field.
Therefore, $q_i=\deg{c_i(x)}=[K_i:F]=[E:F]$.
Says $q_1=q_2=\cdots=q_n=q=[E:F]$.
Then $p=\deg{f(x)}=n\cdot q$ and $n=1$ and $q=p$.
$f(x)=c_1(x)$ is irreducible.
(If $n=p$ and $q=1$, then $\deg{c_i(x)}=1$ and $K_i=F$ and $r\in F$, contrary to the hypothesis.)