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 $F.$

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 $F.$

Please help me. I'm absolutely clueless.

Answer 1

Hint:

Suppose theres a root $\;\omega\in F\;$ of $\;f(x)\;$, so $\;\omega^p=a\;$, but then

$$ x^p-a=(x-\omega)^p =\;\ldots$$

Answer 2

The following should be considered a long comment. I will answer your question in case $F$ is a finite field. Using Don Antonio's answer, it suffices to show that $x^p-a$ is reducible implies that it has a root in $F$. Consider $\phi:F\rightarrow F$ that sends $x$ to $x^p$. One verifies easily that $f$ is a ring homomorphism. One also verifies easily that $Ker(f)$ is trivial, thus $f$ is injective. Since our field is finite, therefore $f$ is surjective. Thus, there exists $w\in F$ such that $w^p=a$.

Note: The above argument shows that $x^p-a$ always has root (hence always reducible) in case $F$ is finite and $Char(F)=p$

Answer 3

Theorem (Capelli)
Let $p$ be a prime and $F$ be a field of any characteristic.
If the polynomial $x^p-a\in F[x]$ has no root in $F$, then it is irreducible.

Proof when $F$ is of characteristic $p$ ( the case actually asked by the OP)
Let $\alpha$ be a root of $x^p-a$ in a splitting field $K$ of that polynomial, so that $x^p-a=(x-\alpha)^p\in K[x]$, and let $f(x)=Irr(\alpha, F,x)$ be the (irreducible !) minimal polynomial of $\alpha$ over $F$.
Then $f(x)$ divides $x^p-a$ (in $F[x]$ or $K[x]$) and so is of the form $(x-\alpha)^r$ with $1\leq r\leq p$, as is any divisor of $x^p-a=(x-\alpha)^p$ .
But since $f(x)\in F[x]$, we have $f(x)=(x-\alpha)^r=x^r-r\alpha x^{r-1}+\cdots \in F[x]$ so that $r\alpha\in F$ which [since $1\leq r\leq p$ and $\alpha\notin F$] is only possible if $r=p$.
But this means that $f(x)=(x-\alpha)^r=(x-\alpha)^p=x^p-a$ and so $x^p-a$ is indeed irreducible since $f(x)$ , being a minimal polynomial, is irreducible.

Proof (much more difficult) when $F$ is not of characteristic $p$
Lang, Algebra Chapter VI, §9, Theorem 9.1 .

Answer 4

Let $E$ be the splitting field of $f(x)$ over $F$ and let $r$ be a root of $f(x)$ in $E$. Then $f(r)=r^p-a=0$ and $r^p=a$ and $f(x)=x^p-a=x^p-r^p=(x-r)^p$ by $char F=p$ and Freshman's Dream.

Case I: If $r\in F$, then $f(x)=(x-r)^p$ splits in $F[x]$.

Case II: If $r\notin F$. Since $F[x]$ is a UFD, suppose that $f(x)=g_1(x)g_2(x)\cdots g_n(x)$, where $g_i(x)$ is irreducible and monic in $F[x]$ for each $i=1, 2, ..., n$. (Note that $f(x)$ is monic.) In addition, $\deg{g_i(x)}\geq 2$ for each $i=1, 2, ..., n$. For if $\deg{g_i(x)}=1$, then $g_i(x)=(x-s)\in F[x]$ and $s\in F$ is a root of $f(x)$. Which is the Case I.

Then we have $f(x)=g_1(x)g_2(x)\cdots g_n(x)=(x-r)^p$. For each $i=1, 2, ..., n$, since $\deg{g_i(x)}\geq 2$, it follows that $g_i(x)$ has multiple root in $E$. Recall that $g_i(x)$ is irreducible over $F$. Which implies that $g_i(x)=h_i(x^p)$ for some $h_i(x)\in F[x]$. (Here I use a theorem.) Then $f(x)=x^p-a=h_1(x^p)h_2(x^p)\cdots h_n(x^p)$. Compare the degree of this equation, it must be $\deg{h_1(x)}=1$ and $h_2(x), h_3(x), ..., h_n(x)$ all are constant. It follows that $g_2(x)=g_3(x)=\cdots =g_n(x)=1$ and $g_1(x)=f(x)$. Therefore, $f(x)=g_1(x)$ is irreducible over $F$.

This question appears at the Gallian's Contemporary Abstract Algebra 8/e (Exercise 20.15).