How can one go from "no roots" to "is irreducible" in this case?

236 Views Asked by At

This problem is paraphrased from an old version of an exam that
I will be taking, and I have no idea how one would do solve it.

Let $p$ be a prime number, let $F$ be a field of characteristic $p$, and let $c$ be a field element
such that $\:x^{\hspace{.025 in}p}\hspace{-0.04 in}+(-c)\:$ has no roots in $F$. $\;\;\;$ Show that $\:x^{\hspace{.025 in}p}\hspace{-0.04 in}+(-c)\:$ is irreducible over $F$.

How would one solve that problem?

1

There are 1 best solutions below

5
On BEST ANSWER

Let $K$ be the splitting field of $f(x)=x^p-c$ over $F$, and $a$ a root. Thus, $a^p=c$ and so $x^p-c=(x-a)^p$ over $K$. Hence every factor of $f$ has the form $(x-a)^k$ for some $k\le p$. Suppose there is some $k\lt p$ with $g(x)=(x-a)^k\in F[x]$, since $k\lt p, (k,p)=1$, so there are integers $m,n$ with $mp+nk=1$, but this implies $f^m(x)g^n(x)=(x-a)^{mp+nk}=x-a\in F[x]$, contradiction.