3

If $F$ a field with $char(F)=p$. Prove: If $x^p -x -a$ is reducible in $F[x]$ , then this it splits in distinct factors in $F[x]$.

I know if for hypothesis $x^p -x -a = P(x)Q(x)$ with $P(x),Q(x) \in F[x]$. Then $deg(P(x))<p$ and $deg(Q(x))<p$. But i don't know as contiuning. To remember that is not necessary that $x^p -x -a$ have roots in $F$.

On the other hand, if $\alpha$ is a root of $x^p -x-a$ then $\alpha +1$ is also a root. Hence, $\mathbb{F}_{p}(\alpha)=\mathbb{F}_{p}(\alpha +k)$ for $k \in \{1,2,...,p-1\}$ including $0$. So $a=0$ and $x^p -x -a= x^p -x$.

Angelo
  • 189
  • Hint: If you have one root $\alpha$, can you find another (all of the other) root of $x^p-x-a$? The freshman's dream may be helpful here... – bzc Apr 19 '15 at 02:43
  • If $\alpha$ is root of $x^p -x -a$ then $\alpha + 1$ is also root. For induction all roots are of the form $\alpha +k $. Hence, $F_{p}(\alpha) = F_{p}(\alpha + k)$ – Angelo Apr 19 '15 at 14:04
  • Exactly! (need more characters) – bzc Apr 19 '15 at 14:22

1 Answers1

1

If $\alpha$ is a root of $f$, then also $\alpha+1,\dots,\alpha+p-1$ are roots and so $f$ splits as $$ f(X)=\prod_{k=0}^{p-1}(X-\alpha-k) $$ So, if $f$ has a root in $F$, it has all roots in $F$.

Suppose it has no root in $F$ and that $f=gh$, with monic $g$ and $h$. Then in some extension of $F$ we have that $g$ is a product of $d$ of the above factors, where $d$ is the degree of $g$. Suppose $0<d<p$ and $g(X)=X^d+b_{d-1}X^{d-1}+\dots+b_0$. Then, by Viète's formulas, $$ b_{d-1}=-\bigl((\alpha+k_1)+(\alpha+k_2)+\dots+(\alpha+k_d)\bigr)= -d\alpha+h $$ with $h$ an integer (modulo $p$) and $d\ne0$ in $F$, so that $$ \alpha=\frac{h-b_{d-1}}{d}\in F, $$ contrary to the assumption that $f$ has no root in $F$.

(From Lang, Algebra, proof of theorem VI.6.4.)

egreg
  • 238,574