Relationship between $x^{p-1} - a$ and $x^p - ax - b$

Question

I am in a finite field with characteristic $p$. I also have a polynomial $g = x^p - ax - b$ where $a$ is not the identity. The question is, if $E$ is the split field of the polynomial $g$, is it also a split field of $h = x^{p-1} -a$? I tried to divide, but I don't think $h$ divides $g$. I also tried to see if $g$ can be reduced, but that depends on $a$. Besides these two, I also know that $x^{q-1} = 1$ where $q$ is the number of elements in the field. But that doesn't work here. What else should I try?

Answer 1

Presumably I can assume that $a\neq0$.

Yes, the splitting field of $h$ is included in the splitting field of $g$.

The key is that the polynomial $$ f(x)=x^p-ax= xh(x)=g(x)+b $$ is a so called linearized polynomial. In other words, $$ f(x+y)=(x+y)^p-a(x+y)=x^p+y^p-ax-ay=f(x)+f(y) $$ for all $x,y$ in any extension field. So if $x_0,x_1,x_2,\ldots,x_{p-1}$ are the zeros of $g(x)$ (they are distinct, because $g'(x)=-a\neq0$), then $$ f(x_i-x_0)=f(x_i)-f(x_0)=[g(x_i)-b]-[g(x_0)-b]=-b-(-b)=0 $$ for all $i=1,2,\ldots,p-1$. As $x_i-x_0\neq0$ and $$ h(x_i-x_0)=\frac{f(x_i-x_0)}{x_i-x_0} $$ it follows that the zeros of $h(x)$ are the elements $x_i-x_0$, $i=1,2,\ldots,p-1$, that manifestly belong to any splitting field of $g(x)$.