For $p$ prime, how can we show that the splitting field of $f(x)=x^{p-1}-1$ over $\mathbb{Z}_p$ coincides with $\mathbb{Z}_p$?
The roots of $f(x)$ are the $p-1$ roots of unity. I think I am correct in saying that $\mathbb{Z}_p(\zeta)$ is a splitting field for $f(x)$ over $\mathbb{Z}_p$ where $\zeta$ is a primitive root $(p-1)^{\text{th}}$ root of unity. I am not quite sure where to go from here however. Any suggestions? Thanks!
Best Answer
If by $\Bbb{Z}_{p}$ you mean the field $\frac{\mathbb{Z}}{p\mathbb{Z}}$ that is the field of integers modulo $p$. Then do as follows:-
for any non-zero $a\in\frac{\mathbb{Z}}{p\mathbb{Z}}$ you have $a^{p-1}=1$ by Fermat's Little Theorem.
But this polynomial itself has degree $p-1$ and hence can have atmost $p-1$ roots.
So all the $p-1$ elements of $\Big(\frac{\mathbb{Z}}{p\mathbb{Z}}\Big)^{\times}$ are itself the roots of this polynomial.
And since $\frac{\mathbb{Z}}{p\mathbb{Z}}$ does not have any proper subfield of order $p-1$. , it follows that it is the splitting field.
More generally a field of order $p^{n}$(denoted by $\mathbb{F}_{p^{n}}$) is precisely the splitting field of the polynomial $x^{p^{n}}-x$.
You can argue like above result to prove this. See that all elements of the field are roots of this polynomial and the smallest subfield containing $p^{n}$ elements of $\mathbb{F}_{p^{n}}$ is $\mathbb{F}_{p^{n}}$ hence it is the splitting field.
Consequently you can use the uniqueness of splitting field to conclude that any finite field of order $p^{n}$ is precisely the splitting field of the polynomial $x^{p^{n}}-x\in \frac{\mathbb{Z}}{\mathbb{pZ}}[x]$. (Note that any field of characteristic $p$ contains $\frac{\mathbb{Z}}{\mathbb{pZ}}$ as it's prime subfield.
Also a slight piece of advice regarding notation. Most people use $\mathbb{F}_{p}$ to denote the field of integers modulo $p$ in order to avoid confusion with the p-adic integers.