Let $n > 0$ be some integer and let $G$ be a splitting field of the set of all polynomials of degree at most $n$ over a field $F$. I need to prove that $G$ is not an algebraic closure of $F$ in the case $F = \mathbb{Z}_{p}$

I've actually found a solution for this, but I would really appreciate it if someone could fill in some of the details because I am so confused:

There are only finitely many polynomials of degree $\leq n$ over a finite field $F$, so $G/F$ will wind up having finite degree. Then $G^\times$ will be a cyclic group of some finite order, and a generator $a$ will not have an $m^{th}$ root for any nontrivial divisor d of |$G^\times$|, so $xd−a$ cannot split in G

First off all, why is $G/F$ going to have finite degree? Also what exactly does $G^\times$ denote? Just the cyclic group $G$? Also, where did the fact that there is no $m^{th}$ root come from?


1 Answer 1


First, $G/F$ has a finite degree since $G$ is obtained by adjoining roots of polynomials of $\deg\leq n$ over $F$, which are all algebraic and there are finitely many such elements. (There are finitely many polynomials of $\deg\leq n$ over finite fields.)

$G^{\times}$ stands for the multiplicative group of the field $G$, i.e. $G^{\times} = G\backslash\{0\}$. It is known that $k^{\times}$ is cyclic for any given finite field $k$. (Here $G$ is finite since it is a finite extension of $F$.)

Now we choose generator $a$ of the cyclic group $G$. If $d\neq 1$ is a divisor of $|G^{\times}|$, then $x^{d}=a$ cannot have a solution - if it has, then it contradicts to the assumption that $a$ is a generator of $G^{\times}$. Hence $G$ is not algebraically closed.


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