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Algebraic closure of $mathbb F_p$

Mathematics Asked by SamoGrecco on January 14, 2021

I’m proving that $overline{mathbb{F}}_p = bigcuplimits_{i=1}^{infty} mathbb{F}_{p^i}$ is an algebraic closure of $mathbb{F}_p$ where $p$ is a prime. I think I’ve gotten down how to prove that $overline{mathbb{F}}_p$ is a field and that it is algebraic over $mathbb{F}_p$.

I have some difficulties with proving that $overline{mathbb{F}}_p$ is algebraically closed.

My attempt is as following:

Suppose $f$ is a non-constant polynomial in $overline{mathbb{F}}_p [X]$. If $overline{mathbb{F}}_p$ contains a root of $f$, then it is algebraiclly closed. Per definition there must exist a $mathbb{F}_{p^k}$ for a certain positive integer $k$ that contains all the coefficients of $f$. Take a root $alpha$ of $f$ and consider the extension $mathbb{F}_{p^k}(alpha)$. How is this now a field of the form $mathbb{F}_{p^l}$ for a certain positive integer $l$?

One Answer

Take a polynomial $p(x) in overline{mathbb{F}}_p [x]$. Then, $p(x) in mathbb{F}_{p^k}[x]$ for some $k$. The splitting field is a finite extension of characteristic $p$, so it is isomorphic to $mathbb{F}_{p^l}$ for some $l$ (by characterisation of finite fields). Hence, $p(x)$ must split over $overline{mathbb{F}}_p$.

Answered by marlasca23 on January 14, 2021

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