Question

Prove that no finite field is algebraically closed.

Short answer

It is proved that no finite field is algebraically closed.

Step-by-step solution

Definition of algebraically closed

An algebraically closed field $F$ contains a root for every non constant polynomial. In $F\left(x\right)$ the right of polynomial in the variable with coefficient in $F$.

Showing that no finite field is algebraically closed

Let$F$be a finite field and$a_1,a_2\mathrm{.......}{a}_n\in F,a_1\ne 0$.then

$f\left(x\right)=a_1+(x-a_1)(x-a_2)\mathrm{..........}(x-a_n)\in F\left(x\right)$

Find the value of$f\left(x\right)$at$a_1\ne 0$as follows

$\begin{array}{c}f\left(x\right)=a_1+(a_1-a_1)(a_1-a_2)\mathrm{..........}(a_1-a_n)\in F\left(x\right)\\ =a_1\end{array}$

This means$a_1$is not the root of$f\left(x\right)$. Since there is a polynomial with coefficient in$F$that does not have$a_1$as a root. This implies that$F$is not an algebraically closed.

Thus no finite field is algebraically closed.