No finite field is algebraically closed

Question

my lecturer used this theorem in the title without proof. I didn't find a proper proof by myself. Could anyone help me out, please?

Answer 1

Let's suppose that $ K $ is finite and write $ K=\{\alpha_{1}, \ldots , \alpha_{n}\}$. Now take the polynomial $ p (x)=(x-\alpha_{1})\ldots (x-\alpha_{n}) +1\in K[x]$. It's easy to see that $ p (x) $ doesn't have any roots in $ K $. Hence, $ K $ is not algebraically closed.

Answer 2

A different approach is to consider the fundamental theorem of Algebra. Since $F$ is algebraically closed, it will contain every root of every polynomial in $F[x]$, even those with degree larger than $|F|$.

Of course you should now proof that there exists such a polynomial with $m > n := |F|$ distinct roots, whereas @Xammm solution constructs a contradiction right away, so this is not a practical proof but it may give you another point of view.