algebraic closure of $\mathbb{F}_5$ contains $1$ field of $5^2$ elements

Question

Prove that the algebraic closure of $\mathbb{F}_5$ contains only 1 field of $5^2$ elements.

If we denote by $K$ the algebraic closure of $\mathbb{F}_5$, and we take $\alpha \in K$ such that $\alpha^2=2$, we know that $[\mathbb{F}_5(\alpha):\mathbb{F}_5]=2$.

Now if we take another element $\beta \in K$, we must prove $ \mathbb{F}_5(\beta) = \mathbb{F}_5(\alpha)$.

I don't know much information about K, maybe if Icompute its cardinal, I can use Sylow theorems to show the unicity of extension of degree 2 over $\mathbb{F}_5$?

Thank you.

Answer 1

Hint:

1. Note that $(2\alpha)^2=3$, and so every element of $\Bbb{F}$ has a square root in $\Bbb{F}(\alpha)$.

2. Use the quadratic formula.

Answer 2

You can use that for every prime number $p$ and every natural number $n$, a field $\mathbb{F}_{p^n}$ of characteristic $p$ and cardinality $p^n$ is a splitting field of the polynomial $f(X)\in\mathbb{F}_p[X]$ given by $f(X)=X^{p^n}-X$. This follows from the Lagrange's Theorem applied to the group of units $\mathbb{F}_{p^n}^{\times}$. So if you fix an algebraic closure $\overline{\mathbb{F}_p}$ the uniqueness of such a subfield follows.