Square roots of $\Bbb F_p$

Question

Can anyone please help me to show that $\Bbb F_{p^2}$ contains all the square roots of $\Bbb F_p$ where $p$ is a prime? Thanks for any help.

Answer 1

Let $a\in\Bbb F_p$ Either it is a square or it isn't. Assume, WLOG, that $a$ is not a square. The $\Bbb F_p[x]/(x^2-a)$ is a field of size $p^2$ which contains a square root of $a$ (namely the image of $x$ in the quotient). But by the uniqueness of finite fields of a given degree, this field, $\Bbb F_p[x]/(x^2-a)=\Bbb F_{p^2}$.

Answer 2

The argument of Adam Hughes is the one that springs to mind, but here’s one of a different flavor, seemingly independent. It depends only on the fact that a finite subgroup of the multiplicative group of a field is cyclic.

Any square root of an element of $\Bbb F_p^*$ will have order dividing $2(p-1)$. But the multiplicative group of $\Bbb F_{p^2}$ has order $(p-1)(p+1)$, which is divisible by $2(p-1)$.