$F$ is a finite field of size $s$. Prove that if $s=2^m$ for some $m>0$, then all elements of F have a square root.

Question

$F$ is a finite field of size $s$. Prove that if $s=2^m$ for some $m>0$, then all elements of $F$ have a square root.

(Hint: For some integer $i$, ($1\leq i \leq s-1$), $i$ is odd if and only if $(s-1)+i$ is even.)

My attempt:

If $s=2^m$, then $s$ is even. This means $s-1$ is odd. So the multiplicative group has an odd number of elements.

Let $x$ be the primitive element of $F$ such that: $F/\{0\}=\{x, x^2, x^3, ... , x^{s-1}\}$.

I tried to arrive at a contradiction assuming there existed an element without a square root, but I couldn't think of anything. Also I'm not quite sure how to use the hint.

Answer 1

I don’t see the point of the hint, and I think it is leading you rather far afield. Rather, let $a$ be an element of your field of cardinality $2^m$. Since $a^{2^m-1}=1$, you have $a^{2^m}=a$. Stare at this, and the square root of $a$ pops out at you.

Answer 2

Consider the Frobenius map $\sigma:x\mapsto x^2$. Then since $(x+y)^2=x^2+y^2 \pmod{2}$, $\sigma$ is an injective field homomorphism. The kernel of $\sigma$ is the set of all elements in our field satisfying $x^2=1$. However, since the multiplicative group of our field has size $2^m-1$ which is relatively prime to $2$, it follows that there are no elements of order $2$, and so the kernel of $\sigma$ is trivial. Thus $\sigma$ is surjective, and the claim follows.

Remark: In general it is a good idea to examine the Frobenius map when given such a problem.