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$F$ 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.

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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.

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    $\begingroup$ +1 but the hint is not as misleading as you think. The nonzero elements can be expressed as $\alpha^i$, $0 \leq i \leq s-2$, and so if $i$ is even, $\alpha^i$ has square root $\alpha^{i/2}$, while if $i$ is odd, then $s-1+i$ is even, and since $\alpha^{s-1} = 1$, we have that $\alpha^i = \alpha^{s-1+i}$ has square root $\alpha^{(s-1+i)/2}$. $\endgroup$ – Dilip Sarwate Apr 7 '14 at 2:34
  • $\begingroup$ @DilipSarwate, right you are! I was perhaps too concentrated on my own method. $\endgroup$ – Lubin Apr 7 '14 at 2:38
  • $\begingroup$ Wonderful, and I was hoping for an explanation of the hint. Thanks guys! $\endgroup$ – Bobby Lee Apr 7 '14 at 2:43
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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.

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  • $\begingroup$ Why are you looking for $2^m-1$ to be coprime with $2$? And how does this imply that there are no elements of order $2$? $\endgroup$ – Bobby Lee Apr 7 '14 at 2:05
  • $\begingroup$ It’s a result in basic group theory that in a finite group, the order of an element divides the order of the group. Since $2^m-1$ is odd, it is not divisible by $2$. So there are no elements of order $2$. $\endgroup$ – Lubin Apr 7 '14 at 2:15
  • $\begingroup$ Oh yes, Lagrange's Theorem! I understand up to the part where you claim that the function is surjective. This function that maps an element of the multiplicative group to its square is injective. And only $1$ maps to $1$, so no element has order $2$. So each nonidentity element has a square that is not the identity. Now, do you claim that the function is surjective because the kernel of $\sigma$ is trivial, so each square root is unique? $\endgroup$ – Bobby Lee Apr 7 '14 at 2:29
  • $\begingroup$ Alternatively: since $\sigma:F \to F$ is injective on a finite set, it is also surjective. $\endgroup$ – sdcvvc Mar 29 '15 at 18:02

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