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If $F$ is a finite field such that every element is a square, why must $char(F)=2$?

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up vote 22 down vote accepted

If every element is a square, then the map $x\mapsto x^2$ is a surjection from $F$ to itself. Since $F$ is a finite field, the map is a bijection, which means that since $-1$ and $1$ have the same image, then $-1=1$, hence the characteristic is $2$.

In fact, the converse holds as well. Suppose $F$ is a finite field of characteristic $2$: then it has $2^n$ elements, for some $n$, hence the multiplicative group of nonzero elements is $F$ of order $2^n-1$, which is odd. In particular, there is no element of order $2$ by Lagrange's theorem. The map $f\colon F-\{0\}\to F-\{0\}$ given by $f(x)=x^2$ is a group homomorphism whose kernel consists of all elements of exponent $2$; by the comment just made, the kernel is trivial, hence $f$ is one-to-one and hence onto, so every nonzero element of $F$ is a square. And of course, $0$ is a square.

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Ah, nice. Thank you. – user6495 Apr 1 '11 at 16:33

If every elt a finite field of order $\rm\: 2\:n+1\:$ is a square then $\rm\ x^{\:n} - 1\ $ has $\rm\: 2\: n\:$ roots (all elts $\ne 0$) $\ \Rightarrow\Leftarrow$

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