On every other finite field at least one of −1, 2 and −2 is a square, because the product of two non squares is a square

Question

[Except on field extensions of $\mathbb{F}_2$] On every other finite field at least one of $−1$, $2$ and $−2$ is a square, because the product of two non squares is a square.

I don't see why this is so. Please explain. This statement is from the Wikipedia article on factorisation of polynomials over finite fields.

Answer 1

Looking at the Wikipedia article, I think you meant to say "On every finite field of characteristic other than $2$" rather than "On every other finite field apart from $\mathbf{F}_2$". Also, the Wikipedia article doesn't mean to imply that the statement isn't true in finite fields of characteristic $2$—just that the characteristic $2$ statement is not needed for the argument in that particular passage. In fact, every element in a field of characteristic $2$ is a square. In particular, $-1=1$ and $2=-2=0$, both of which are obviously squares.

In the odd characteristic case, if you are asking how the statement that at least one of $-1$, $2$, and $-2$ is a square follows from the statement that the product of two non-squares is a square, then suppose that all three of $-1$, $2$, and $-2$ were non-squares. Then there's an immediate contradiction, since $(-1)(2)=-2$, and therefore $-2$ should be a square.

Others have given nice proofs that the product of two non-squares is a square. Also, once you know the fundamental fact that all nonzero elements of a finite field are powers of some primitive element $r$, it's straightforward to see this: if an element were a non-square, it would have to be an odd power of $r$. But the product of two odd powers of $r$ is an even power of $r$.

Answer 2

The statement is true in $ \mathbb{F}_2 $ as well, so I am not sure why it was excluded.

The idea is that the quadratic character is multiplicative in a finite field. Note that $ \phi : x \to x^2 $ is an endomorphism of the multiplicative group of any finite field $ K $, and its kernel is $ \ker \phi = \{ -1, 1 \} $ since the equation $ x^2 = 1 $ has precisely these solutions in any finite field. First, we assume that the characteristic of $ K $ is not 2, so that $ -1 \neq 1 $ and the kernel has $ 2 $ elements. In that case, $ \operatorname{Im} \phi = H $, which is the subgroup of all quadratic residues in $ K^{\times} $, is of index $ 2 $ in $ K^{\times} $. We may then consider the quotient group $ K^{\times}/H $ which has $ 2 $ elements, and consider the natural epimorphism $ \mu $ of $ K^{\times} $ onto its quotient. For any quadratic residue $ x $, we have $ \mu(x) = xH = H $. For a quadratic nonresidue $ x $, $ \mu(x) = xH $ cannot be $ H $ since $ x \in xH $ while $ x\notin H $, therefore $ \mu $ maps quadratic nonresidues to the other element of the quotient group $ K^{\times}/H $. Now, note that

$$ H^2 = H, H(xH) = xH, (xH)^2 = H $$

where $ xH $ denotes the other element of $ K^{\times}/H $, therefore there is an isomorphism $ \varphi : K^{\times}/H \to \{ 1, -1 \}^{\times} $. We define the quadratic character $ \chi $ as the composition $ \varphi \circ \mu $. It is clearly an epimorphism $ \chi : K^{\times} \to \{ 1, -1 \}^{\times} $ whose kernel is $ H $, the subgroup of quadratic residues. Therefore, for two quadratic nonresidues $ x $ and $ y $, we have

$$ \chi(xy) = \chi(x)\chi(y) = (-1)(-1) = 1 $$

Since $ \ker \chi = H $, we conclude that $ xy \in H $, i.e $ xy $ is a quadratic residue.

If $ K $ has characteristic $ 2 $, then the kernel of $ \phi $ is trivial, therefore it is actually an automorphism of $ K^{\times} $, and $ \operatorname{Im} \phi = K^{\times} $. In other words, every element is a quadratic residue.

Answer 3

The multiplicative group of any finite field is cyclic, so if it has even order half the elements are squares and if it has odd order all the elements are squares. In any case at least half the (non-zero) elements are squares.

If $y$ runs through the non-zero elements and $x$ is a non-square, then $xy$ also runs through the non-zero elements. But if $y$ is a square then $xy$ cannot be (because $xa^2=b^2$ implies $x=(a^{-1}b)^2$). So all the elements $xz$ as $z$ runs through the non-squares must be squares (in order to give us at least half the elements squares).