I came across one problem in a finite field as follows:
Let $F$ be a finite field. Show that if $a, b\in F$ both are non-squares, then $ab$ is a square.
I wanted to prove it by using the idea of Biquadratic field extension. But there is no biquadratic extension over finite fields. Please, any hints for proving above fact? Thanks.
Best Answer
If the characteristic of $F$ is $2$, then $x\mapsto x^2$ is an automorphism of $F$ and therefore every element is a square. So we can assume the characteristic is an odd prime.
Consider the multiplicative group $F^*$ of nonzero elements; the map $x\mapsto x^2$ is a group endomorphism of $F^*$ with kernel $\{1,-1\}$. Therefore the image $H$ of this map is a subgroup satisfying $|H|=|F^*|/2$; this amounts to saying that $H$, which is the set of all squares in $F^*$, has index $2$. Therefore $F^*/H$ is a two-element group and the statement follows.