Let $F_q$ be a finite field of $\operatorname{char}F\neq2$. suppose $a_1$,$a_2$,...,$a_q$ are the elements in $F_q$.
- show there exists $i$ such that for any $j$ $a_i\neq a_j^2$.
- compute the degree of the extension $F_q(\sqrt{a_1},...,\sqrt{a_q})/F_q$.
for the first question, I supposed not, then for any $i$ there exists $j$ s.t. $a_i=a_j^2$. then $f:F_q\to F_q, f(x)=x^2$ is an automorphism. so we can arrange $F_q$ elements to couples (except 0,1-so consinder them a couple). but then $2\mid q$ and this is a contradiction. Does it work?
I'm stuck with the second question.
Thank you in adavance.
Thank you, then the function I mentioned is surjective (by assumption) and a bijection since Fq is finite. But then f(1)=f(-1)=1, and it's a contradiction because in Fq 1 isn't equals to -1. for the second question, F_q(√ai) isomorphic to Fq iff sqrt(ai) is in F, else it's isomorphic to Fq^2, which also has the same char.