Suppose that $2\nmid q$ and $a\in \mathbb{F}_{q^2}$ (more precisely: $a \in \mathbb{F}_{q^2} \setminus \mathbb{F}_q$) such that $a^2 \in \mathbb{F}_q$. Why is it true that actually $a = -a^q$?
An element of a $a\in \mathbb{F}_{q^2}$ such that $a^2 \in \mathbb{F}_q$.
0
$\begingroup$
finite-fields
-
0What is $F_q2$? – 2017-02-03
-
0@terrace A field with $q^2$ elements, $q = p^k$. – 2017-02-03
1 Answers
2
Since $a^2 \in \mathbb{F}_q$, we have $a^{2q} = a^2$. This means $(a^{q}-a)(a^q+a)=0$. The first bracket is nonzero since $a \not\in \mathbb{F}_q$, hence the result.
-
0That's a good way! The same argument in different words is that the minimal polynomial of $a$ over $\Bbb{F}_q$ is $x^2-a^2$. The zeros of that are $\pm a$. But, by Galois theory, they are also $a$ and $a^q$. I do like your way more. – 2017-02-03