If $[K:F] = 3$, show a non square in $F$ is a non square in $K$

Question

Let $F \subset K$ be finite fields with $[K:F] = 3$. Show that if $\alpha \in F$ is not a square in $F$, it is not a square in $K$.

My attempt:

$[K:F] = 3 \implies \forall x \in K$, $x = \beta_1a_1 + \beta_2a_2+\beta_3a_3$, where $\beta_1, \beta_2, \beta_3 \in F$ are basis for K. So if $\alpha \in F$ not a square then $x^2-\alpha = 0$ has no solution in F. So then we have to show that $(\beta_1a_1 + \beta_2a_2+\beta_3a_3)^2 - \alpha = 0$ also has no solutions.

I am not sure if I am even headed in the correct direction. Any help is appreciated.

Arthur · Accepted Answer · 2019-04-11T11:49:24.440

1

Assume for contradiction that $\alpha$ is a non-square in $F$, but there is a $\beta\in K$ such that $\beta^2 = \alpha$. Then because $\beta$ is a root of $X^2 - \alpha$, we have $[F(\beta):F] = 2$.

But $[K:F] = [K:F(\beta)]\cdot [F(\beta):F]$, which is impossible.

edited Apr 11 '19 at 11:49

answered Apr 11 '19 at 11:46

Arthur

199,419

Does that imply then this is only possible in extension of even degrees and not odd? – mlimber Apr 11 '19 at 11:49
@mlimber Yes, that is what it means. An extension that adds any square root to any non-square element must have even degree. – Arthur Apr 11 '19 at 11:50
Also, the $\beta$ that you use is not related to the $\beta_i$ that I am using as the basis right? – mlimber Apr 11 '19 at 11:50
@mlimber No, it's not. We had $\alpha$ established as an element of the field already, so I used $\beta$ because of that. – Arthur Apr 11 '19 at 11:51
Got it, thanks a ton – mlimber Apr 11 '19 at 11:51

If $[K:F] = 3$, show a non square in $F$ is a non square in $K$

1 Answers1