TITLE: Elements in finite field extensions
QUESTION [5 upvotes]: Let $A,K$ be finite fields with $K\supset A$. If $[K:A]=3$, I would like clarification as to why, if $x$ is not a square in $A$, then $x$ is not a square in $K$. My notes just mention this fact, but without proof.

REPLY [7 votes]: I agree with Dylan and Dilip that this problem is best approached without using the fact that the fields in question are finite. But if you really want something that uses the properties of finite fields, may I propose the following.
Assume that $|A|=q$, so $|K|=Q=q^3$, where $q$ is a power of a prime. The index of the multiplicative group $A^*$ in the multiplicative group $K^*$ is $(Q-1)/(q-1)=q^2+q+1$. This is always an odd number. So if the square of an element $z\in K^*$ is in $A^*$, then we must have $z\in A^*$, for otherwise the order of the coset $zA^* \in K^*/A^*$ would be two contradicting the fact that $|K^*/A^*|$ is odd. The claim follows from this.