# Elements in finite field extensions

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

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The minimal polynomials in $A[z]$ of elements of $K-A$ have degrees that are $[K:A] = 3$ or proper divisors thereof. So if $\sqrt{x} \in K-A$ is a square root of $x \in A$, then $z^2-x$ would be irreducible in $A[z]$ but split in $K[z]$ as $(z-\sqrt{x})(z+\sqrt{x})$. – Dilip Sarwate Feb 16 '12 at 14:45

I don't think we need assume that the fields are finite. If there exists a $y \in K$ such that $y^2 = x$, then I claim that you can write down the minimal polynomial for $y$ over $A$, and hence determine the degree of the field extension $A(y)/A$. Now recall the tower law for the degrees of field extensions.
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.