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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.

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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
up vote 8 down vote accepted

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.

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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.

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