When is a number square in Galois field $p^n$ if it's not square mod $p$?

Question

Here is the problem, that I'm stuck on.

There is no square root of $a$ in $\mathbb{Z}_p$. Is there square root of $a$ in $GF(p^n)$?

Well, it's certainly true that $$x^{p^n}=x$$ and $$x^{p^n-1}=1$$ for nonzero $x$.

If there is such an $x$ that $a=x^2$ then $$a^{\frac{p^n-1}{2}}=1$$ Now this actually is an equation in $\mathbb{Z}_{p}$ only, so I can write $$a^{\frac{p^n-1}{2}}\equiv 1\mod p$$

Now the answer seems to be just a touch away because I know that $x^2\equiv a \mod p$ iff $a^{(p-1)/2}\equiv 1 \mod p$. How can I tie the ends together, what am I missing?

Answer 1

If $a$ is not a square in $\mathbb F_p$, then it is a square in $\mathbb F_{p^n}$ precisely when $n$ is even. Indeed, $\mathbb F_p(\sqrt a)$ is a quadratic extension of $\mathbb F_p$, hence $\mathbb F_p(\sqrt a) = \mathbb F_{p^2}$. Now, we know that $\mathbb F_{p^m} \subseteq \mathbb F_{p^n}$ if and only if $m \mid n$. It follows that $a$ is a square in $\mathbb F_{p^n}$ if and only if $\mathbb F_p(\sqrt a) \subseteq \mathbb F_{p^n}$ if and only if $2 \mid n$.

Answer 2

Because finite fields are uniquely determined by their order, you know that if $a\in\Bbb Z$, then one of two things is true, either $a=x^2$ for some other $x\in\Bbb Z/p=\Bbb F_p$, or not. In the case it is not, then $x^2-a$ is irreducible over $\Bbb F_p$. If so then

$$\Bbb F_p[x]/(x^2-a)\cong\Bbb F_{p^2}$$

and this is independent of choice of non-square $a\in\Bbb F_p$. Since there is only one field of order $p^2$, all integers are squares iff

$$\Bbb F_{p^n}\supseteq\Bbb F_{p^2}\iff 2|n$$