Are nonsquares actually squares in extensions of even degree?

Question

I was playing around with finite fields, and noticed that if $F$ is a finite field, and $E$ an extension of odd degree, then every nonsquare $a\in F$ is again a nonsquare in $E$. I found this out by looking at the order of the coset of a possible root of $a$ in the quotient group $E^\times/F^\times$.

I now wonder if $E$ is an extension of even degree, does this lead every element of $F$ to have a square root in $E$? (If possible, is there a way to do so without heavy use of Galois theory?) Thanks.

Answer 1

If $F$ is finite, yes. The multiplicative group of a finite field of order $q$ is cyclic of order $q-1$, so if $[E:F]=r$, $F^\times$ is embedded in $E^\times$ as the subgroup of $(q^r-1)/(q-1)$th powers. If $q$ is odd and $r$ is even, $(q^r-1)/(q-1)$ is even, so every element of $F^\times$ is a square in $E^\times$. If $q$ is even, $q^r-1$ is odd, so every element of $E^\times$ is a square.

Answer 2

It's easy, for finite fields, if you're allowed to use the fact that for each prime power $q$ there is exactly one field of $q$ elements. For if $F$ has $m$ elements, then every extension of $F$ by the square root of a nonsquare in $F$, being an extension of degree $2$, has $m^2$ elements, so all these extensions must be the same, so an extension having any one square root of an element of $F$ must have all of them.

It's also true for the reals but not for, say, the rationals.