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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.

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+1 for showing your thoughts, and working out (correctly) many of the answers. Indeed, it is easy to see that $|E^*/F^*|$ is odd, when $[E:F]$ is, so your claim of no new square roots emerging within an odd degree extension follows. The existence of square roots in even degree extension has been adequately covered in the answers below. –  Jyrki Lahtonen Jan 24 '13 at 12:00

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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.

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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.

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