What are the finite fields for which $-1$ is not a square? Of course they are of the form $\mathbb{F}_q$, with $q=p^r$, where $p$ prime, such that $p \neq 2$ and $p \equiv 3\pmod 4$. This, I remember from my good old Algebra courses. But for which values of $r$ is $-1$ not a square? For instance, if $r=2$, then by adjoining a root of $-1$ to $\mathbb{F}_p$, we get a field isomorphic to $\mathbb{F}_{p^2}$ containing a square root of minus $1$. So $r \neq 2$ also. I can probably rule out more cases this way, but, what is the general answer please?


3 Answers 3


A square root of $-1$ is an element of order $4$ in the group of units, $\mathbb{F}_q^*$. What do you know about this group, how many elements does it contain, which structure does it have? This will allow you to answer then exactly there is an element $i$ of order $4$. Once you have that, it will be easy to show $i^2 = -1$ as claimed.

Remark: Don't forget about the tricky case of $p=2$. Here, $-1=1 = 1^2$...

  • $\begingroup$ ah yes. Thank you! $\endgroup$
    – Malkoun
    Jun 19, 2017 at 12:34

Such a field must not contain $\mathbf F_{p^2}$. Hence it is necessarily a field $\mathbf F_{p^r}$ with $r$ odd (and $p\equiv 3\mod 4$), since $\mathbf F_{p^r}\subset\mathbf F_{p^s}$ if and only if $r\mid s$.

  • $\begingroup$ yes but is the condition r odd sufficient for minus 1 not to be a square? $\endgroup$
    – Malkoun
    Jun 19, 2017 at 12:24
  • $\begingroup$ I think so: if $\mathbf F_{p^r}$ contains a square root of $-1$ it contains (a copy of) $\mathbf F_{p^2}$ , so $r$ is even. $\endgroup$
    – Bernard
    Jun 19, 2017 at 12:37

Yes, you are right.

Here is how you can prove for $q=p$ prime.

Consider the morphism $x\mapsto x^2\in \mathbb F_p$ for $p\ne 2$, considering its kernel, you can show that there is


squares in $\mathbb F_p^*$.

Then let's define

$$X:=\{x\in \mathbb F_p^*,\ x^{(p-1)/2}=1\}.$$

You can show that all squares are in $X$, and since $\mathbb F_p$ is a field, $\vert X\vert \leqslant \frac {p-1}2$.

So $X$ contain all the non-null squares.

And you have:

$$-1\in X\iff \frac{p-1}2\equiv 0\pmod 4\iff p\equiv 1\pmod 4.$$

  • 1
    $\begingroup$ Nice proof! Thank you. Somehow these things used to be more complicated and longer when I was a student. It is a nice short proof. $\endgroup$
    – Malkoun
    Jun 19, 2017 at 18:09
  • $\begingroup$ @Malkoun Yes, bad typo indeed. I made myself the same comment when I first discovered this proof! $\endgroup$
    – E. Joseph
    Jun 19, 2017 at 18:18
  • $\begingroup$ nice proof. Thanks for sharing. $\endgroup$
    – Malkoun
    Jun 19, 2017 at 18:19
  • $\begingroup$ By the way, does your method give something interesting, for the map $x \mapsto x^n$, where $n | (p-1)$? $\endgroup$
    – Malkoun
    Jun 19, 2017 at 18:22
  • $\begingroup$ @Malkoun No idea, I will think about it. $\endgroup$
    – E. Joseph
    Jun 19, 2017 at 18:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.