# What are the finite fields for which -1 is not a square?

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?

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 $r\mid s$.

• yes but is the condition r odd sufficient for minus 1 not to be a square? – Malkoun Jun 19 '17 at 12:24
• 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. – Bernard Jun 19 '17 at 12:37

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, $-1=1 = 1^2$...

• ah yes. Thank you! – Malkoun Jun 19 '17 at 12:34

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

$$\frac{p-1}2$$

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:

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

• 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. – Malkoun Jun 19 '17 at 18:09
• @Malkoun Yes, bad typo indeed. I made myself the same comment when I first discovered this proof! – E. Joseph Jun 19 '17 at 18:18
• nice proof. Thanks for sharing. – Malkoun Jun 19 '17 at 18:19
• By the way, does your method give something interesting, for the map $x \mapsto x^n$, where $n | (p-1)$? – Malkoun Jun 19 '17 at 18:22
• @Malkoun No idea, I will think about it. – E. Joseph Jun 19 '17 at 18:29