Show that $\sqrt{-1} \notin \mathbb{F}_7$

Question

I'm doing some group theory tasks and I need to show that $\sqrt{-1} \notin \mathbb{F}_7$. I'm really not sure how to achieve this, and this is my attempt so far:

$\mathbb{F}_7=\{0, 1, 2, 3, 4, 5, 6\}$

$x = \sqrt{-1} \\ x^2 = -1 \\ x^2 + 1 = 0$

I'm not sure where to go from here, because I can't factor $x^2+1$ any further. I hope someone can help me with this, thanks a lot in advance!

Answer 1

First of all

$$\mathbb{F}_7 = \{0,1,2,3,4,5,6\}$$

Now just take any of these numbers and compute $x^2+1$. Show that nothing equals zero

Answer 2

$\mathbb{F}_7$ has seven elements: $\{0,1,2,3,4,5,6\}$. In this case, given the small number of elements, it is simpler to observe that $-1=6$ since $1+6=0$, and then just square all the elements to show $x^2 \neq 6$ for all $x \in \mathbb{F}_7$.

Answer 3

A prerequisite to group theory is elementary number theory. So we might know that the Legendre symbol is $$ \left( \frac{-1}{7} \right)=(-1)^{\frac{7-1}{2}}=-1. $$ Hence $-1$ is not a square in $\Bbb F_7$.