$x^2-3$ is separable over $\mathbb Q$ but not separable over $F_2$

Question

$x^2-3=(x-\sqrt{3})(x+\sqrt{3})$ over $\mathbb Q$, so that part makes sense.

Now, when it says $x^2-3$ is a polynomial over $F_2$, I imagine it means all the coefficients are calculated mod $3$, so $p(x)=x^2-3=x^2$ in $F_2$.

$x^2$ doesn't have a second solution other than $0$ in $F_2=\{0,1,2\}$, but how do I know there is not some algebraic extension that it will have two roots in?

Also, is this discussion so far accurate?

Answer 1

$X^2-3$ in $F_2$ means that the coefficients are calculated $mod$ $2$ and not $mod$ $3$.

So $x^2-3=x^2+1=(x+1)^2$ mod $2$. You cannot have an extension with two different roots since the factorization over a field is unique. If you assume that the coefficient of higher degree of factors of degree $>1$ is $1$.