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$x^2-3=(x-\sqrt{3})(x+\sqrt{3})$ 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?

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    $\begingroup$ Because a polynomial of degree $n$ has at most $n$ Distinct roots counting multaplicities. Moreover, any field is integral domain, so $x^n = 0$ if and only if $x=0$ in any field extension of any ground field. $\endgroup$ – Adam Higgins Jan 30 at 0:54
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    $\begingroup$ Also, $F_2 = \{0,1\}$, not $\{0,1,2\}$. $\endgroup$ – Adam Higgins Jan 30 at 0:55
  • $\begingroup$ Thus, if you have a polynomial $f$ of degree $n$ over a field $K$ that has $n$-roots in $K$ counting multaplicities, then if $K’$ is any exntesion of $K$, then if an element of $k\in K’$ is a root of $f$, then $k\in K$. $\endgroup$ – Adam Higgins Jan 30 at 0:57
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$X^2-3$ 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 $1$.

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