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

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

• 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. – Adam Higgins Jan 30 at 0:54
• Also, $F_2 = \{0,1\}$, not $\{0,1,2\}$. – Adam Higgins Jan 30 at 0:55
• 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 $k\in K$. – Adam Higgins Jan 30 at 0:57

$$X^2-3$$ 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 $$1$$.