# Splitting field of $x^2+[1]$ over $\mathbb Z_2$

While finding splitting field of $x^2+[1]$ over $\mathbb Z_2$ I found that all roots of $x^2+[1]$ lies in $\mathbb Z_2$ than how can I find splitting field ?

• It is $\mathbf Z_2$ itself: in a field of characteristic $2$, $\;x^2+1=(x+1)^2$. – Bernard Apr 30 '18 at 6:55
• Provided you mean $\mathbb{Z}_2 = \mathbb{Z}/2\mathbb{Z}$, $x^2+1$ factors as $(x+1)^2$, so already splits into linear factors – Mark Apr 30 '18 at 6:55
• Yes, I think one of the characterization of splitting field of $f$ over $F$ is the smallest field extension of $F$ containing all the roots of $f$. More precisely, the intersection of all subfields of $F$ where $f$ splits. – Alex Vong Apr 30 '18 at 8:06

$x^2+1 = (x+1)^2$, so it actually already splits. Thus $\Bbb{Z}_2$ is in fact the splitting field. Remember that the $K$ is a splitting field of $f(x) \in F[x]$ if $f(x)$ splits completely over $K$ and if $K = F(\alpha_1,\cdots,\alpha_n)$ where $\alpha_1,\cdots,\alpha_n$ are the roots of $f(x)$ (We then refer to the splitting field since any two splitting fields are isomorphic). In this case, the splitting field would be $\Bbb{Z}_2(1)=\Bbb{Z}_2$.