Find a splitting field of $x^2 + 1$ over $\mathbb{Z}_3$

Question

We know that $x^2 -3$ is irreducible in $\mathbb{Q}[x]$. We also know that $\sqrt{3}$ solves $x^2 - 3$. As a result, $x^2 - 3 = (x-\sqrt{3})(x+\sqrt{3})$. This implies that the splitting field of $x^2 - 3$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt{3})$.

At the same time, we also know that the field $\mathbb{Q}[x]/(x^2 - 3)$ has element that solves $x^2 - 3$. So I suppose $\mathbb{Q}[x]/(x^2-3)$ is isomorphic to $\mathbb{Q}(\sqrt{3})$.

Now I have the following question

Find a splitting field of $x^2 + 1$ over $\mathbb{Z}_3$

One answer would be $\mathbb{Z}_3[x]/(x^2 + 1)$. But does it make sense to say one splitting field is $\mathbb{Z}_3 (\alpha)$ where $\alpha$ solves $x^2 + 1$?

Answer 1

let $\alpha$ is root of $f(x)=x^2+1$ in Extended field of $\mathbb{Z}_3$,therefore $$f(\alpha)=f(-\alpha)=\alpha^2+1=0$$ on the other hand $$-\alpha=2\alpha$$ as a result $$\mathbb{Z}_3(\alpha)=\{a+b\alpha|\,a,b\in\mathbb{Z}_3\}\cong \frac{{{\mathbb{Z}}_{3}}[x]}{({{x}^{2}}+1)}=a+bx+\left\langle {{x}^{2}}+1 \right\rangle$$

Answer 2

Yes of course. It's a basic theorem in field theory that for each field $F$ and irreducible polynomial $f(x) \in F[X]$, there exists some field extension $E$ of $F$ such that $f$ has a root (say $\alpha$) in $E$. Since your polynomial has degree $2$, $\mathbb{Z}_3(\alpha)$ is necessarily the splitting field of $x^2+1$ over $\mathbb{Z}_3$

Answer 3

The idea is that if $ f $ is irreducible in $ K[x] $ where $ K $ is a field, then the ring homomorphism $ \phi : K[x] \to K[\alpha] $ ($ \alpha $ is a root of $ f $) defined by $ \phi(p(x)) = p(\alpha) $ has kernel $ (f) $, so by the first isomorphism theorem we have $ K[x]/(f) \cong K[\alpha] $. Since $ (f) $ is maximal, this quotient is actually a field, which gives $ K[x]/(f) \cong K(\alpha) $.

Since $ x^2 + 1 $ is irreducible in $ \mathbb{F}_3[x] $, we conclude that $ \mathbb{F}_3[x]/(x^2 + 1) $ is a field which contains a root of $ x^2 + 1 $, and it contains the other root since the two roots sum to zero. Since it is the smallest such field, we conclude that it is in fact the splitting field of $ x^2 + 1 $ over $ \mathbb{F}_3 $.