# Extension of Splitting Fields over An Arbitrary Field

Let $$F$$ be a field in which $$0 \neq2$$ in $$F$$, and consider $$f=x^4+1$$. If $$E$$ is the splitting field for $$f$$ over $$F$$, it turns out that $$E$$ is a simple extension of $$F$$. How does one realize this fact? I'm not so sure as to what field element I can adjoin to $$F$$ to allow $$f$$ to split into linear factors. Finding the splitting field over something like $$\mathbb{Q}$$ is straight forward and easy in comparison, but I'm having trouble working with any general field $$F$$.

Also, if we indeed did have that $$0=2$$ in our field $$F$$, then $$f=x^4+1=(x+1)^4$$, so $$F$$ is its own splitting field, is this correct reasoning?

• Letting $\alpha$ be any root, then $f$ splits as $(x-\alpha)(x+\alpha)(x-\alpha^3)(x+\alpha^3)$ in $F[\alpha]$. – Mike Earnest Mar 14 at 1:53

If $$\theta$$ is a root of $$x^4+1$$, then so are $$\theta,\theta^3,\theta^5,\theta^7$$. These are all different because $$\theta$$ has order $$8$$ in $$E^\times$$, since $$\theta^4=-1\ne1$$. Therefore, $$x^4+1$$ splits in $$F(\theta)$$.
If $$\operatorname{char}(F)=2$$, then $$x^4+1=(x+1)^4$$ and $$E=F=F(1)$$, also a simple extension.