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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?

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

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