Prove that $\sqrt{-1}$ is not an element of $\mathbb{Q}(\sqrt[4]{-7})$ I know this result seems obviously true but I'm not sure how to start.
 A: Okay, so it seems that you know that $$\Bbb Q\left(\sqrt[4]{-7}\right)=\left\{q+r\alpha+s\alpha^2+t\alpha^3:q,r,s,t\in\Bbb Q\right\},$$ where $\alpha^4=-7$ is given.
Now, if we had $\sqrt{-1}\in\Bbb Q\left(\sqrt[4]{-7}\right),$ then $\sqrt{-1}=q+r\alpha+s\alpha^2+t\alpha^3$ for some $q,r,s,t\in\Bbb Q,$ yes? Consequently, we have $$-1=\left(q+r\alpha+s\alpha^2+t\alpha^3\right)^2,\tag{$\star$}$$ or put another way, $$-1=\sum_{k=0}^3f_k(q,r,s,t)\alpha^k,$$ where each $f_k(q,r,s,t)$ is a polynomial in $q,r,s,t.$ I leave the calculation of these polynomials to you. Just expand the square in $(\star)$ and simplify using the fact that $\alpha^4=-7$.
Since $-1\in\Bbb Q\subset\Bbb Q\left(\sqrt[4]{-7}\right)$ and $\left\{1,\alpha,\alpha^2,\alpha^3\right\}$ is a basis for $\Bbb Q\left(\sqrt[4]{-7}\right)$ over $\Bbb Q,$ then we have the following system of equations: $$\begin{cases}f_0(q,r,s,t)=-1\\f_1(q,r,s,t)=0\\f_2(q,r,s,t)=0\\f_3(q,r,s,t)=0\end{cases}$$ While cumbersome, we can conclude from this system of equations that at least one of $q,r,s,t$ is not rational, yielding a contradiction. If you take this route, I would recommend that you proceed by cases, dealing with the case that $q=0$ (with subcases $r=0$ and $r\neq 0$), then the case that $q\neq 0$ and $t=0$, then the case that $qt\neq 0$ (so that each of $q,r,s,t\neq 0$). That last case is the least obvious, but I managed to prove that $s$ was irrational on two post-it notes in that case, so hopefully it isn't too much of a struggle.
Let me know if you have any trouble, any more questions about my answer, or just want to bounce your work off of somebody.
