1

I know this result seems obviously true but I'm not sure how to start.

Jean Marie
  • 88,997
Eu Meng Chong
  • 135
  • What definitions of $\Bbb Q\left(\sqrt[4]{-7}\right)$ do you have? – Cameron Buie Nov 12 '16 at 18:20
  • @CameronBuie, Let a = quarticroot(-7). Then the minimal polynomial of a over Q is x^4 + 7, which has degree 4. Therefore [Q(a):Q] = 4. So a basis for Q(a) is 1, a, a^2, a^3 – Eu Meng Chong Nov 12 '16 at 19:07

1 Answers1

2

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.

Community
  • 1
Cameron Buie
  • 105,149
  • I had the same idea but I thought it was more-or-less operative. Is there another less computational solution? – Xam Nov 12 '16 at 22:59
  • @Eu: What have you come up with so far? – Cameron Buie Nov 13 '16 at 19:19
  • @I got -1 = a^2 -14bd -7c^2, 0 = 2ab - 14cd, 2ac + b^2 -7d^2 = 0, 0= 2ad + 2bc, where -1 = (a + bx + cx^2 + dx^3)^2 where x^4 = -7. – Eu Meng Chong Nov 13 '16 at 19:26
  • @Eu: That's much easier than what I did (in part because I made a calculation error. If you've proved that $ad\neq0$ implies that $b,c\neq 0,$ then you can proceed as you say and conclude $7d^2+b^2=0.$ This isn't (on its own) enough to conclude that $b=d=0,$ though. Rather, we must also use our assumption that each of $a,b,c,d$ is rational (and so real), from which we get the contradiction you say. Nicely done! :-) – Cameron Buie Nov 13 '16 at 20:21
  • @Charter: I imagine there may be, but this isn't too involved, computationally. – Cameron Buie Nov 14 '16 at 13:00