Prove that $\mathbb{Q}[\sqrt[4](7), i] = \mathbb{Q}[\sqrt[4](7)+ i]$

Question

I want to prove that $\mathbb{Q}[\sqrt[4](7), i] = \mathbb{Q}[\sqrt[4](7)+ i]$.

Of course $"\supseteq"$ is clear. Do you have any hints how I can show that $\sqrt[4](7), i \in \mathbb{Q}[\sqrt[4](7)+ i]$

Answer 1

Let $f(x) = x^2+1$. Now consider $g(x) = f(\sqrt[4]{7} + i - x) = (\sqrt[4]{7} + i - x)^2 + 1 \in \mathbb{Q}(\sqrt[4]{7}+i)[x]$. It's obvious that $\sqrt[4]{7}$ is a root of it. By Vieta's Formula the second root is $\sqrt[4]{7} + 2i$, which is not a root of $x^4 - 7$ and so the minimal polynomial of $\sqrt[4]7$ over $\mathbb{Q}(\sqrt[4]{7}+i)$ is $x - \sqrt[4]{7}$, so $\sqrt[4]7 \in \mathbb{Q}(\sqrt[4]{7}+i)$.

Then trivially $i \in \mathbb{Q}(\sqrt[4]{7}+i)$, so $\mathbb{Q}(\sqrt[4]{7},i) \subseteq \mathbb{Q}(\sqrt[4]{7}+i)$

Answer 2

Here’s a much more pedestrian argument that $\Bbb Q(\sqrt[4]7,i)\subset\Bbb Q(\sqrt[4]7+i)$ than that of @Stefan4024, but you may find the method more easily generalized.

You see that $x^4-7$ is $\Bbb Q(i)$-irreducible, by Eisenstein. This shows that $[\Bbb Q(\sqrt[4]7,i):\Bbb Q]=8$. You only need to show that $\sqrt[4]7+i$ satisfies an irreducible $\Bbb Q$-polynomial of degree eight, to show that $\Bbb Q(\sqrt[4]7+i)$ is also an extension of degree eight over $\Bbb Q$.

Well, since $g(x)=x^4-7$ is irreducible over $\Bbb Q(i)$, so is $h(x)=g(x-i)=(x-i)^4-7$, and this polynomial has $\sqrt[4]7+i$ as a root. The conjugate polynomial $\bar h(x)=(x+i)^4-7$ is similarly $\Bbb Q(i)$-irreducible. Now $H(x)=h(x)\bar h(x)\in\Bbb Q[x]$, a $\Bbb Q$-polynomial of degree eight, has $\sqrt[4]7+i$ as a root, and I need only show that $H$ is $\Bbb Q$-irreducible.

Indeed, if a monic $k(x)\in\Bbb Q[x]$ divides $H$ properly, then of course it’s a divisor of the $\Bbb Q(i)$-polynomial $h\bar h$, which is the factorization of $H$ into $\Bbb Q(i)$-irreducibles. Thus $k=h$ or $k=\bar h$, both of which are impossible, because $k$ has $\Bbb Q$-coefficients, while $h,\bar h$ don’t. So uniqueness of factorization in $\Bbb Q(i)[x]$ gives us irreducibility of $H$ as a $\Bbb Q$-polynomial.

Answer 3

In the spirit of Galois theory, things are more "visible" when dealing (directly or indirectly) with groups rather then with polynomials. The field $K=\mathbf Q(i, \sqrt [4] 7)$ is obviously the splitting field of the polynomial $X^4-7$, which is irreducible over $\mathbf Q$ by Eisenstein criterion. Hence $K/\mathbf Q$ is galois, of degree $8$ because $i$ cannot belong to the real subfield $\mathbf Q (\sqrt [4] 7)$. Consider the ascending chain of successive quadratic extensions $\mathbf Q < \mathbf Q(i) < \mathbf Q(i,\sqrt 7) < K$. Note that $ \mathbf Q(i+\sqrt 7)=\mathbf Q(i,\sqrt 7)$ : there is an obvious inclusion, and $i+\sqrt 7$ cannot belong to $\mathbf Q(i)$ because $\sqrt 7$ is real, not rational. Analogously, $\mathbf Q(i+ \sqrt [4] 7)=K$: there is an obvious inclusion, and $i+ \sqrt [4] 7$ cannot belong to $\mathbf Q(i,\sqrt 7)$ because $\sqrt [4] 7 - \sqrt 7$ is real, not in $\mathbf Q(\sqrt 7)$ (the three quadratic subfields of $\mathbf Q(i,\sqrt 7)$ are $\mathbf Q(\sqrt 7),\mathbf Q(i), \mathbf Q(i\sqrt 7)$).

Galois groups played the role of "éminence grise" in the whole argument above : actually, it is easily checked that $Gal(K/\mathbf Q)$ is the dihedral group $D_8$, and what we did was merely to follow a particular resolution of $D_8$ (in the sense of solvable groups).

Answer 4

In general such problems do not need any concepts from modern algebra. Usual algebraic manipulation in the style of classical algebra suffices. Thus if $a=\sqrt[4]{7}+i$ then $(a-i) ^4=7$ or $$a^4-4a^3i-6a^2+4ai+1=7$$ ie $$i=\frac{a^4-6a^2-6}{4a^3-4a}$$ so that $i\in\mathbb{Q} (a) $ and then $\sqrt[4]{7}=a-i\in\mathbb{Q}(a)$ and you are done.