Show that $[\Bbb{Q}:\Bbb{Q}(\rho+i)]=8$ without using Primitive element theorem

Question

Let $\rho=p^{\frac{1}{4}}$ with $p$ prime. The proof that

$$\Bbb{Q}(\rho,i)=\Bbb{Q}(\rho+i)$$ follows directly from the Primitive element theorem. However, I'm curious if there is a explicit way of proving this by computing the minimal polynomial of $\rho+i$ over $\Bbb{Q}$ or at least show that

$$[\Bbb{Q}:\Bbb{Q}(\rho+i)]=8$$

since then by the transitivity property of the degree, we would get that

$$[\Bbb{Q}(\rho+i):\Bbb{Q}(\rho,i)]=1$$

and hence both extensions are equal.

I've tried it by brute force, that is letting $\alpha=\rho+i$ and squaring, substracting, etc to find a polynomial that has $\alpha$ as a root, but I always get things that are too messy to handle with degree $16$, so I'm need of new paths to solve the problem. My question the would be the following:

Is there any simple way of showing that $$[\Bbb{Q}:\Bbb{Q}(\rho+i)]=8$$ without using then Primitive element theorem, that is, computing the minimal polynomial of $\rho+i$ over $\Bbb{Q}$ or using any other tool?

Thank you in advance for your time!

Answer 1

Well, when I try to compute the minimal polynomial with $\alpha := \sqrt[4]{p} + \mathrm{i}$, I get:

\begin{align} \alpha & = \sqrt[4]{p} + \mathrm{i} & \implies \\ (\alpha - \mathrm i )^4 &= p & \implies\\ (\alpha^2-2\alpha\mathrm i-1)^2 &= p & \implies \\ \alpha^4-6\alpha^2+1-p & = \mathrm i \cdot (4\alpha^3 - 4\alpha) & (\mathrm 1) \end{align}

Square the last line, and (Wolfram Alpha is your friend) this leads to the polynomial

$$f(X) = X^8+4X^6+(6-2p)X^4+(4+12p)X^2+(p-1)^2$$

You can verify here, that $\sqrt[4]{p} + \mathrm i$ is a root of $f$.

Hmm, well now you could try to show that $f$ is irreducible. On a first glance this isn't obvious to me and may involve heroic computations.

You say that you know that $[\Bbb{Q}(\rho,\mathrm i):\Bbb{Q}] = 8$, so $[\Bbb{Q}(\rho+\mathrm i):\Bbb{Q}]$ divides 8 (so it's either 2,4 or 8). So another thing you could do now is show that it can't be 2 neither 4. In order to show that $[\Bbb{Q}(\rho+\mathrm i):\Bbb{Q}] \neq 2$, suppose it is. Then the minimal polynomial is of the form $X^2+aX+b \in \Bbb{Q}[X]$. Put in your $\alpha$ and get \begin{align} a \sqrt[4]{p}+\mathrm i a+b+\sqrt[4]{p}+2 \mathrm i \sqrt[4]{p}-1 & = 0 \implies \\ a & = -2\sqrt[4]{p}\ \ \text{(because complex part of $b$ must be 0)} \end{align} , which cannot be. So $[\Bbb{Q}(\rho+\mathrm i):\Bbb{Q}] \neq 2$. In a similar way, you could show that it can't be 4 either. But I don't think this is any fun.

On the other hand, ($1$) shows at once what Matt B had been talking about in the comments: it shows that $\mathrm i = \displaystyle \frac{\alpha^4-6\alpha^2+1-p}{4\alpha^3-4\alpha} \implies \mathrm i \in \mathbb{Q}(\alpha) \implies \mathbb{Q}(\rho, \mathrm i) \subset \mathbb{Q}(\rho + \mathrm i)$ and you're done.

PS: I'd rather write $[\Bbb{Q}(\rho+i):\Bbb{Q}]$ than $[\Bbb{Q}:\Bbb{Q}(\rho+i)]$ and put the bigger field on the left hand side of the ":".

Answer 2

Let $\alpha= \rho+i$. Now $p=\rho^4= (\alpha-i)^4$, and the right hand side will be of the form $f(\alpha)+ i g(\alpha)$ for $f,g$ polynomials, and $g(\alpha)\neq 0$ since $\alpha^2-1\neq 0$ because $\rho^2 -1 \neq 0$ is its real part). This shows that $i$ lies in the subfield generated by $\alpha$. Thus $\rho$ does too, and you're done.

Answer 3

Here is an answer using a bit of Galois theory, as requested by @user314159. I realised whilst typing that this was a bit longer than expected, but there is a lot less guesswork and we never need to find the minimal polynomial.

First note $K=\mathbb{Q}(\rho,i)$ is the splitting field of the polynomial $f=x^4-p$ over $\mathbb{Q}$ and hence is a Galois extension. We can directly determine that $[K:\mathbb{Q}]=8$ by the tower law since $f$ is irreducible and also has both real and complex roots.

This means that the Galois group $G=Gal(K/\mathbb{Q})$ has order $8$. Furthermore, we note that the subextension $\mathbb{Q}(\rho)/\mathbb{Q}$ is not Galois since $\rho$ has minimal polynomial $f$ and the splitting field of that is $K$. By the Galois correspondence theorem, $G$ must have a non-normal subgroup which means it must be isomorphic to the dihedral group $D_8$.

If we order the roots of $f$ to be $r_1=\rho, r_2=i\rho, r_3=-\rho, r_4=-i\rho$, then we can see that $G$ is generated by the elements $\sigma=(r_1,r_2)(r_3,r_4)$ and $\tau=(r_1,r_2,r_3,r_4)$.

Recall that $D_8$ has $10$ subgroups and so by the Galois correspondence theorem, if we can determine the number of elements of $G$ that fix $\rho+i$ (these form a subgroup), then we can find the degree of the extension.

Note that $5$ subgroups contain the element $\tau^2$ (this is the other element in the centre). Now $\rho+i=r_1+\frac{r_2}{r_1}$ so $\tau^2(\rho+i)=r_3+\frac{r_4}{r_3}=-\rho +i \neq \rho+i$. This means $\tau^2$ does not lie in the stabiliser subgroup and hence we've ruled out those $5$ options.

We are just left to check the elements $\sigma, \sigma\tau, \sigma\tau^2,\sigma\tau^3$ to rule out other subgroups with similar calculations. Having done this, we discover that the only element that fixes $\rho+i$ is the identity so the subgroup we're looking for has order $1$.

By the correspondence theorem again, this shows that $[\mathbb{Q}(\rho,i):\mathbb{Q}(\rho+i)]=1$ and so by the Tower Law, $[\mathbb{Q}(\rho+i):\mathbb{Q}]=8$ as required.