A splitting field extension of $x^4 – 5$ over $\mathbb{Q}$ is $K = \mathbb{Q}(\sqrt[4]{5}, i)$, and $[K:\mathbb{Q}] = 8$.
Let $\alpha = \sqrt[4]{5}$ and $\gamma = \alpha + i$. I'm trying to show $\gamma$ is a primitive element, and it is getting complicated. On a computer, I get
$$\begin{aligned}
{\gamma}^2 &= {\alpha}^2 + 2\alpha i – 1 \\
{\gamma}^3 &= {\alpha}^3 + 3{\alpha}^2i – 3\alpha – i \\
{\gamma}^4 &= 4{\alpha}^3i – 6{\alpha}^2 – 4{\alpha}i + 6 \\
{\gamma}^5 &= -10{\alpha}^3 – 10{\alpha}^2i + 10\alpha + 26i \\
{\gamma}^6 &= -20{\alpha}^3i + 20{\alpha}^2 + 36{\alpha}i – 76 \\
{\gamma}^7 &= 40{\alpha}^3 + 56{\alpha}^2i – 112\alpha – 176i
\end{aligned}$$
Next, I look at the coefficients of $\gamma, {\gamma}^3, {\gamma}^5$, and ${\gamma}^7$ to form a $4 \times 4$ matrix. On a computer again, I then check that the matrix is invertible, and so there are linear combinations to solve for $\alpha$ and $i$. Thus $\mathbb{Q}(\sqrt[4]{5}, i) \subseteq \mathbb{Q}(\sqrt[4]{5} + i)$.
Is there an easier solution?
By the way, this problem is in Chapter 7 of the book on Galois theory by Garling, and Galois groups are not yet covered. A more elementary solution is expected.
Best Answer
I would do it this way, without any linear algebra, because this is a rather simple situation:
Consider $\alpha$ as an irrationality over the Gaussian numbers, $\Bbb Q(i)$. You know that $X^4-5$ is still irreducible as a $\Bbb Z[i]$-polynomial, because $(5)$, though not a prime, is product of two distinct primes, $5=(2+i)(2-i)$, and Eisenstein applies. That is, $f(X)=\text{Irr}\bigl(\alpha,\Bbb Q(i)[X]\bigr)=X^4-5$ and by translation, $\text{Irr}\bigl(\alpha+i,\Bbb Q(i)[X]\bigr)=f(X-i)=X^4-4iX^3-6X^2+4iX-4=g(X)$. Now your desired polynomial is just $g(X)\overline g(X)$, where the bar is complex conjugation of each coefficient.