Determine the splitting field for a polynomial

Question

Let $P(X)=X^4+1 \in \mathbb{Q[X]}$. Find the splitting field for $P$ over $\mathbb{C}$ and determine the degree of it over $\mathbb{C}$.

My attempt: Roots of $P$ are $\alpha_1 = \sqrt{i},\alpha_2=-\sqrt{i},\alpha_3=i^{3/2},\alpha_4=-i^{3/2}$

Now the splitting field is $\mathbb{Q}(\sqrt{i},i)$. Since, $i$ has minimal polynomial of degree 2, $\sqrt{i}$ has minimal polynomial of degree 4, thus $[\mathbb{Q}(\sqrt{i},i):\mathbb{Q}]=[\mathbb{Q}(\sqrt{i}):\mathbb{Q}]=4$

Is there a more elegant argument? Can the roots of $P$ be expressed in a better form (analogue to roots of unity for $X^n-1$)?

Answer 1

Let $\alpha=\dfrac1{\sqrt 2}(1+i)$. The roots of $P$ are $\alpha$, $\alpha^3$, $\alpha^5$ and $\alpha^7$. So the splitting field over $\Bbb Q$ is $\Bbb Q(\alpha)$.

Answer 2

I'd use the decomposition $$X^4+1 = (X^2-i)(X^2+i).$$ Then $$X^2-i = (X-\alpha)(X+\alpha)\quad\mbox{and}\quad X^2+i = (X+\beta)(X-\beta),$$ where $i=\alpha^2$ and $-i=\beta^2$. So $\alpha$ and $\beta$ are 8-th roots of unity. Take the primitive 8-th root of unity $\xi=e^{2\pi i/8}$. Then $\alpha=\xi$ and $\beta=\xi^3$.

Added: There is a simpler way to arrive at the decomposition. For this, consider $$X^8-1 = (X^4+1)(X^4-1).$$ The decomposition of $X^4-1$ are the 4th roots of unity: $$X^4-1 = (X-1)(X+1)(X-i)(X+i).$$ So in view of the above notation, $\xi^0=1$, $\xi^2=i$, $\xi^4=-1$, $\xi^6=-i$, and so $$X^4+1 = (X-\xi)(X-\xi^3)(X-\xi^5)(X-\xi^7).$$ The zeros are all the primitive 8th roots of unity. Cheers!