Find a primitive element for the extension $\mathbb Q(i,\sqrt3,e^{{\pi i/4}})/\mathbb Q$.
I was guessing the primitive element is $i+\sqrt3+e^{{\pi i/4}}$. It is obvious that $\mathbb Q(i+\sqrt3+e^{{\pi i/4}})\subset \mathbb Q(i,\sqrt3,e^{{\pi i/4}})$.
To show that the converse inclusion, we need to show $i,\sqrt3$ and $\text{}e^{{\pi i/4}}$ are contained in $Q(i+\sqrt3+e^{{\pi i/4}}).$
I have no clue how to do it. Can any one help me on this question?
One can see that $$\mathbb Q(i,\sqrt{3},e^{{\pi i/4}})=\mathbb Q(e^{{2\pi i/3}},e^{{\pi i/4}})=\mathbb Q(e^{{2\pi i/3}})\cdot \mathbb Q(e^{{\pi i/4}}) \subseteq \mathbb{C}.$$ That is, we have the composite of the cyclotomic extensions $\mathbb Q(\zeta_3)$ and $\mathbb Q(\zeta_8)$ (where $\zeta_n$ denotes a $n$-th primitive root of unity). The theory of cyclotomic extensions (see p. 6 in here: http://www.math.uconn.edu/~kconrad/blurbs/galoistheory/cyclotomic.pdf ) tells us that this composite is generated by $\zeta_{lcm(3,8)}=\zeta_{24}$. Therefore
$$\mathbb Q(i,\sqrt{3},e^{{\pi i/4}})=\mathbb Q(e^{{\pi i/12}}).$$