I try to show that $\Bbb Q(u,i)$ with u root of the polynomial $t^4-3t^2+4$ cannot be reduced to $\Bbb Q(u)$, that is, that $i$ does not belong $\Bbb Q(u)$, but I have not been able to prove it and I would like someone to help me. So far I have tried something that does not seem to be successful: Suppose that I can be written as a linear combination of ${1, u, u ^ 2, u ^ 3}$ and, through operations on the equation $x^2+1=0$, arrive at a contradiction with respect to the coefficients, and then demonstrate the irreducibility of the polynomial $f$ in $\Bbb Q(i)$ to see that the degree of $[\Bbb Q(u,i):\Bbb Q(i)] = 4$.
PS:I am new to this page and I am learning, so I hope you take it into consideration. Thanks in advance.
Solving $$u^4-3u^2+4=0$$ gives $$u^2=\frac{3\pm\sqrt{-7}}2$$ and $$u=\sqrt{\frac{3\pm\sqrt{-7}}2}.$$ As an element of the number field $\Bbb Q(\sqrt{-7})$, $u^2$ has norm $4$ so we might hope that $u^2$ is a square there. But instead we find that $$\frac{3\pm\sqrt{-7}}2=-\left(\frac{1\mp\sqrt{-7}}2\right)^2.$$ So $$u=\pm i\frac{1\mp\sqrt{-7}}3$$ which means that $i$ lies in the Galois closure $K$ of $\Bbb Q(u)$.
We should be in luck if we find that $K$ has degree $4$ over $\Bbb Q$. Let's fix one of the four values for $u$, so suppose $$2u=i+\sqrt7.$$ Then $$4u^2=6+2i\sqrt 7$$ so $$2u^2-3=i\sqrt7.$$ Therefore $$2u(2u^2-3)=i\sqrt7(i+\sqrt7)=-\sqrt7+7i$$ and then $$2u(2u^2-3)+2u=8i.$$