Minimal polynomial of $\alpha+i$ over $\mathbb{Q}$ $\alpha$ is a root of the polynomial $x^3-x+1$ over $\mathbb{Q}$.
 The task is to find the minimal polynomial of $\alpha+i$
I have shown that $\mathbb{Q}(\alpha+i)=\mathbb{Q}(i,\alpha)$ and that $|\mathbb{Q}(i,\alpha):\mathbb{Q}|=6$. So I'm expecting this to be a degree $6$ polynomial. I have tried getting various powers of $\alpha+i$.
Since complex conjugation is a field automorphism and maps roots to roots, I know that $\alpha-i$ must also be a root.
I do no know what else to try.
 A: First find the polynomial that has the root $\beta=\alpha+ i$ over $\Bbb{Q}(\alpha)\subset\Bbb{R}$ in-terms of $\alpha.$  
Then use the fact that $\alpha^3-\alpha+1=0$ to get rid of $\alpha$ in previous polynomial. 
OR:
Note that $$(\beta-i)^3-(\beta-i)+1=0$$
A: This is the systematic approach (why am I doing this):
Pick $\Bbb Q$-basis $\{1, \alpha, \alpha^2, i, i\alpha, i\alpha^2\}$. Note that $\alpha^3 = \alpha - 1$.
Look at the map $\varphi : \Bbb Q(\alpha, i) \to \Bbb Q(\alpha, i)$ sending $x$ to $(\alpha + i)x$.


*

*$\varphi(1) = \alpha + i$

*$\varphi(\alpha) = \alpha^2 + i\alpha$

*$\varphi(\alpha^2) = \alpha^3 + i\alpha^2 = -1 + \alpha + i\alpha^2$

*$\varphi(i) = -1 + i\alpha$

*$\varphi(i\alpha) = -\alpha + i\alpha^2$

*$\varphi(i\alpha^2) = -\alpha^2 + i\alpha^3 = -\alpha^2 - i + i\alpha$


So the matrix representing the map is:
$$\begin{bmatrix}
 0 &  1 &  0 &  1 &  0 &  0 \\
 0 &  0 &  1 &  0 &  1 &  0 \\
-1 &  1 &  0 &  0 &  0 &  1 \\
-1 &  0 &  0 &  0 &  1 &  0 \\
 0 & -1 &  0 &  0 &  0 &  1 \\
 0 &  0 & -1 & -1 &  1 &  0
\end{bmatrix}$$
whose characteristic polynomial is $t^6 + t^4 + 2t^3 + 4t^2 - 8t + 5$ (thanks, Wolfram Alpha), which is also the minimal polynomial of $\alpha + i$ thanks to Cayley-Hamilton.
A: My different systematic approach:
If $\alpha$ is a root of the $\Bbb Q$-irreducible polynomial $f(X)$, then of $f$ is $\alpha$’s minimal (and characteristic) polynomial. Now $f(X)$ is still irreducible over $k=\Bbb Q(i)$, because $k$ contains no cubic quantities and therefore $f$ has no roots there. It follows that $f(X-i)$ is also $k$-irreducible, and is therefore the $k$-minimal polynomial for $\alpha+i$. Mutiply this by its conjugate to get $g(X)=f(X-i)(f(X+i)\in\Bbb Q[X]$. It is a $\Bbb Q$-polynomial of the right degree, with $\alpha+i$ as a root. And:
By an easy hand computation, we find that $g=X^6 + X^4 +2X^3 + 4X^2 - 8X+5$, same as @KennyLau got, but without prayer or supplication to Wolfram.
