Abstract Algebra - Minimal Polynomials 
Let $\alpha=\frac{\sqrt{2}+\sqrt{2}i}{2}$.  Find the minimal polynomials over the fields $\mathbb Q$, $\mathbb Q(\sqrt{2})$, $\mathbb Q(i)$, and $\mathbb R$.

My solutions are as follows:
The minimal polynomial of $\alpha$ over $\mathbb Q$ is $\alpha^4+1=0$.
The minimal polynomial of $\alpha$ over $\mathbb Q(\sqrt{2})$ is $\alpha^2-\sqrt{2}\alpha+1=0$.
The minimal polynomial of $\alpha$ over $\mathbb Q(i)$ is $\alpha^2-i=0$.
The minimal polynomial of $\alpha$ over $R$ is the same as the minimum polynomial of $\alpha$ over $\mathbb Q(\sqrt{2})$.
Am I way off or on track?
 A: You've found the minimal polynomials, but often you're expected to prove that they are correct as well as find them. If that's the case, you're half-way done.
As requested in the comments, I'll prove that these polynomials are correct. I'll use a wonderful theorem to do so:

Theorem: Let $F$ be a field, $\alpha\in\overline{F}$, and $p_\alpha$ the minimal polynomial for $\alpha$ over $F$. Then if $q\in F[x]$ is a polynomial such that $q(\alpha)=0$, it follows that
  $p_\alpha|q$. Moreover, the only polynomials with this property
  besides the minimal polynomial are scalar multiples of the minimal
  polynomial.

This theorem is very important, and one of its uses is to prove that a given polynomial is the minimal polynomial of a number. All you have to do is give a monic polynomial with $\alpha$ as a root and prove that that polynomial is irreducible. If it's irreducible, then no proper divisor exists, and so for or theorem to hold, it must be that the given polynomial is the minimal polynomial.
In your problem, for all the cases except $\mathbb{Q}$ it is trivial to see that the polynomial is irreducible because it has degree $2$ and so the only proper divisor would be of the form $x-c=0$. For this to have $\alpha$ as a root, $\alpha=c$ must hold, but $\alpha$ isn't in the base field, contradiction. Thus the polynomial is irreducible.
For $F=\mathbb{Q}$, first find the other four roots of the polynomial and note that none of them are rational numbers. Thus there are no degree $1$ divisors, so any factorization must be into two quadratic factors. If we look at the polynomial over $\mathbb{C}$ instead of $\mathbb{Q}$ we see that the only way that it factors into linear factors is $x^4+1=(x^2+i)(x^2-i)$. However, $i\not\in\mathbb{Q}$ so that can't be a factorization of $p$ over $\mathbb{Q}$ and so it's irreducible.
Note: When writing a proof, you need to show some computational steps I'm skipping, specifically the ones that show that $p$ has $\alpha$ as a root, that none of the roots of the minimal polynomials lie in the base fields, and that $x^4+1=0$ only has the one factorization into quadratics.
