Construct field Extension given that $\alpha$ has the following minimal polynomial I apologize I have nothing to show that I have really attempted this question, but it's simply because I am struggling to get used to the terms and ideas.
I randomly chose a question from a past paper for my own practice and not very sure what the answer is.

If $\alpha$ has the minimal polynomial $x^4+x^3+x^2+x+1$over $\mathbb{Q}$, construct the extension $\mathbb{Q}(\alpha)$.

I suppose I am to solve the polynomial, to determine $\alpha$ but over $\mathbb{Q}$, I believe there is no solution. There are 4 distinct solutions in $\mathbb{C}$ though. So what exactly is $\alpha$ and how should I answer this question?
 A: The solution are the non-real fifth roots of unity: $\mathrm e^{\tfrac{2\mathrm i k\pi}5}$. Any of these roots is a power of any other root (this is because $5$ is prime), so $\alpha$ can be any of these values.
As $\alpha$ is algebraic over $\mathbf Q$, and its minimal polynomial is $x^4+x^3+x^2+x+1$, the extension is simply
$$\mathbf Q(\alpha)=\mathbf Q[\alpha]\simeq \mathbf Q[x]/(x^4+x^3+x^2+x+1),$$
so that it is made up of the polynomials in $\alpha$, of degree $\le 3$. As the other roots of the minimal polynomial of $\alpha$, are powers of  $\alpha$, the extension is normal, and, as we're in characteristic $0$, is a Galois extension of degree $4$.
A: The abstract algebra approach is to verify that $p(x) = x^4 + x^3 + x^2 + x + 1$ is irreducible over $\mathbb{Q}$, and then construct the quotient ring $\mathbb{Q}[x]/\langle p(x) \rangle$. [Note that being irreducible over $\mathbb{Q}$ is equivalent to $p(x)$ being irreducible over $\mathbb{Z}$.]
The "value" of $\alpha$ can then be identified with residue class of $x$ in the quotient ring $\mathbb{Q}[x]/\langle p(x) \rangle$.
This turns out to be isomorphic to the ring extension $\mathbb{Q}[\alpha]$ that "lives" inside the complex numbers $\mathbb{C}$.  Since $p(x)$ is irreducible, the quotient ring is an integral domain, and the field extension $\mathbb{Q}(\alpha)$ is isomorphic to the quotient field of $\mathbb{Q}[x]/\langle p(x) \rangle$.
A: I feel like if you are struggling with the terms and ideas of field extensions and how polynomials are related to this, you should try reading a different source of explanation. Try reading Chapter 21 from this book on abstract algebra. I know it's a 20-30 page chapter, but you don't necessarily need to read the whole chapter to learn about extension and splitting fields.
The most important thing you have to understand about $\alpha$ is that it's defined as one of the solutions to $x^4+x^3+x^2+x+1=0$. It's not a specific solution to that polynomial, since, as you pointed out, there are actually four different solutions, but $\alpha$ can be any one solution. Using that information, we have to construct our field extension.
Now, from the chapter above, you'll find that $\mathbb{Q}(\alpha)$ is the same thing as $\mathbb{Q}[x]/\langle x^4+x^3+x^2+x+1 \rangle$ with the isomorphism between them being the evaluation homomorphism. The important thing to notice about this is that this means that all elements of $\mathbb{Q}(\alpha)$ are in the form of $a\alpha^3+b\alpha^2+c\alpha+d$ for $a, b, c, d \in \mathbb{Q}$. We never reach $\alpha^4$ because we know that since $\alpha^4+\alpha^3+\alpha^2+\alpha+1=0$, then $\alpha^4=-\alpha^3-\alpha^2-\alpha-1$, so we can just substitute $\alpha^4$ with lesser powers.
Now, two important properties about $\mathbb{Q}(\alpha)$ is if the field extension is normal and if the field extension is separable. If the extension is separable, then the minimal polynomial of $\alpha$ of degree $n$ has $4$ distinct roots. Now, in our case, this means that $x^4+x^3+x^2+x+1$ has $4$ distinct roots and none of them are repeated. As @Bernard wrote, the roots of this polynomial are all distinct roots of unity, which means that yes, this extension is separable. One handy thing to note is that all field extensions of $\mathbb{Q}$ are separable because $\mathbb{Q}$ has a characteristic of $0$. You can find a proof of this here. (NOTE: I haven't verified this proof or read this whole document, but I have read other proofs of this and I know that it is true.)
We also need to find if this field extension is normal. If the field extension is normal, then this field extension is the splitting field of the minimal polynomial of $\alpha$, which is $x^4+x^3+x^2+x+1$ in our case. This means that all of the roots of $x^4+x^3+x^2+x+1$ is inside $\mathbb{Q}(\alpha)$ and that we don't need to add any more roots to fully factor the polynomial. Now, as @Bernard wrote, since all of the roots of this polynomial are roots of unity, we can generate the roots of unity with powers of $\alpha$ and factor this polynomial into linear factors as follows: $x^4+x^3+x^2+x+1=(x-\alpha)(x-\alpha^2)(x-\alpha^3)(x-\alpha^4)$. Thus, since $x^4+x^3+x^2+x+1$ fully factors, this extension is normal.
Since the extension is both normal AND separable, it is a Galois extension which means that it's really useful and has a lot of special properties.
I hope this explanation of this field extension helps you understand field extensions better!
