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I have some trouble understanding an example from my reader$^1$. We have the fifth root of unity $zeta_5=e^2pi i/5$, so by definition $zeta_5^5=1$.

Then I want to find its minimal polynomial over $mathbbQ$. I know that $zeta_5$ is zero of $(X^5-1)$. I see that $1, zeta_5,zeta_5^2, zeta_5^3, zeta_5^4$ are zeros of this polynomial. My reader however gives $(X^5-1)/(X-1)=X^4+X^3+X^2+X+1$ as the minimal polynomial. Why? I think I had to see that $(X^5-1)$ was reducible or something? How do I see that. And once I notice it, how do I confirm that $zeta_5$ is a zero of $X^4+X^3+X^2+X+1$, and how do I confirm that also $zeta_5^2, zeta_5^3, zeta_5^4$ are zeros of it? These zeros belong to $mathbbQ(zeta_5)$.

Once I understand this:
By some theorem I know that the number of zeros of a minimal polynomial of $alpha$ is the order of the group $|textGal(K(alpha)/K)|$ for some field K. So we have $|textGal(mathbbQ(zeta_5)/mathbbQ$)|=4.

We then see that $textGal(mathbbQ(zeta_5)/mathbbQ)=mathbbtextid,sigma,sigma^2,sigma^3$ where $sigma$ is a $K$-automorphism$^1$ given by $sigma:zetamapstozeta^3$. How and why did they pick/find this $K$-automorphism?

$1:$ Example 8.2.11 from Rings and Galois Theory, 2018, Department of Mathematics, Utrecht University

$2:$

Let $K$ be a field and $L$ a finite extension. An automorphism $sigma: Lto L$ is called a $K$-automorphism if it is a field isomorphism with the additional property that $sigma(x)=x$ for all $xin K$.

The only observation you need for all of these conclusions is the very basic one that

For $fin k[X]$ and $alphain k$ you have $f(alpha)=0$ if and only if $x-alpha$ divides $f$.

Because $zeta^5=1$ by construction, clearly the minimal polynomial of $zeta$ divides $X^5-1$. As you note, you can see that $1$, $zeta$, $zeta^2$, $zeta^3$ and $zeta^4$ are roots of this polynomial. By the observation above this means
$$X^5-1=(X-1)(X-zeta)(X-zeta^2)(X-zeta^3)(X-zeta^4)Q,$$
for some $QinBbbQ[X]$, and comparing degrees and leading coefficients shows that $Q=1$. This also immediately shows that $X^5-1$ is not irreducible over $BbbQ$, because it has a linear factor $X-1$. It follows that the minimal polynomial of $zeta$ divides
$$fracX^5-1X-1=X^4+X^3+X^2+X+1,$$
and by looking at the factorization above we see tht
$$X^4+X^3+X^2+X+1=(X-zeta)(X-zeta^2)(X-zeta^3)(X-zeta^4).$$
This shows that $zeta$, $zeta^2$, $zeta^3$ and $zeta^4$ are all zeros of $X^4+X^3+X^2+X+1$. To conclude that this is the minimal polynomial, it remains to prove that it is irreducible; this follows from the fact that
$$(X+1)^4+(X+1)^3+(X+1)^2+(X+1)+1=X^4+5X^3+10X^2+10X+5,$$
is Eisenstein at $p=5$.

As for picking this field automorphism; because a field automorphism is $BbbQ$-linear and $K:=BbbQ(zeta)$ is spanned by the powers of $zeta$ as a $BbbQ$-vector space, a field automorphism $sigma$ of $K$ is uniquely determined by $sigma(zeta)$. Because $zeta$ is a zero of $X^4+X^3+X^2+X+1$ it follows that
$$sigma(zeta)^4+sigma(zeta)^3+sigma(zeta)^2+sigma(zeta)+1=sigma(zeta^4+zeta^3+zeta^2+zeta+1)=sigma(0)=0,$$
which shows that $sigma(zeta)$ is also a zero of $X^4+X^3+X^2+X+1$, so $sigma(zeta)=zeta^k$ for some $kin1,2,3,4$.

First, the minimal polynomial over $mathbbQ$ of some element $alphainmathbbC$ must be, by definition, monic and irreducible. If we have $p(alpha)=0$ for some polynomial $pinmathbbQ[X]$ then we know that the minimal polynomial is a factor of $p$. In your case, $p(X):=X^5-1$ is the obvious candidate: if it were irreducible it would be the minimal polynomial of $zeta_5$, but it is not as $1inmathbbQ$ is also a fifth root of unity, so we have at least to discard $1$. We divide $p(X)$ by $X-1$ and get $q(X)=X^4+X^3+X^2+X+1$, which may be irreducible or not. We know that $p(X)=(X-1)q(X)$, hence if $betaneq1$ is a root of $p$ then $0=p(beta)=(beta-1)q(beta)$ implies $q(beta)=0$, so the roots of $q$ are the $5$-roots of unity which are not $1$. None of them is a rational number, so $q$ is not the product of a linear factor and a third degree polynomial over $mathbbQ$, but it could yet be the product of two polynomials of degree $2$. Eisenstein's criterion allows to prove the following result: if $p$ is prime, then $X^p-1+cdots+X+1$ is irreducible over $mathbbQ$ (i.e., the irreducible polynomial of a $p$-root of unity is found dividing $X^p-1$ by $X-1$).

What happens with $n$-roots of unity when $n$ is not necessarily prime? Perhaps you'd like to take a look at cyclotomic polynomials.