Why is $(X^5-1)/(X-1)$ the minimal polynomial for $e^{2\pi i/5}$, the fifth root of unity, over $\Bbb Q$?

Question

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

Then I want to find its minimal polynomial over $\mathbb{Q}$. 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 $\mathbb{Q}(\zeta_5)$.

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 $|\text{Gal}(K(\alpha)/K)|$ for some field K. So we have $|\text{Gal}(\mathbb{Q}(\zeta_5)/\mathbb{Q}$)|=4.

We then see that $\text{Gal}(\mathbb{Q}(\zeta_5)/\mathbb{Q})=\mathbb\{\text{id},\sigma,\sigma^2,\sigma^3\}$ where $\sigma$ is a $K$-automorphism$^1$ given by $\sigma:\zeta\mapsto\zeta^3$. 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 $\sigma: L\to L$ is called a $K$-automorphism if it is a field isomorphism with the additional property that $\sigma(x)=x$ for all $x\in K$.

Answer 1

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

For $f\in k[X]$ and $\alpha\in k$ 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 $Q\in\Bbb{Q}[X]$, and comparing degrees and leading coefficients shows that $Q=1$. This also immediately shows that $X^5-1$ is not irreducible over $\Bbb{Q}$, because it has a linear factor $X-1$. It follows that the minimal polynomial of $\zeta$ divides $$\frac{X^5-1}{X-1}=X^4+X^3+X^2+X+1,$$ 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 $p=5$.

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As for picking this field automorphism; because a field automorphism is $\Bbb{Q}$-linear and $K:=\Bbb{Q}(\zeta)$ is spanned by the powers of $\zeta$ as a $\Bbb{Q}$-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 $k\in\{1,2,3,4\}$.

Answer 2

First, the minimal polynomial over $\mathbb{Q}$ of some element $\alpha\in\mathbb{C}$ must be, by definition, monic and irreducible. If we have $p(\alpha)=0$ for some polynomial $p\in\mathbb{Q}[X]$ 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 $1\in\mathbb{Q}$ 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 $\beta\neq1$ 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 $\mathbb{Q}$, 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 $\mathbb{Q}$ (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.