# Cyclotomic polynomial as minimal polynomial

I'm in the process of learning Galois theory and got stuck on Wikipedia's alternative definition of the $$n$$th cyclotomic polynomial as the "minimal polynomial over the field of the rational numbers of any primitive nth-root of unity". Is this really enough to prove that the $$n$$th cyclotomic polynomial is in fact the one whose roots are all the primitive $$n$$th roots of unity?

I can prove that the primitive $$n$$th roots of unity are roots of the $$n$$th cyclotomic polynomial. Proof: All primitive $$n$$th roots of unity $$\zeta_n$$ share the same minimal polynomial, i.e. the $$n$$th cyclotomic polynomial $$\Phi_n$$, so the roots of $$\Phi_n$$ at least include all the $$\zeta_n$$.

But proving the other direction -- that these are in fact all the roots of $$\Phi_n$$ -- seems more difficult. In particular, we don't know things like:

• The degree of the $$n$$th cyclotomic polynomial (or the extension it generates) is $$\varphi(n)$$
• The $$nth$$ cyclotomic polynomial contains only roots of unity as roots
• The $$n$$th cyclotomic extension is a Galois extension

Am I missing some information to prove this, or is Wikipedia the one lacking information?

• The definitely are equivalent because they're just the exact same idea articulated using different vocabulary. The product definition constructs a minimal polynomial that contains all the roots. So they're always the minimal polynomial that contains all the roots. Commented Dec 15, 2023 at 23:09
• My understanding is that taking the minimal polynomial for a given root $\alpha$ might include other roots that are unrelated to $\alpha$. Though the minimal polynomial contains all the primitive $n$th roots of unity for sure, how do you know it doesn't also include roots unrelated to the primitive $n$th roots of unity? Commented Dec 15, 2023 at 23:14
• If there is another root unrelated to the $n$-th roots it's not minimal. Commented Dec 16, 2023 at 1:11
• In my opinion the most difficult part is proving that $\Phi_n(x)$ is irreducible over $\Bbb{Q}$. The fact that all the primitive $n$th roots of unity are Galois conjugates hinges on that. If $n$ is prime, it follows from Eisenstein, but the general case is a bit delicate. It is not at all obvious that ALL the primitive roots are needed as zeros to produce a polynomial with rational coefficients. See the top rated "Related" question on the right margin. Commented Dec 16, 2023 at 4:34

To see that these are all the roots, I think it would suffice for you if you knew that the $$\deg \Phi_n=\phi(n)$$ as you have stated that you know that the $$\phi(n)$$ primitive $$n$$th roots of unity are clearly roots of the minimal polynomial of $$\zeta_n$$ (a primitive $$n$$th root of unity). One way to see this would be to claim that $$x^n-1 = \prod_{d\mid n}\Phi_d(x)$$. To see this we note that the roots of the polynomial on the left hand side are all $$n$$th roots of unity which is exactly all primitive $$d$$th roots of unity for all $$d\mid n$$, so we have that these two polynomials are the same. Now we use the fact that the degree map is additive, so $$n=\deg(x^n-1)=\deg(\prod_{d\mid n}\Phi_d(x))=\sum_{d\mid n}\deg\Phi_d(x)$$ Now you have observed that $$\deg \Phi_d(x)\geq \phi(d)$$, but as there is the well known formula that $$\sum_{d\mid n}\phi(d)=n$$, if for any $$d$$, we have that $$\deg\Phi_d>\phi(d)$$, then we would have that $$n=\sum_{d\mid n}\phi(d)<\sum_{d\mid n}\deg\Phi_d(x) = n$$ which is a contradiction, so you can conclude that $$\deg\Phi_n(x)=\phi(n)$$ for all $$n$$. Hence, you may conclude that $$\Phi_n(x)$$ has exactly $$n$$ roots (those being the primitive $$n$$th roots of unity). I believe this should resolve your problem.