A few months ago, someone told me there existed a scheme theoretic proof of the irreducibility of cyclotomic polynomials. I've tried coming up with a proof, and when that didn't really yield anything $($just the scattered thoughts I have below$)$, I searched online for a reference to a proof, to no avail. Can anyone provide a proof or a reference to one? Does a proof even exist via schemes, or was this someone trolling me? Would this question be a better fit on MathOverflow?
Here are my thoughts so far on such a proof, if it were to exist.
The standard linear algebraic $($or in context of "modern" frameworks, the "$p$-adic"$)$ formulation of the irreducibility of cyclotomic polynomials is as follows.
Theorem. Let $n \in \mathbb{N}$. The degree of $\mathbb{Q}(\zeta_n)/\mathbb{Q}$ is $[\mathbb{Q}(\zeta_n):\mathbb{Q}] = \phi(n)$.
This is saying is that the vector space generated by the powers of $\zeta_n$ over $\mathbb{Q}$ is of dimension $\phi(n)$. The most important case in the proof is where $n = p$, where one shows that the $\mathbb{Q}$-vector space $\mathbb{Q}(\zeta_p)$ has dimension $p-1$. This is done by showing $$\dim_\mathbb{Q}(\mathbb{Q}(\zeta_p)) = \dim_\mathbb{Z}(\mathbb{Z}[\zeta_p]) = \dim_{\mathbb{F}_p}(\mathbb{Z}[\zeta_p]/p),$$and that $\dim_{\mathbb{F}_p}(\mathbb{Z}[\zeta_p]/p) = p-1$.
Now, the purpose of scheme theory is to make geometry out of rings. In the case of this elusive proof, I imagine the scheme theoretic perspective claims that the extension of rings $\mathbb{Z}[\zeta_n]/\mathbb{Z}$ is like a ramified cover of topological spaces. One imagines $\mathbb{Z}$ as space populated by its primes, and similarly for $\mathbb{Z}[\zeta_n]$. The scheme theoretic restatement of the above theorem is that this is a connected cover of degree $\phi(n)$. In the standard proof of the linear algebraic formulation above one sees this by looking above the prime $p$, where one sees "nilpotency" of order $p-1$, which implies that the cover must have degree $p-1$.
Is this viable? Should I use some sort of monodromy or something? $($But that's harder to make sense of in the arithmetic setting...$)$