# Maximal ideal lying above $p\mathbb{Z}$?

I've been pondering the following situation for a little while now. Let $p$ be a prime number, and I denote $\zeta:=\zeta_p$ to be a primitive $p^{\text{th}}$ root of unity.

Consider the rings $\mathbb{Z}$ and $\mathbb{Z}[\zeta]$. Now $\mathbb{Z}[\zeta]$ is a finitely generated $\mathbb{Z}$ module, hence $\mathbb{Z}[\zeta]$ is integral over $\mathbb{Z}$. Of course $p\mathbb{Z}$ is a prime ideal of $\mathbb{Z}$, hence it known that there exists a prime ideal $\mathfrak{P}$ lying above $p\mathbb{Z}$, that is, $\mathfrak{P}\cap\mathbb{Z}=p\mathbb{Z}$. Moreover, $p\mathbb{Z}$ is maximal since $\mathbb{Z}/p\mathbb{Z}$ is a field, and thus $\mathfrak{P}$ is maximal in $\mathbb{Z}[\zeta]$.

Can we say something stronger though, that there exists a unique maximal ideal $\mathfrak{M}$ of $\mathbb{Z}[\zeta]$ such that $\mathfrak{M}\cap\mathbb{Z}=p\mathbb{Z}$ with the additional property that $\mathfrak{M}^{p-1}=p\mathbb{Z}[\zeta]$?

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Yes; in fact, the ideal you're looking for is just $(1 - \zeta)$. This is classical. The main observation is that $$x^{p-1} + x^{p-2} + ... + 1 = \prod_{i=1}^{p-1} (x - \zeta^i)$$

and substituting $x = 1$ gives $$p = \prod_{i=1}^{p-1} (1 - \zeta^i).$$

Now observe that $\frac{1 - \zeta^i}{1 - \zeta} = 1 + ... + \zeta^{i-1}$ lies in $\mathbb{Z}[\zeta]$. Moreover, for $i$ relatively prime to $p$ we can find $j$ such that $ij \equiv 1 \bmod p$, hence $$\frac{1 - \zeta}{1 - \zeta^i} = \frac{1 - \zeta^{ij}}{1 - \zeta^i} \in \mathbb{Z}[\zeta].$$

So $\frac{1 - \zeta^i}{1 - \zeta}$ is a unit in $\mathbb{Z}[\zeta]$, and it follows that $(1 - \zeta^i) = (1 - \zeta)$ as ideals for $1 \le i \le p-1$, hence $$(p) = \prod_{i=1}^{p-1} (1 - \zeta^i) = (1 - \zeta)^{p-1}$$

as desired.

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Sorry to ask one more thing, but is this $\mathfrak{M}$ indeed unique? I couldn't tell if it was implicit in your answer. – Coralin Jan 20 '12 at 22:20
@Coralin: yes. Note that $\mathbb{Z}[\zeta]/p$ has characteristic $p$, so its only possible residue fields are finite fields of characteristic $p$. Since $(1 - \zeta)^p = 0$ in this ring, we necessarily have $\zeta = 1$ in any residue field, so in fact $\mathbb{F}_p$ is the only possible residue field and the morphism $\mathbb{Z}[\zeta]/p \to \mathbb{F}_p$ is unique (since we already know what happens to $\zeta$). – Qiaochu Yuan Jan 20 '12 at 22:24

In fact, $\mathfrak{M}$ is precisely the principal ideal generated by $(1-\zeta_p)$. That is,

$$(1-\zeta_p)^{p-1}\mathbb{Z}[\zeta_p]=p\mathbb{Z}[\zeta_p]$$

As your tag indicates, this falls squarely within the purview of algebraic number theory. The proof of the above equality is quite straight-forward (edit: as in Qiaochu's answer), but the algebraic number theory machinery behind the factorization of primes in number fields, e.g., generalization of your observation that $\mathfrak{P}$ must maximal, are of tremendous importance. I'd recomment reading through (at least) the first couple of chapters of Washington's Cyclotomic Fields.

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Thank you, Cam. – Coralin Jan 20 '12 at 21:49