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Let $p$ be a prime number, $K=\mathbb{Q}(\zeta_p)$ be the cyclotomic field extension of $\mathbb{Q}$ by adding a $p$-th root of unity.

There is a notation called different ideal, which is defined to measure the (possible) lack of duality in the ring of integers. I want to know what is the different ideal of the cyclotomic field extension $K/\mathbb{Q}$?

ZZP
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    Welcome to MSE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please [edit] the question. This will help you recognize and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. – José Carlos Santos Feb 12 '24 at 09:35
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    Fields have no nontrivial ideals. Do you mean the (fractional) ideals of the ring of integers? I can't remember the standard lingo in number theory. – pancini Feb 12 '24 at 09:47
  • @JoséCarlosSantos Ok, thanks for your reminding! – ZZP Feb 12 '24 at 09:48
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    @pancini this abuse of terminology is very common in algebraic number theory. – Lukas Heger Feb 12 '24 at 09:51
  • @LukasHeger I suspected that; thanks. – pancini Feb 12 '24 at 09:52

1 Answers1

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It's well-known that $p$ is totally ramified in $K$ and that it is the only ramified prime in this extension. The unique prime ideal above $p$ is $(1-\zeta_p)$, so we only need to figure out the exponent of $(1-\zeta_p)$.

The norm of the different is the (absolute value of) discriminant. The discriminant in this case is also well-known:

We have $|\Delta_{K/\Bbb Q}|=p^{p-2}$.

Now if $\mathfrak{d}_{K/\Bbb Q}=(1-\zeta_p)^m$, then using $N((1-\zeta_p))=p$ we get

$p^{p-2}=|\Delta_{K/\Bbb Q}|=N(\mathfrak{d}_{K/\Bbb Q})=p^m$

So the exponent $m$ is $p-2$ and we have $\mathfrak{d}_{K/\Bbb Q}=(1-\zeta_p)^{p-2}$.

Another approach: If the ring of a number field is $\Bbb Z[\alpha]$ with $\alpha$ having minimal polynomial $f$, then the different is generated by $f'(\alpha)$. Here $\alpha=\zeta_p$, $f=\Phi_p$.

We have $(X-1)\Phi_p(X)=X^p-1$. Differentiating yields.

$$(X-1)\Phi'_p(X)+\Phi_p(X)=pX^{p-1}$$ Plugging in $\zeta_p$ gives

$$(\zeta_p-1)\Phi'_p(\zeta_p)=p\zeta_p^{p-1}$$

The factor $\zeta_p^{p-1}$ doesn't bother us, being a unit. We know that $(\zeta_p-1)^{p-1}=(p)$ (as, again, $p$ is totally ramified and $(\zeta_p-1)$ is the unique prime ideal above it) Thus we get an equality of ideals:

$(\zeta_p-1)(\Phi'_p(\zeta_p))=(p)=(\zeta_p-1)^{p-1}$. Which yields the result upon diving by $(\zeta_p-1)$.

Third method (This one is quite a bit more advanced):

Consider the extension of local fields $\Bbb Q_p(\zeta_p)/\Bbb Q_p$. Because $p$ is totally ramified in $K/\Bbb Q$, the problem is purely local.

Generally if $E/F$ is an extension of non-archimedean local fields, then the different $\mathfrak{d}_{E/F}$ is some power of the maximal ideal $\mathfrak{m}_E$ of the valuation ring $\mathcal O_E$ of $E$. If $E/F$ is Galois with Galois group $G$, then the exponent of $\mathfrak{d}_{E/F}$ may be expressed in terms of higher ramification groups. Namely, we have $\mathfrak{d}_{E/F}=\mathfrak{m}_E^w$, where $w$ is $\sum_{i=0}^\infty(|G_i|-1)$.

Now the higher ramification groups of the field $\Bbb Q_p(\zeta_{p^k})=:E_k$ may be computed as $G_i=\operatorname{Gal}(E_k/E_e)$, where $e$ is chosen such that $p^{e-1}\leq i <p^e$. For a proof, see e.g. Serre Local Fields. In our situation $k=1$. And so we get a really simply ramification filtration:

$G_{-1}=G_0=\operatorname{Gal}(\Bbb Q_p(\zeta_p)/\Bbb Q_p)\cong (\Bbb Z/p\Bbb Z)^\times$ and $G_i={1}$ for $i \geq 1$.

Thus the formula simply yields as an exponent $|G_0|-1=p-2$.

Lukas Heger
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