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I'm currently reading Kummer's famous paper on Fermat's Last Theorem (if anyone wants the link, I'll post it, but the paper is in German). There's the following statement in there, which should be "very easy to prove":

Let $\xi$ be a primitive $p$-th root of unity with $p$ an odd prime. Any element $x \in \mathbf{Z}[\xi]$ can be multiplied with a power $\xi^r$ such that it is congruent to an integer mod $(1-\xi)^2$.

I've tried to prove it, but I need help. I thought one could write up a basis $1, (1-\xi)^2, ..., (1-\xi)^{p-1}$, but this is wrong, as has been shown in an answer to an earlier question of mine.

Thanks!

Added Bounty-question: I went through the paper again and saw that I missed the hypothesis that $x$ does not lie in the ideal $(1 - \xi)$. It didn't really help me proving it though. Can anyone find a proof under this stronger condition?

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  • $\begingroup$ I guess you mean the paper Allgemeiner Beweis des Fermat'schen Satzes, dass die Gleichung $x^\lambda+y^\lambda = z^\lambda$ durch ganze Zahlen unlösbar ist, für alle diejenigen Potenz-Exponenten $\lambda$, welche ungerade Primzahlen sind und in den Zählern der ersten $\frac12(\lambda-3)/2$ Bernoulli'schen Zahlen als Factoren nicht vorkommen, which is available at GDZ. Also here: dx.doi.org/10.1515/crll.1850.40.130 - but this links is behind paywall (for subscribers). $\endgroup$ – Martin Sleziak Sep 24 '16 at 0:16
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This is false - $ 1 - \xi $ can never be multiplied by a power of $ \xi $ to be an integer modulo $ (1 - \xi)^2 $. Let $ \mathfrak p = (1 - \xi) $ throughout the post.

To see this, note that $ p \equiv 0 \pmod{\mathfrak p^2} $ since the ideal $ (p) $ totally ramifies as $ (p) = \mathfrak p^{p-1} $, so if $ \xi^k (1 - \xi) $ is an integer modulo $ \mathfrak p^2 $, without loss of generality it can be taken to be in the set $ \{ 0, 1, \ldots, p-1 \} $. It cannot be $ 0 $, because that would imply $ 1 - \xi \in \mathfrak p^2 $ and $ \mathfrak p = \mathfrak p^2 $, but $ 1 - \xi $ is not a unit: it has norm $ \pm p $. It cannot be anything else in this set, because $ \mathbf Z \cap \mathfrak p = p\mathbf Z $, therefore all of the nonzero integers in the set are invertible modulo $ \mathfrak p $, and thus modulo $ \mathfrak p^2 $. Thus, such a congruence would imply that $ 1 - \xi $ is invertible modulo $ \mathfrak p^2 $, which is absurd, since the ideals $ (1 - \xi) = \mathfrak p $ and $ \mathfrak p^2 $ are not coprime.

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Perhaps you dropped some hypothesis in your statement, which should be: "Any unit $u$ in $\mathbf Z[\zeta]$ can be multiplied, etc." By Dirichlet's unit theorem (or by direct computation), $u$ is of the form $\zeta^r .v$, where $v$ is a unit in $\mathbf Z[\zeta +\zeta ^{-1}]$ (= the ring of integers of the maximal totally real subfield). But $(1-\zeta)(1-\zeta ^{-1}) = 2 -(\zeta +\zeta ^{-1})$, and $(1-\zeta)/(1-\zeta ^{-1})$ is a unit, hence taking quotients mod $(1-\zeta)(1-\zeta ^{-1})$ shows that every element of $\mathbf Z[\zeta +\zeta ^{-1}]$ is congruent to a rational number mod $(1-\zeta)^2$ QED

I guess that this property is a preliminary in the proof of the so called Kummer lemma, in which case you could fine more information in Washington's book, §5.6.

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Answer to the bounty question:

Denote $ R = \mathbf Z[\xi]/(1 - \xi)^2 $. We make two observations: the map $ \mathbf Z \to R $ has kernel $ p\mathbf Z $, so it descends to an embedding $ \mathbf Z/p \mathbf Z \to R $, and since an integer is invertible modulo $ (1 - \xi)^2 $ if and only if it is not divisible by $ p $ (this was shown in my other answer), it follows that this embedding gives an embedding of groups $ H = (\mathbf Z/p \mathbf Z)^{\times} \to R^{\times} $. On the other hand, $ \xi $ is certainly in $ R^{\times} $, and its order divides $ p $, so it is either congruent to $ 1 $ or has order $ p $ in $ R^{\times} $. The former is impossible since it implies that $ 1 - \xi $ is divisible by $ (1 - \xi)^2 $, so $ \xi $ is an element of order $ p $. Then, the subgroup $ K = \{ x \in R^{\times} : x = \xi^i z, z \in \mathbf Z \} $ is the product of the subgroups $ H $ and $ \langle \xi \rangle $, which intersect trivially since their orders are coprime. It follows that $ |K| = |H| |\langle \xi \rangle| = p(p-1) $.

On the other hand, $ N((1 - \xi)^2) = N(1 - \xi)^2 = p^2 $, so $ |R| = p^2 $, and the submodule $ (1 - \xi) / (1 - \xi)^2 $ of $ R $ has cardinality $ p $, which consists entirely of non-invertible elements in $ R $. Thus, $ |R^{\times}| = p^2 - p = p(p-1) $, and from this it follows that $ K = R^{\times} $. The statement follows upon writing $ x \equiv \xi^i z \pmod{(1 - \xi)^2} $ for $ x \notin (1 - \xi) $ and multiplying both sides by $ \xi^{p-i} $.

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With your additional hypothesis that $x$ does not lie in the prime ideal $(1- \zeta)$, the question becomes much easier, because of its now "local" nature. To simplify notations, put $\pi=1 - \zeta $ throughout and consider the ring $A = \mathbf Z_p[\zeta]$, where $\mathbf Z_p$ denotes the $p$-adic integers. The group $U$ of units (= invertible elements) of $A$ is endowed with a descending filtration $U > U_1 > ... > U_n > ...$, where $U_n = 1+ (\pi^{n})$ . The residual field $k = A/(\pi)$ is isomorphic to $\mathbf F_p$ because $p$ is totally ramified, and the residue map $A \to \mathbf F_p$ induced by $\zeta \to 1$ gives rise to an isomorphism $U/U_1 \cong \mathbf F_p^{*}$ (multiplicative group), whereas for $n\ge 1$ the map $u \to u - 1$ gives rise to $U_n/U_n+1 \cong (\pi^{n})/(\pi^{n+1})\cong \mathbf F_p$ (additive group) (see e.g. Cassels-Fröhlich, chap. 1, propos. 4). One gets in particular an exact sequence $0 \to \mathbf F_p \to U/U_2 \to \mathbf F_p^{*} \to 1$, which is split because the two extreme terms have coprime orders. Given the explicit isomorphisms above, this shows that $U/U_2 = <\zeta>.\mathbf Z_p$ , hence for any $u\in U$ , a certain $u.\zeta^{r}$ will be congruent to $a$ mod $(\pi^{2})$, with $a \in\mathbf Z_p$. Coming back to our $x \in \mathbf Z[\zeta], \notin (\pi)$, and using the density of $\mathbf Z$ in $\mathbf Z_p$, a certain $x.\zeta^{r}$ will be will be congruent to $b$ mod $(\pi^{2})$, with $b \in\mathbf Z$ as desired.

The $p$-adics were unknown at Kummer's time, but the above proof can be adapted to be (probably)more in Kummer's spirit. Consider the ring $B = \mathbf Z[\zeta]= \mathbf Z[\pi] $. Every element $x \in B$ can be uniquely written under the form $x = a_0 + a_1\pi + ... + a_{p - 1}\pi^{p-1}$, with $a_n \in \mathbf Z$. Analogously to above, the natural exact sequence $ 0 \to (\pi)/(\pi^{2}) \to B/(\pi^{2}) \to B/(\pi) \cong \mathbf F_p \to 0$ gives rise to another exact sequence $ 0 \to (\pi)/(\pi^{2})\cong \mathbf F_p \to (B/(\pi^{2}))^{*} \to (B/(\pi))^{*} \cong \mathbf F_p^{*} \to 0$, where $(.)^*$ denotes the units. The hypothesis that $x \notin (\pi)$ means that $a_0$ is not a multiple of $p$, hence by the definition of the residue map, that $x$ mod $(\pi)$ $\in (B/\pi)^{*}$. Analogously to above, one can conclude that a certain $x.\zeta^{r}$ will be will be congruent to $b$ mod $(\pi^{2})$, with $b \in\mathbf Z$ as desired.

NB: This second proof is just a formalization of the one given by @Starfall.

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