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Let $n\in\mathbb N$ and $p$ be any prime. Denote by $\mathbb Q_p$ the $p$-adic numbers. For a field extension $L/K$ denote by $Tr_{L/K}$ the corresponding trace function.

Let $\zeta_n$ be a primitve $n$-th root of unity and $\xi_n$ any $n$-th root of unity. I would like to know the following:

What is $Tr_{\mathbb Q_p(\zeta_n)/\mathbb Q_p}(\xi_n)$?

In particular: Is $Tr_{\mathbb Q_p(\zeta_n)/\mathbb Q_p}(\xi_n)\in\mathbb Q$?

I know that the trace is a coefficient of the minimal polynomial of $\xi_n$ over $\mathbb Q_p$. So if $n=p^l$ the minimal polynomial of $\xi_n$ is the same as over $\mathbb Q$, so it is a cyclotomic polynomial and thus has integer coefficients and above trace is an integer.

But what about the case $n\neq p^l$?

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  • $\begingroup$ The trace is not a coefficient: it's the negative of a coefficient. $\endgroup$ – KConrad Mar 29 '14 at 19:56
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Certainly not in general: If $p \equiv 1 \bmod n$ (but $n \neq \pm 1$) then $\mathbb{Q}_p(\zeta_n) = \mathbb{Q}_p$) so $Tr(\zeta_n) = \zeta_n$ which is not in $\mathbb{Q}$. To give another example, take $n=5$ and $p \equiv 4 \bmod 5$. Then $\mathbb{Q}_p$ contains a square root of $5$ and $x^4+x^3+x^2+x+1 = (x^2 + (1-\sqrt{5})/2 x +1) (x^2 + (1+\sqrt{5})/2 x +1)$ so the trace of a primitive $5$th root of unity will be $(-1 \pm \sqrt{5})/2$.

I am tempted to guess that the only time that this happens is when the $n$-th cyclotomic polynomial is irreducible in $\mathbb{Q}_p$, which is equivalent to asking that $(\mathbb{Z}/n \mathbb{Z})^{\ast}$ is cyclic and $p$ is a generator. But I don't see an easy proof of that. Guess is wrong, see below. But it is right when $n$ is square free!

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    $\begingroup$ Take $n=9$, $p=4$ mod $9$. Then isn't the trace $0$, because the Frobenius conjugates of a root differ by multiplcication by a third root of unity? $\endgroup$ – Will Sawin Mar 26 '14 at 15:28
  • $\begingroup$ Yup, I was just coming back to edit in that example! On the other hand, when $n$ is square free, I can prove the result. According to math.stackexchange.com/questions/87290/… , in that case, the primitive $n$-th roots of unity are linearly independent over $\mathbb{Q}$. So the unique linear relation between the primitive $n$-th roots of unity and $1$ is $\sum_{a \in \mathbb{Z}/n^{\ast}} \zeta^a = Tr_{\mathbb{Q}(\zeta)/\mathbb{Q}} \zeta$ and we see that no proper subset of the primitive roots can have rational sum. $\endgroup$ – David E Speyer Mar 26 '14 at 16:50
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You're asking for the trace down to $\mathbf Q_p$ of any $n$th root of unity $\xi_n$ in a cyclotomic extension of $\mathbf Q_p$. David has interpreted the question in his answer to be the case $\xi_n = \zeta_n$ (i.e, $\xi_n$ has order $n$), which is more or less the general case because $\xi_n$ will be a primitive root of unity for whatever its order happens to be, so by the transitivity formula $$ {\rm Tr}_{\mathbf Q_p(\zeta_n)/\mathbf Q_p}(\xi_n) = {\rm Tr}_{\mathbf Q_p(\xi_n)/\mathbf Q_p}({\rm Tr}_{\mathbf Q_p(\zeta_n)/\mathbf Q_p(\xi_n)}\xi_n) = [\mathbf Q_p(\zeta_n):\mathbf Q_p(\xi_n)]{\rm Tr}_{\mathbf Q_p(\xi_n)/\mathbf Q_p}(\xi_n) $$ we could assume $\xi_n$ generates the cyclotomic extension of $\mathbf Q_p$; we just need to know additionally the degree $[\mathbf Q_p(\zeta_n):\mathbf Q_p(\xi_n)]$, and writing this as $[\mathbf Q_p(\zeta_n):\mathbf Q_p]/[\mathbf Q_p(\xi_n):\mathbf Q_p]$ turns it into a question about degrees of cycltomic extensions of $\mathbf Q_p$ over $\mathbf Q_p$. The degree of cyclotomic extensions of $\mathbf Q_p$ is explained in Serre's Local Fields, for instance.

As for ${\rm Tr}_{\mathbf Q_p(\xi)/\mathbf Q_p}(\xi)$, where $\xi$ is a root of unity, this number is the negative of the second-leading coefficient of the minimial polynomial of $\xi$ over $\mathbf Q_p$. These polynomials are the irreducible factors over $\mathbf Q_p$ of cyclotomic polynomials, and there's no reason to believe these coefficients are given by tidy formulas in general. For example, $\mathbf Q_2(\zeta_7)$ is a cubic extension of $\mathbf Q_2$ and the minimal polynomial of $\zeta_7$ over $\mathbf Q_2$ is one of the two cubic irreducible factors of the 7th cyclotomic polynomial $\Phi_7(x)$ over $\mathbf Q_2$. While the 7th cyclotomic polynomial itself has nice coefficients (they're all 1), its two cubic factors over $\mathbf Q_2$ are going to have irrational coefficients: they "are what they are". The $\mathbf Q_2$-conjugates of $\zeta_7$ are $\zeta_7, \zeta_7^2$, and $\zeta_7^4$, so the minimal polynomial of $\zeta_7$ over $\mathbf Q_2$ is $$ (x - \zeta_7)(x - \zeta_7^2)(x - \zeta_7^4) = x^3 - (\zeta_7 + \zeta_7^2 + \zeta_7^4)x^2 + (\zeta_7^3 + \zeta_7^5 + \zeta_7^6)x + \zeta_7^7, $$ which simplifies to $x^3 + ax^2 + (a-1)x - 1$, where $a = -(\zeta_7 + \zeta_7^2 + \zeta_7^4)$. Over the complex numbers, using $\zeta_7 = e^{2\pi i/7}$, this formula for $a$ is $(1 \pm \sqrt{-7})/2$, so this would be a formula for $a$ over the 2-adics as well, but using 2-adic square roots of $-7$ (there are two square roots, one for each of the two irreducible factors of $\Phi_7(x)$). In particular, ${\rm Tr}_{\mathbf Q_2(\zeta_7)/\mathbf Q_2}(\zeta_7) = -a = \zeta_7 + \zeta_7^2 + \zeta_7^4$.

There is a nice formula (other than its definition) for the trace of roots of unity over $\mathbf Q$: ${\rm Tr}_{\mathbf Q(\zeta_n)/\mathbf Q}(\zeta_n) = \mu(n)$, where $\mu$ is the Moebius function. As far as I know, none of the other coefficients of $\Phi_n(x)$ outside its leading coefficient and constant term have nice formulas for them (the exception is the linear coefficient, which is also $\mu(n)$ when $n > 1$ on account of the symmetry formula $\Phi_n(1/x) = \Phi_n(x)$ when $n > 1$). Therefore if $\Phi_n(x)$ is irreducible over $\mathbf Q_p$ then ${\rm Tr}_{\mathbf Q_p(\zeta_n)/\mathbf Q_p}(\zeta_n) = \mu(n)$, and this happens exactly when $p \bmod n'$ generates $(\mathbf Z/n'\mathbf Z)^\times$, where $n'$ is the $p$-free part of $n$ (e.g., when $n$ is a power of $p$, which is when $n' = 1$). But beyond this case I don't think there's going to be a simple-minded formula for the trace in general; it will usually be some irrational algebraic number.

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  • $\begingroup$ Dear Keith, in your formula for the trace to $\mathbb{Q}$, I don't think you want the minus sign. $\endgroup$ – Alex B. Apr 17 '14 at 14:03
  • $\begingroup$ @AlexB.: Thanks, I've fixed that now. $\endgroup$ – KConrad Apr 17 '14 at 20:41

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