If the integral linear combination of some $n$th roots of unity has magnitude 1, does this necessarily imply that this linear combination is some root of unity as well? More precisely,

Let $\zeta_1, \ldots \zeta_k$ be $n$th roots of unity. If $$|\sum_{i=1}^k n_i \zeta_i| = 1,$$ where $n_i \in \mathbb{Z}$, does this imply that $\sum_{i=1}^k n_i \zeta_i$ is an $n$th root of unity? What about if the $n_i$ are Gaussian integers?

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    $\begingroup$ By sum, do you mean any linear combination? $\endgroup$ – Joel Cohen May 18 '11 at 13:39
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    $\begingroup$ @Qiaochu, @ joriki : Indeed, I meant "integer linear combination". Thanks for pointing the omission. $\endgroup$ – Joel Cohen May 18 '11 at 13:57
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    $\begingroup$ Why does this question have 10 upvotes if no one even knows what it's asking? It was only asked 41 minutes before this comment and it has 10 upvotes already? $\endgroup$ – Graphth May 18 '11 at 13:57
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    $\begingroup$ @Chandru: $\mathrm e^{2\pi\mathrm i/3}+e^{-2\pi\mathrm i/3}=-1$, that is, two cube roots of unity add up to a square root of unity. $\endgroup$ – joriki May 18 '11 at 13:58
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    $\begingroup$ @Numth: Because it's a one-liner that evokes the immediate intuition that the answer must be "yes" but on second thought appears not quite as trivial to prove as one might have thought. Also it seems like something that might be useful in applications. $\endgroup$ – joriki May 18 '11 at 14:00

Yes. To be more specific, I will prove the following: Let $\zeta$ be an $n$-th root of unity and let $\alpha = \sum a_k \zeta^k$ for some integers $a_k$. If $|\alpha|=1$, then $\alpha$ is a root of unity.

The key will be the following theorem of Kronecker: Let $\beta$ be an algebraic integer, and suppose that all the Galois conjugates of $\beta$ have absolute value $1$. Then $\beta$ is a root of unity. We discussed this over on MO. (I will restate the condition on Galois conjugates more explicitly below.)

It is clear that $\alpha$ is an algebraic integer, so the key is to see that all of its Galois conjugates also have norm $1$. Very explicitly, we want to see that, for any $r$ relatively prime to $n$, we have $|\sum a_k \zeta^{kr}|=1$. Let's write $\alpha^{(r)}$ for $\sum a_k \zeta^{kr}$.

Here is the point. We have $$|\alpha|^2 = \alpha \overline{\alpha} = \left( \sum a_k \zeta^k \right) \left( \sum a_{\ell} \zeta^{- \ell} \right).$$ Multiplying this out gives a big polynomial in $\zeta$ with integer coefficients; call this polynomial $N$. The same computation shows that $|\alpha^{(r)}|^2 = N(\zeta^r)$.

So our hypothesis is that $N(\zeta) = 1$ and we want to show that $N(\zeta^r)=1$ as well. Since $N(\zeta)=1$, we see that $\zeta$ is a root of $N-1$. Since the $n$-th cyclotomic polynomial is irreducible, this implies that $\zeta^r$ is also a root of $N-1$, and we are done.

Conceptually, this argument worked because complex conjugation was in the center of $\mathrm{Gal}(\mathbb{Q}(\zeta)/\mathbb{Q})$. An analogous argument shows that, if $K$ is any CM field, and $\alpha$ in $K$ has norm $1$, then all the conjugates of $\alpha$ have norm $1$.

  • $\begingroup$ Allow me to point out that the proof essentially uses the fact that $\mathbb{Q}(\zeta_n)$ is a CM-field, so the complex conjugation map (denote by $\iota$) commutes with all embedding from the field to $\mathbb{C}$. So let $F\subseteq \mathbb{C}$ be a CM field, and $z\in F$ such that $$\lvert z \rvert^2 = z\cdot \iota z=1, $$ then for all embedding $\sigma: F\hookrightarrow \mathbb{C}$, $$ \lvert \sigma z \rvert^2= \sigma z \cdot \iota \sigma z=\sigma z\cdot \sigma \iota z = \sigma (z\cdot\iota z)=1.$$ So all the Archimedean valuation of $z$ is 1. Now we just apply the theorem of Kronecker. $\endgroup$ – Jiangwei Xue May 18 '11 at 14:24
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    $\begingroup$ I think David's proof needs only that the field is cyclic, hence, abelian, hence complex conjugation commutes with all other elements of the Galois group. Perhaps the simplest example of an algebraic integer of modulus 1 not a root of unity would be a non-real root of $x^4+4x^3+4x^2+4x+1=(x^2+(2-\sqrt2)x+1)(x^2+(2+\sqrt2)x+1)$. The two roots of the first quadratic are complex conjugates and multiply to 1 so they have modulus 1, the two roots of the other quadratic are real, one inside, one outside the unit circle. $\endgroup$ – Gerry Myerson May 19 '11 at 7:32

Much the same problem was just posted on mathoverflow (https://mathoverflow.net/questions/75753), and before David Speyer made the connection with this Stackexchange problem I posted an answer using the Dirichlet unit theorem instead of Kronecker's characterization of roots of unity. Like the approach David used, this one generalizes to an arbitrary "CM field" (i.e. a totally imaginary quadratic extension of a totally real number field, which by Dirichlet is the only way that a nontrivial extension of number fields can have the same unit-group rank).


Here's an alternative proof. It seems more elementary to me, but that might just be because I don't see the connections to the results used in David's proof and am reinventing the wheel. If there are such connections, I'd appreciate if someone could point them out in the comments. My proof will use David's result that $|\sum a_k \zeta^{k}|=1$ implies $|\sum a_k \zeta^{kr}|=1$ for $r$ relatively prime to $n$; the only "sophisticated" result he used for that was the irreducibility of the cyclotomic polynomials (and implicitly unique factorization).

First, note that the question has changed and the answer to the new question is "no". As I wrote in the comments, two cube roots of unity sum to a square root of unity, so the restriction to $n$-th roots of unity makes the statement false.

The true statement in the original question was "If the sum of some $n$th roots of unity has magnitude $1$, this implies that this sum is some root of unity as well" (where the same $n$-th root of unity may occur multiple times in the sum, to make this equivalent to the integer linear combination formulation). This is equivalent to the same statement with "$n$-th" deleted, since for every set of roots of unity there is a common denominator $n$ for which they are all $n$-th roots.

Now for the proof, consider the free vector space on the set of the $n$-th roots of unity. Any linear combination of $n$-th roots of unity acts on this vector space by multiplication. The corresponding matrix with respect to the canonical basis is a circulant matrix with the first column given by the coefficients $a_i$ of the linear combination. The eigenvectors of this circulant matrix are the Fourier modes $v_r=(1,\omega^r,\omega^{2r},\dotsc$), with $\omega=\mathrm e^{2\pi\mathrm i/n}$, and the corresponding eigenvalues are the Fourier transforms of the coefficients, $\sum a_k \omega^{-kr}$.

Now consider the natural map $f$ from this vector space to $\mathbb C$ sending each vector to the corresponding linear combination of the roots of unity. All but one of the Fourier modes are in the kernel of this map. Only the $r=-1$ mode isn't, and the corresponding eigenvalue is our number $\alpha=\sum a_k \omega^k$.

Now let's start with the vector $(1,0,\dotsc,0)$, corresponding to unity, and successively multiply by the circulant matrix. This initial vector has equal components $1/n$ of each Fourier mode. Consider what happens if we add the components corresponding to the primitive roots for some divisor $d\mid n$ together. If $d\neq n$, all of these are in the kernel of $f$. The entries in their sum are Ramanujan's sums, which are integers. Thus, by dropping these components, we don't change the complex number that the vector maps to, and we don't change the fact that all entries in the vector are integer multiples of $1/n$.

Now all that's left is the sum of the components corresponding to the primitive $n$-th roots. We know that the eigenvalue for $r=-1$ has magnitude $1$, and by David's result that implies that all the others also have magnitude $1$. So we have a vector whose entries are integer multiples of $1/n$ and we keep multiplying it by an integer matrix whose eigenvalues (as far as this vector is concerned) have magnitude $1$. It follows that there are only finitely many values this vector can take, so it has to eventually get back to where it started, which implies that $\alpha$ is a root of unity.

  • $\begingroup$ I found something that seems very relevant to the connections between this approach and David's: circulants.org/circ/circall.pdf. Any insightful comments would be appreciated. $\endgroup$ – joriki May 19 '11 at 22:39

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