Is an integral sum of periodic vectors always a sum of integral periodic vectors?

Question

Update:

I have found reference to this problem. It is known as "the Rédei-de Bruijn-Schönberg theorem", which is proved in the following papers:

• N. G. de Bruijn: On the factorization of cyclic groups, Indag. Math.15(1953), 370-377.
• L. Rédei: Ein Beitrag zum Problem der Faktorisation von Abelschen Gruppen, ActaMath. Acad. Sci. Hungar.1(1950), 197-207.
• I. J. Schoenberg: A note on the cyclotomic polynomial, Mathematika11(1964), 131-136.

End of update.

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The background of this question is something about cyclotomic fields, but the statement doesn't involve any algebraic number theory. I just get puzzled by this (might be stupid) little question...

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Let $n>1$ be an integer, and consider the vector space $\mathbb{C}^n$.

A vector $v=(v_1, \cdots, v_n) \in \mathbb{C}^n$ is called

• periodic, if there is a proper divisor $d$ of $n$, such that $v_i = v_{i + d}$ for all $i$;
• integral, if every $v_i$ is an integer.

Question: if an integral vector can be written as a finite sum of periodic vectors, then is it true that it can always be written as a finite sum of integral period vectors?

It should be clear that the field $\mathbb{C}$ could be replaced with any field of characteristic zero (e.g. $\mathbb{Q}$).

I would guess that the claim is true, but cannot convince myself with a proof...

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So far I can only prove the case when $n$ has at most $2$ different prime factors, which doesn't help much in the general case.

I've also tried to adopt a point of view from cyclotomic fields, or representation theory, or doing some Fourier transform - but again I'm not intelligent enough to morph the question to something known...

Therefore I add all the possibly relevant tags.

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EDIT: I add here a proof when $n = pq$ is the product of $2$ different primes. The more general case $n = p^r q^s$ is morally the same.

Since $n = pq$, every periodic vector $a$ either satisfies $a_i = a_{i + p}$ for all $i$, or satisfies $a_i = a_{i + q}$ for all $i$.

Hence any finite sum of periodic vectors can be written as $a + b$, where $a_i = a_{i + p}$ and $b_i = b_{i + q}$.

Now suppose that $v = a + b$ is such a sum, which is integral. This means that $a_i + b_i = 0\mod\mathbb{Z}$, which then gives: $$a_i = -b_i = -b_{i + q} = a_{i + q} \mod \mathbb{Z}.$$ Together with $a_i = a_{i + p}$, we conclude that all $a_i$ are equal $\mod\mathbb{Z}$, hence all $b_i$ also, and we can adjust them with a constant vector so as to make them both integral.

This trick doesn't work for more than $2$ prime factors, though...

Answer 1

A beautiful question! Though I don't have the time or the space to fill in all the details, I think one can answer the question in the following way. The answer is yes.

We can reformulate the question as follows: let $I \subset \mathbb{Z}[x]$ be the ideal generated by the polynomials $$ \frac{x^n-1}{x^d-1}, \quad d \mid n, \quad d \neq n. $$ We claim that $I$ is $\mathbb{Z}$-saturated, that is, within $\mathbb{Q}[x]$, $$ \mathbb{Q} I \cap \mathbb{Z}[x] = I. $$ (Attach to each polynomial $f \in \mathbb{Q}[x]$ the sequence given by the coefficients of $f$ mod $x^n - 1$. The passage from $\mathbb{C}$ to $\mathbb{Q}$ is effected by some easy linear algebra.)

We claim, indeed, that $$ I = J := (\Phi_n) $$ is the ideal generated by the $n$th cyclotomic polynomial. Note that $J$ is visibly $\mathbb{Z}$-saturated.

1. Clearly $$ (x^n-1) \subseteq I \subseteq J. $$
2. Also, $\mathbb{Q}I = \mathbb{Q}J$, because the only common roots of the generators of $I$ are at the primitive $n$th roots of unity and are all simple. (This step uses the Nullstellensatz, which, since we have just one variable, isn't all that deep.)
3. So if $I \neq J$, the discrepancy can be found locally, that is, there is a prime $p$ such that the images $\bar{I}, \bar{J}$ of $I$ and $J$ in $\mathbb{F}_p[x]$ are unequal. (Pass to the finite-dimensional $\mathbb{Q}$-vector space $\mathbb{Q}[x]/(x^n-1)$. The images $L_I$, $L_J$ are $\mathbb{Z}$-lattices with the same $\mathbb{Q}$-span; if they are unequal, pick $p | [L_J : L_I]$.)

4. If $p \nmid n$, we are done by the Nullstellensatz again, because the $n$th roots of unity remain distinct in $\bar{\mathbb{F}}_p$.

5. Now $p \mid n$. Let $$ \bar K = \left\{\frac{f}{\Phi_n(x)} : f \in \bar I \right\}. $$ We wish to prove that $\bar K$ is the unit ideal, that is, that the polynomials in $\bar K$ have no common roots. Suppose there is a common root $\lambda \in \bar{\mathbb{F}}_p$. Then $\lambda^n = 1$, since $x^n - 1 \in \bar I$. Let $\lambda$ be a primitive $r$th root of unity; note that $r \mid n$ and $p \nmid r$, so $r \mid n'$, where $n = p^k n'$ with $p \nmid n'$. We get a contradiction as follows:

   • If $r \neq n'$, then the element $$ \frac{x^n - 1}{\Phi_n(x) (x^{p^k r} - 1)} = \prod_{d|n,\, d\nmid p^k r,\, d \neq n} \Phi_d \in \bar K $$ has no root at $\lambda$ (because the factors $\Phi_r, \Phi_{pr}, \ldots, \Phi_{p^k r}$ have all been eliminated).
   • If $r = n'$, then the element $$ \frac{x^n - 1}{\Phi_n(x) (x^{n/p} - 1)} = \prod_{d|n,\, p^k \parallel d,\, d \neq n} \Phi_d \in \bar K $$ likewise has no root at $\lambda$.