Torsion units of the ring $\mathbb{Z}[x]/(1+x+x^2+\cdots+x^{n-1})$

Question

What are the torsion units of the ring $R_n:=\mathbb{Z}[x]/(1+x+x^2+\cdots+x^{n-1})$? Since $x^n = 1$ in $R_n$ it is clear that all elements of the form $\pm x^i$ are torsion units. Is this all of them?

Of course, if $n$ is prime then $R_n$ is the ring of integers of the $n^{th}$ cyclotomic field and the result is true in this case.

For small composite $n$ is it not hard to verify the statement numerically. The Chinese Remainder Theorem gives an injective ring map $$ \Psi: R_n\rightarrow \prod_{\substack{k|n \\ k\neq 1}}\mathbb{Z}[\zeta_k] $$ where $\mathbb{Z}[\zeta_k]$ is the ring of integers of the $k^{th}$ cyclotomic field. For small $n$ one can enumerate all of the torsion units on the right-hand side, choose $\mathbb{Z}$-bases for both sides, and via a simple matrix calculation deduce which torsion units on the right-hand side are in the image of $\Psi$. I've carried out this calculation for all composite $n\leq 30$ and verified that the only torsion units in $R_n$ are indeed the elements $\pm x^i$. I've been unable to turn this into a proof for all $n$ as the inverse of the matrix of $\Psi$ has no obvious general form.

The corresponding statement is indeed true for the similar ring $S_n:=\mathbb{Z}[x]/(x^n-1)$, which is the group ring of the cyclic group of order $n$. The result in this case goes back to at least Higman [1], and is not difficult to show. Indeed, let $\phi: S_n\rightarrow M_n(\mathbb{C})$ denote the left regular representation of $S_n$, and suppose $u = a_0 + a_1x + \cdots + a_{n-1}x^{n-1}$ with $a_i\in\mathbb{Z}$ satisfies $u^m = 1$. Multiply $u$ by an appropriate power of $x$ so that $a_0 \neq 0$. We have $\mathrm{tr}(\phi(x^i)) = 0$ for all $i \neq 0$, and hence $\mathrm{tr}(\phi(u)) = a_0n$. On the other hand, since $\phi(u)$ is an $n\times n$ matrix of finite order we must have $|\mathrm{tr}(\phi(u))|\leq n$ and therefore $a_0 = \pm 1$. This implies that we have equality $|\mathrm{tr}(\phi(u))| = n$ and therefore $\phi(u)$ is a scalar matrix, from which we see that $\phi(u) = a_0I_n$. Since $\phi$ is injective we must have $u = a_0$, which completes the proof.

The difficulty in applying this argument to the ring $R_n$ is that the left regular representation results in a much weaker inequality. Specifically, if $\rho:R_n\rightarrow M_{n-1}(\mathbb{C})$ denotes the left regular representation of $R_n$ we now have $\mathrm{tr}(\rho(x^i)) = -1$ if $i\neq 0$. So if $u = a_0 + a_1x + \cdots + a_{n-2}x^{n-2}$ with $a_i\in\mathbb{Z}$ satisfies $u^m = 1$, we can only say that $$ |a_0(n-1) - \sum_{i = 1}^{n-2}a_i|\leq n - 1 $$ from which I can deduce very little about the individual $a_i$.

[1] Higman, G., The units of group-rings., Proc. London math. Soc. (2) 46, 231-248 (1940). ZBL66.0104.04.

Answer 1

I think, they are all of them. Let me be more concrete and accurate than in the initial answer. But this also makes the answer sometimes boring. Shortcuts are welcome.

Let $f$ be a torsion unit, that is, $f$ is represented by a polynomial with integer coefficients (again denoted by $f$) of degree at most $n-2$ such that $f^N-1$ is divisible by $1+x+\ldots+x^{n-1}$ for certain positive integer $N$.

Denote by $\Theta(n)$ the set of primitive roots of unity of degree $n$, and by $T(n)=\sqcup_{1<d,d|n} \Theta(d)$ the set of all roots of unity of degree $n$ other than 1. Thus, we are given that $f(\xi)$ is a root of unity for all $\xi\in T(n)$.

Lemma 1. If $w\in \Theta(a)$, $g(z)\in \mathbb{Z}[z]$ and $g(w)\in \Theta(b)$, then either $b$ divides $a$ or $a$ is odd and $b$ divides $2a$.

Proof. Denote $c={\rm LCM}(a,b)$ and choose $\eta\in \Theta(c)$. Then we may find integers $k,\ell$ such that $w=\eta^k$, $g(w)=\eta^\ell$. So, the polynomial $P(x):=g(x^k)-x^\ell$ has a root $x=\eta$. Thus any element of $\Theta(c)$ is a root of $P$, since cyclotomics are irreducible. For any integer $t$ coprime to $c$ we have $\eta^t\in \Theta(c)$, so $g(w^t)-(g(w))^t=P(\eta^t)=0$. Try to choose $t$ such that $w=w^{t}$ but $g(w)^t\ne g(w)$, this would yield a contradiction. Our requirements for $t$ are, in other words: $a$ divides $t-1$; but $b$ does not divide $t-1$. Assume that $b$ does not divide $a$, and, if $a$ is odd, then $b$ does not divide $2a$. This allows to choose a prime $p$ such that $s:=\nu_p(b)$ (notation means that $p^s$ is the maximal power of $p$ which divides $b$) satisfies $s>\nu_p(a)$ and either $p$ is odd, or $p=2$ and $s\geqslant 2$. Using Chinese remainders theorem, choose $t$ be congruent to $1+p^{s-1}$ modulo $p^s$ and $t$ be congruent to 1 modulo $c/p^{s}$. Then $t$ is coprime with $c$, $a$ divides $t-1$ but $b$ does not divide $t-1$. This yields a necessary contradiction.

Let now $a>1$ be a divisor of $n$, choose $w\in \Theta(a)$. Let $g(w)\in \Theta(b)$. If $a$ is odd and $b$ is an even divisor of $2a$, then $-g(w)\in \Theta(b/2)$. Thus in all cases of Lemma 1 we have $f(w)=\pm w^m$ for certain $m$. Then this equation $f(x)=\pm x^m$ holds for all $x\in \Theta(a)$, since cyclotomics are still irreducible.

So, for each divisor $a>1$ we have a sign $\varepsilon(a)=\pm 1$ and an exponent $m(a)$ such that $f(x)\equiv \varepsilon(a) x^{m(a)} \pmod{\Phi_a(x)}$. Now we need to prove that these signs and exponents for different $a$ are consistent. The idea (not very deep) is to prove this by induction on $n$. Let $p_1<p_2<\ldots<p_r$ be all prime divisors of $n$. By induction, we may suppose that for certain signs $\delta_i$ and exponents $m_i$ we have $\varepsilon(a)x^{m(a)}=\delta_i x^{m_i}$ for all divisors $a>1$ of $n/p_i$ and all $x\in \Theta(a)$ (in other words, that $f(x)\equiv \delta_ix^{m_i} \pmod{1+x+\ldots+x^{n/p_i-1}}$).

We proceed with proving the consistency of signs $\delta_i$ and exponents $m_i$. Let's start with signs. Assume that $\delta_i=-\delta_j$ for some $i<j$. This yields that $x^{m_i}+x^{m_j}$ is divisible by $1+x+\ldots+x^{n/(p_ip_j)-1}$. Substituting $x=1$, we get that $2$ is divisible $n/(p_ip_j)$, that is, $n=p_ip_j$ or $n=2p_ip_j$. In the latter case, if $1<i<j$, then replacing the pair $i<j$ to $1<i$ or $1<j$ in the above argument, we get a contradiction. So, $n=p_ip_j$ or $n=4p_j$. If $p_1=2$ and $n=2p_2$, then the signs may be chosen consistently because $-1=x$ on $T(2)$. The cases $n=pq$ with odd primes $p\ne q$ and $n=4p$ with odd prime $p$ are to be considered separately, this is done at the end of this answer.

Now for the exponents, assuming that the signs are consistent (say, $\delta_i=1$ for all $i$). We know that $x^{m_i}-x^{m_j}$ is divisible by $\tau:=1+x+\ldots+x^{n/(p_ip_j)-1}$. That is, $m_i$ and $m_j$ are congruent modulo $n/(p_ip_j)$ (powers of $x$ are periodic modulo the polynomial $\tau$ with period $n/(p_ip_j)$). For every $i=1,\ldots,r$, denote $\alpha_i:=\nu_{p_i}(n)$. Then there exists an integer $T$ congruent to $m_j$, $j\ne i$, modulo $p_i^{\alpha_i}$, and congruent to $m_i$ modulo $p_i^{\alpha_i-1}$. For such $T$ we have $f(x)\equiv x^{T} \pmod {1+x+\ldots+x^{n/p_i-1}}$ for all $i=1,\ldots,r$.

So, the signs and exponents are consistent for all proper divisors of $n$. Without loss of generality, $f(\xi)=1$ for all $\xi\in \sqcup_{1<d<n,d|n} \Theta(d)$. It remains to prove that if also $f(\xi)=\varepsilon \xi^a$ for $a\in \Theta(n)$, this is also consistent (not necessarily by trivial $\varepsilon=1$, $a=0$: it may be $n=p^\alpha$, $\varepsilon=1$, $\nu_p(a)=\alpha-1$, or $n=4$, $f(x)=-x=x^3$, in the latter case the formula $-x$ works on the whole $T(4)$, but $x^3$ does not.)

If $n$ is even, we may replace $\varepsilon=-1$ to $x^{n/2}$ modulo $\Phi_n(x)$, so, if $n$ is even, we may suppose that $\varepsilon=1$. If $n$ is odd and $\varepsilon=-1$, then the polynomial $f^n \pmod {1+x+\ldots+x^{n-1}}$ (which is of course again a torsion with integer coefficients) takes the values $-1$ on $\Theta(n)$ and 1 on $T(n)\setminus \Theta(n)$.

If $\varepsilon=1$ and $n$ is not a prime power, then the only consistent variant is $a$ divisible by $n$. Assuming that this is not the case, we may a prime $p$ for which $\nu_p(a)<\nu_p(n)$ and taking integers $u,v$ such that $au+nv=n/p$ (they exists since $n/p$ is divisible by a GCD of $n$ and $a$) replace $f$ to $f^ux^{nv}$, this allows to assume that $a=n/p$. If $n=p^\alpha$ is a prime power, then $a$ divisible by $p^{\alpha-1}$ is consistent, and, analogously, we may suppose that $a=p^{\alpha-2}$ for odd $p$ or for $p=2$, $\alpha\geqslant 3$ (in the case of $n=4$ everything is consistent, see above).

So, we should exclude three cases for $f(x)$ restricted to $\Theta(n)$ subject to $f(x)=1$ on $T(n)\setminus \Theta(n)$:

1. $f(x)=-1$ for odd $n$;

2. $f(x)=x^{n/p}$, when $p$ is a prime divisor of $n$ and $n$ is not a prime power;

3. $f(x)=x^{p^{\alpha-2}}$ for $n=p^\alpha$, $n>4$.

We use the interpolational formula $$ 1-f(x)=\frac1n\left(\sum_{\xi\in \Theta(n)}(1-f(\xi))\xi\cdot \frac{x^n-1}{x-\xi}- \sum_{\xi\in \Theta(n)}(1-f(\xi))\xi( 1+x+\ldots+x^{n-1})\right). $$ And we use that the sum of elements of the set $\Theta(n)$ is $\mu(n)$, and that the $k$-th powers of $\Theta(n)$ uniformly cover the set $\Theta(n/{\rm GCD}(n,k))$.

Look at $1-f(0)$.

In the case 1), we get $$1-f(0)=\frac{2}n(\varphi(n)-\mu(n))$$ which is not an integer (since $n$ is not prime).

In the case 2), we get $$1-f(0)=\frac{1}n\left(\varphi(n)+\frac{\varphi(n)}{p-1}-\mu(n)+\sum \xi^{n/p+1}\right).$$ Since ${\rm GCD}(n/p+1,n)$ is either 1 or $p$, the sum $\sum \xi^{n/p+1}$ equals either to $\mu(n)$, or to $(p-1)\mu(n/p)$. In the first case
$$ \varphi(n)+\frac{\varphi(n)}{p-1}-\mu(n)+\sum \xi^{n/p+1}= \frac{p}{p-1}\varphi(n)=n\prod_{q\, \text{is prime},q\ne p,q|n}(1-1/q) $$ which clearly is not divisible by $n$ (recall that $n$ is not a prime power, so the product over $q$ is strictly smaller than 1). In the second case, since $\mu(n)+\mu(n/p)=0$, we need $$n\prod_{q\, \text{is prime},q\ne p,q|n}(1-1/q)+p\mu(n/p)$$ to be divisible by $n$. This is the case only if $n=2p$ (otherwise it is too small). But $p$ divides $n/p+1=3$, so $p=3$, $n=6$. In this case the coefficient of $x^4$ in $1-f(x)$ is not an integer.

In the case 3), the number $p^{\alpha-2}+1$ is coprime to $n=p^\alpha$, thus $\sum \xi(1-f(\xi))=0$. We have $\sum \xi^j=0$ for $j$ not divisible by $p^{\alpha-1}$, but if $\nu_p(j)=\alpha-1$ then $\sum \xi^j=-p^{\alpha-1}$. Thus, between the coefficients of $1-f(x)$ we may find $\pm\frac1n\cdot p^{\alpha-1}=\pm 1/p$, a contradiction.

Finally, the cases of not consistent signs with $n=4p$ and $n=pq$. Well, they are similar to the above analysis.

If $n=pq$ with distinct odd primes $p$, $q$, and $f(x)=x^a$ on $\Theta(p)$, $f(x)=-x^b$ on $\Theta(q)$, $f(x)=x^c$ on $\Theta(pq)$, then replacing $f$ to $f^n$ we may suppose that $f$ equals 1 on $\Theta(p)$ and $\Theta(pq)$ and $f=-1$ on $\Theta(q)$. Thus $$ 1-f(x)=\frac2{pq}\left(\sum_{\xi\in \Theta(q)} \xi\cdot \frac{x^n-1}{x-\xi}+(1+x+\ldots+x^{n-1})\right) $$ and $1-f(0)=2/p$, a contradiction.

Finally, if $n=4p$, $f(x)=x^a$ on $T(2p)=\Theta(2)\sqcup \Theta(p)$, $f(x)=-x^b$ on $T(4)=\Theta(4)\sqcup \Theta(2)$, $f(x)=x^c$ on $\theta(4p)$ (for $\Theta(4p)$ the sign may be chosen as we wish by the already used "$-1=x^{2p}$ on $\Theta(4p)$" trick). As before, we may suppose that $a=0$ and that $c$ is divisible by $p$ (otherwise replace $f$ to $(f/x^a)^p$). Also, $b$ is odd since $1=f(-1)=(-1)^{b+1}$. Next point is that $c$ must be even. Assume that, on the contrary, $c$ is odd multiple of $p$. Choose $w\in \Theta(4p)$, then $-w\in \Theta(2p)$ and $\rho:=\frac{f(w)-f(-w)}{w-(-w)}$ must be an algebraic integer. We have $f(-w)=1$ and $f(w)=\pm i$, thus $w\rho=(-1\pm i)/2$ is an algebraic integer, but it is not. So, $c$ is divisible by $2p$. Then the polynomial $g(x):=x^{c}-f(x)$ equals 0 on $T(2p)\sqcup \Theta(4p)$, and takes values $\pm 1\pm i$ at two elements of $\Theta(4)$. Therefore (reducing as usually $g$ modulo $1+x+\ldots+x^{n-1}$) we get $$ g(x)=\frac1{4p}\left(\sum_{\xi\in \Theta(4)} g(\xi)\cdot \xi\cdot \frac{x^{4p}-1}{x-\xi}-\sum_{\xi\in \Theta(4)}\xi g(\xi)(1+x+\ldots+x^{4p-1})\right). $$ This does not have integer coefficients simply because the coefficients are too small.