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Consider the ring $\mathbb{Z}[t,t^{-1}]$ of Laurent-polynomials over $\mathbb{Z}$. The abelian group $M:=\prod_\mathbb{Z}\mathbb{Z}$ becomes a module over this ring via $t\cdot (x_*):=x_{*+1}$.

Is there a irreducible polynomial $p\in \mathbb{Z}[t,t^{-1}]$ with leading coefficient not equal to $\pm 1$ such that $M$ has $p$-torsion ?

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  • $\begingroup$ I just wanted to add the consequences. I would like to define a minimal polynomial of an element $m\in M$ as a generator of its Annihilator. However since $R$ is not a PID, that ideal might a priori not have a single generator. Using Davids answer we know that whenever $p\in Ann(m)$, so is the product of its irreducible factors with leading coefficient $\pm 1$. Although division with rest is in gerenal not possible in $R$, it is possible it one devides by a polynomial with leading coefficient $\pm 1$. With this one can show that $Ann(m)$ is indeed principal and hence we can pick a generator. $\endgroup$ Commented Mar 20, 2012 at 10:28
  • $\begingroup$ Maybe the term "minimal polynomial" is not good. It might not be irreducible and so on. $\endgroup$ Commented Mar 20, 2012 at 10:54

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No. Minor note: I assume that your definition of irreducible includes not being divisible by an integer $>1$, so that $5x-5$ annihilating the all ones sequence is not an example.

I'm going to show that the constant term of $f$ is $\pm 1$, since that lets me use power series in positive powers of $t$. Obviously, this is equivalent to looking at the leading coefficient by the symmetry $t \mapsto t^{-1}$.

Suppose to the contrary that $f(t)$ annihilates the sequence $(a_i)$ and $f(t)$ is irreducible. Let $p$ be a prime which divides the constant term of $f$ but does not divide every term of $f$. Let $\mathbb{C}_p$ be the completion of the algebraic closure of the $p$-adics. The $p$-adic Newton polygon of $f$ has a line segment with positive slope, so $f$ has a root $\rho$ in $\mathbb{C}_p$ with $|\rho|_p<1$.

Let $a(t) = \sum_{i \geq 0} a_i t^i$. Since the $a_i$ are integers, they have $|a_i|_p \leq 1$, and thus the sums converges for any $t \in \mathbb{C}_p$ with $|t|_p<1$. The condition that $f$ annihilates $(a_i)$ means that $f(t) a(t) = g(t)$ for $g$ a polynomial in $\mathbb{Z}[t]$ of degree less than that of $f$. In particular, $g(\rho)=0$.

But $f$ is irreducible over $\mathbb{Q}$, so it must be the minimal polynomial of $\rho$, and $\deg g < \deg f$. This is a contradiction.

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  • $\begingroup$ I'm afraid this is going to be vague, but I remember thinking about problems like this when I was trying to read arxiv.org/abs/math/0104261 . $\endgroup$ Commented Mar 19, 2012 at 16:15

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