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 ?



Torsion in some specific module over the Laurent polynomials
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.


