Primitive Laurent annihilators preserve nonzero rational and prime reductions

Suppose $x\in T_{\mathrm{\Delta }}$ maps to zero in the rational quotient. It has a representative $r\in {\mathrm{\Lambda }}_{\mathbb{Q}}$, and $\mathrm{\Delta }r\in \mathrm{\Lambda }$. Let $m\ge 1$ be the least positive integer clearing the finitely many denominators of $r$ and put $f=mr\in \mathrm{\Lambda }$. If $m>1$, take a prime divisor $p$ of $m$; the least divisor greater than one is prime, so this requires no choice principle. Not every coefficient of $f$ is divisible by $p$, since otherwise $m\mathrm{/}p$ would also clear the denominators of $r$, contrary to minimality. Primitivity likewise says $\bar{\mathrm{\Delta }}\mathrm{\ne }0$ modulo $p$. Thus $\bar{\mathrm{\Delta }}\bar{f}\mathrm{\ne }0$ in the field Laurent polynomial ring. But $\mathrm{\Delta }f=m(\mathrm{\Delta }r)$ has all coefficients divisible by $p$, a contradiction. Hence $m=1$, so $r\in \mathrm{\Lambda }$ and $x=0$. This proves the asserted injection.

Every $x\in T_{\mathrm{\Delta }}$ has a representative $a\mathrm{/}\mathrm{\Delta }$ with $a\in \mathrm{\Lambda }$: take any rational-function representative $r$ of $x$ and use $a=\mathrm{\Delta }r$, which is in $\mathrm{\Lambda }$ by the annihilation condition. For two such representatives $a\mathrm{/}\mathrm{\Delta },a^{\mathrm{\prime }}\mathrm{/}\mathrm{\Delta }$, equality of their classes means $a^{\mathrm{\prime }}-a=\mathrm{\Delta }b$ for some $b\in \mathrm{\Lambda }$. If $\bar{\mathrm{\Delta }}\mathrm{\ne }0$, their reduced fractions differ by $\bar{b}$ and so define the same class in ${\mathbb{F}}_p(t)\mathrm{/}{\mathrm{\Lambda }}_p$. The additive and $\mathrm{\Lambda }$-scalar identities follow by reducing their actual finite numerators. This constructs precisely the claimed map and uses no reduction of arbitrary rational-function coefficients or denominators.

Fix a nonzero $x$. Multiply $\mathrm{\Delta }$ by a power of $t$ to obtain a polynomial $d(t)$ with nonzero constant and leading coefficient; this unit change does not change its annihilated module. Adjust its numerator accordingly, then multiply $x$ by a power $t^m$ so that $y=t^{m}x=c(t)\mathrm{/}d(t)$ with $c\in \mathbb{Z}[t]$. The class $y$ is nonzero, and step 1.1 implies $c\mathrm{/}d\mathrm{\notin }{\mathrm{\Lambda }}_{\mathbb{Q}}$. Ordinary finite polynomial division over $\mathbb{Q}$ gives $c=dQ+R$, with $\mathrm{deg}R<\mathrm{deg}d$. Its remainder is nonzero, since a zero remainder would give $c\mathrm{/}d\in \mathbb{Q}[t]$. Clear the finitely many coefficient denominators: $qc=dA+B,q\in {\mathbb{Z}}_{>0},A,B\in \mathbb{Z}[t],B\mathrm{\ne }0,\mathrm{deg}B<\mathrm{deg}d\mathrm{.}$ In particular $\mathrm{deg}d>0$. Choose the first nonzero coefficient $C$ of $B$ in increasing exponent order and set $D=qd(0)\mathrm{lead}(d)C\mathrm{\ne }0$. All choices here are finite coefficient operations and polynomial division.

If $p\nmid D$, then $\bar{d}$ has its original positive degree and nonzero constant term, $\bar{B}\mathrm{\ne }0$ and $\mathrm{deg}\bar{B}<\mathrm{deg}\bar{d}$, and $\bar{q}$ is invertible. The reduced identity gives $[\bar{c}\mathrm{/}\bar{d}]={\bar{q}}^{-1}[\bar{B}\mathrm{/}\bar{d}]\text{in }{\mathbb{F}}_p(t)\mathrm{/}{\mathrm{\Lambda }}_p\mathrm{.}$ The latter class cannot be zero. If $\bar{B}=\bar{d}L$ for $L\in {\mathrm{\Lambda }}_p$, the nonzero constant term of $\bar{d}$ and the polynomial nature of $\bar{B}$ force the lowest exponent of $L$ to be nonnegative: lowest exponents add and their coefficients do not cancel. Hence $L$ is a polynomial, contradicting $\mathrm{deg}\bar{B}<\mathrm{deg}\bar{d}$. Thus ${\rho }_{p,\mathrm{\Delta }}(y)\mathrm{\ne }0$, with the unit changes in $\mathrm{\Delta }$ cancelled between numerator and denominator. By step 1.2, ${\rho }_{p,\mathrm{\Delta }}(y)=t^m{\rho }_{p,\mathrm{\Delta }}(x)$; the unit $t^m$ preserves nonzero classes. This proves the nonzero reduction assertion for the explicit finite exceptional integer $D$. If $\mathrm{\Delta }$ is a Laurent unit, $T_{\mathrm{\Delta }}=0$ and the nonzero case is vacuous. The argument used only finitely many integer and polynomial operations, proving the full choice scope.