Compute $T(M)$, where $M=\mathbb{Z}[x]/(x^2-9)$

Question

Given an $R$-module $M$, let

$$T(M):=\{ x\in M \ | \ rx=0, \ \text{for some} \ r\ne 0 \ \text{in} \ R\}$$.

Let $R=M=\mathbb{Z}[x]/(x^2-9)$. Compute $T(M)$.

Here is what I am thinking, let $x=f(x)+(x^2-9)\in M$, and let $r=g(x)+(x^2-9)\in R-\{0\}$, since $r\ne 0$, then $g(x)$ is not in $(x^2-9)$. We want to find $f(x)$ such that

$0=rx=f(x)g(x)+(x^2-9)$, this happen if $f(x)g(x)\in(x^2-9)$, then $(x^2-9)$ divides $f(x)g(x)$, however, I am not sure how to complete.

Answer 1

You are looking for zero divisors in $R$. (It's a bit weird to phrase this in terms of $T(M)$, which usually refers to torsion submodules for modules over domains.)

Setting $0$ aside, the definitions amount to asking when $f,g\in \mathbb Z[x]$ happen to satisfy $fg\in (x^2-9)$. Since $\mathbb Z[x]$ is a UFD, things are pretty simple. You have the factorization $(x-3)(x+3)$ to use with the relation $x^2-9|fg$. Now $x-3$ is irreducible/prime, so it has to divide one of $f$ or $g$, and something similar can be said for $x+3$. If they both happen to divide $f$, $f$ is already zero. So the interesting case is when a single one of these divides each of $f$ and $g$.

With that in mind, $T(M)=(x-3)R\bigcup (x+3)R$. Notice how this is not a submodule at all (as the notation $T(M)$ is usually used.)

Answer 2

u are on the right track. choose representatives of $f$ and $r$,which I still call them $r$ and $f$,with $r\notin <x^{2}-9>$,now $rf=0$,in the quotient means $x^{2}-9=(x-3)(x+3)\lvert rf$,now we know that $(x-3);(x+3) $ cannot both divide $r$,So at least one of them divides $f$ (because $x-3,x+3$ are primes in the ring $\mathbb{Z}[X]$ ,So $T(M)\subset \pi<x-3> \cup <x+3>$,the revers inclusion is easier .we conclude that $T(M)=\pi(<x-3> \cup <x+3>)$,where $\pi:\mathbb{Z}[X]\to \mathbb{Z}[X]/(x^{2}-9)$ is the natural surjection. just a reamrk;your ring $\mathbb{Z}[X]/(x^{2}-9)$ isn't an integral domain, so the torsion submodule is defined with the extra condition that $r$ must be regular,but u did'nt mention this in your post.over integral domains $T(M)$ is a submodule but this is not the case for arbitrary (commutative)rings.