Direct sum of p-primary components of an R-module counterexample?

[kalish1]

Let $x \in R - \{0\},$ where $R$ is a domain.

Define $T_x(M) = \{m \in M \ | \ x^n m=0 \ \ \mathrm{for \ some} \ n \in \mathbb{N}\}$ as the $x$-torsion of $M.$

I know that $T_x(M \oplus N) = T_x(M) \oplus T_x(N)$ for $R$-modules $M,N$ only if $R$ is a PID.

But I can't think of a counterexample for $R$ an integral domain.

Any ideas?

 

[kalish1]

Let $x \in R - \{0\},$ where $R$ is a domain.

Define $T_x(M) = \{m \in M \ | \ x^n m=0 \ \ \mathrm{for \ some} \ n \in \mathbb{N}\}$ as the $x$-torsion of $M.$

I know that $T_x(M \oplus N) = T_x(M) \oplus T_x(N)$ for $R$-modules $M,N$ only if $R$ is a PID.

But I can't think of a counterexample for $R$ an integral domain.

Any ideas?

Here is the modified problem:

Let $x \in R - \{0\},$ where $R$ is a domain.

Define $T_x(M) = \{m \in M \ | \ x^n m=0 \ \ \mathrm{for \ some} \ n \in \mathbb{N}\}$ as the $x$-torsion of $M.$

I need to show that $T_x(M \oplus N) = T_x(M) \oplus T_x(N)$ for $R$-modules $M,N$ or show that there is a counterexample.

Why is this proof wrong?

Take $(a,b)\in T_x(M\oplus N)$. Then there exist $m,n \in \mathbb{N},$ not necessarily equal, such that $x^m a + x^n b=0.$ Take $(a,b)\in T_x(M) \oplus T_x(N)$. Then there exist $m,n \in \mathbb{N}$ such that $x^m a=0$ and $x^n b=0.$ So if $(a,b)\in T_x(M) \oplus T_x(N),$ then $(a,b)\in T_x(M\oplus N).$ But the converse is not true. Thus $T(M \oplus N) \neq T(M) \oplus T(N).$