Prove $T_\mathfrak{p} (M)$ submodule of $M$

Question

Let $M$ be $R$-module and $\mathfrak{p}$ be maximal ideal of $R$.

Prove

$$T_\mathfrak{p} (M)=\{m\in M \mid\text{there exist }r\in R\text{ and }q\in \mathfrak{p} \text{ such that } r+q=1\text{ and }rm=0\}$$ is submodule of $M$.

This is my attempt.

Proof.

Let $a,b\in T_\mathfrak{p}(M)$, $r\in R$. We have

$$a\in M, \text{there exist }r_1\in R\text{ and }q_1\in \mathfrak{p} \text{ such that } r_1+q_1=1\text{ and }r_1a=0$$ and $$b\in M, \text{there exist }r_2\in R\text{ and }q_2\in \mathfrak{p} \text{ such that } r_2+q_2=1\text{ and }r_2b=0.$$

To prove $T_\mathfrak{p}(M)$ is a submodule of $M$, we must show

(1) $a-b\in T_\mathfrak{p}(M)$.

(2) $ra\in T_\mathfrak{p}(M)$.

To prove (1) we must show $$\text{there exist }r_3\in R\text{ and }q_3\in \mathfrak{p} \text{ such that } r_3+q_3=1\text{ and }r_3(a-b)=0.$$

Now I confused to prove $r_3+q_3=1\text{ and }r_3(a-b)=0.$ What the connection between $$r_1+q_1=1\text{ and }r_1a=0,$$ $$r_2+q_2=1\text{ and }r_2b=0$$ to obtain $$r_3+q_3=1\text{ and }r_3(a-b)=0?$$

To prove (2) we must show $ra\in T_\mathfrak{p}(M)$, i.e. $$\text{there exist }r_4\in R\text{ and }q_4\in \mathfrak{p} \text{ such that } r_4+q_4=1\text{ and }r_4(ra)=0.$$

I confused to prove this part.

What the connection between $$r_1+q_1=1\text{ and }r_1a=0$$ to obtain $$r_4+q_4=1\text{ and }r_4(ra)=0?$$

Answer 1

For (1), the idea should be that $(r_1+q_1)(r_2+q_2)=1$ so $$ r_1r_2+(\underbrace{r_1q_2+r_2q_1+q_1q_2}_{\in\mathfrak{p}})=1 $$ and $$ (r_1r_2)(a-b)=r_2(r_1a)-r_1(r_2b)=r_20-r_10=0 $$

For (2), just use the same $r_1$ and $q_1$, because $r_1(ra)=r(r_1a)=r0=0$.

Answer 2

Throughout the answer I suppose your ring $R$ is commutative:

$1)$ Define $r_3:=r_1r_2$ and then $$r_3(a-b)=r_1r_2(a-b)=r_1r_2a-r_1r_2b=r_2r_1a-r_1r_2b=r_2 \cdot 0 - r_1 \cdot 0=0$$

Now take $q_3:=r_1q_2+r_2q_1+q_1q_2$ (clearly $q_3 \in \mathfrak{p}$) and we have $$ r_3+q_3=r_1r_2+r_1q_2+r_2q_1+q_1q_2=(r_1+q_1)(r_2+q_2)=1 \cdot 1=1$$

$2)$ Now take $r_4=r_1$ and $q_4=q_1$; then $$r_4ra=rr_4a=rr_1a=r\cdot 0=0$$ and $$r_4+q_4=r_1+q_1=1$$