Let $R$ be a ring with $1$. If there exist disjoint comaximal ideals $I, J$, then for any $R$-module $M$, $M = IM \oplus JM$.

Question

This questions is from a past year paper.

Let $R$ be a ring with $1$. Suppose there exist distinct ideals $I, J$ of $R$ such that $R = I + J$ and $I \cap J = \{ 0 \}$.

(a) Show that there exists $a \in I$ and $b \in J$ such that $ax = x $ for all $x \in I$ and $by = y$ for all $y \in J$.

(b) Show that for any $R$-module $M$, $M = IM \oplus JM$.

For (a), $I$ and $J$ are comaximal, so there exists $a,b$, such that $1 = a + b$. Then for any $x \in I$, we have $x = ax + bx$. Now, $b$ is in $J$, and $x$ is in $I$. So $bx$ is in both $I$ and in $J$, and hence $bx=0$. Therefore, $ax=x$ for all $x \in I$. By symmetry, $bx=x$ for all $x \in J$.

For (b), I can show that $M = IM + JM$. So it remains to show that $IM \cap JM = \{ 0 \}$. So suppose $m \in IM \cap JM$. Then $m = \sum s_i n_i$ for $s_i \in I$ and $n_i \in M$. Then by (a), we have $am = \sum a s_i n_im = \sum s_i n_i = m$. Similarly, $bm = m$. But then I don't know what to do from here. How can I show $m=0$? Maybe $0 = m-m = am - bm = (a-b)m$, but I don't see how to carry on.

Answer 1

This answer would probably be better as a hint but I don't have enough reputation to comment.

So far you have elements $a \in I$ and $b \in J$ such that $am =m$ and $bm = m$. If we use the fact that $IJ \subseteq I\cap J = 0$ we get that $ab = 0$. Then, $0=(ab)m = a(bm)= am = m.$

Answer 2

Use part (a):

$m = \sum i_k n_k=\sum j_tn_t$ for $i_k \in I$, and $j_t\in J$ and $n_i \in M$.

Notice that (for your $a+b=1$) $am=m=bm$. But then $m=1\cdot m=(a+b)\cdot m=m+m$, but from $m+m=m$ you can conclude $m=0$.