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Let $R$ be a commutative ring, and let $M$ be a module with submodules $A,B \subset M$. Show that if $A \cap B=0$ and $A+B=M$, then there are $R$-module isomorphism $M/A \cong B$ and $M/B \cong A$.

Here is what I have:

Since $M=A+B$, then every $m$ in $M$ can be written as $a+b$ for some $a\in A$ and $b\in B$. Define

$$f:M \to B$$

$$m=a+b \to b$$

This is well defined: Assume that $a_1+b_1=a_2+b_2$, then $b_1=a_2+b_2-a_1$, and then we have $b_1=f(a_1+b_1)=f(a_1+(a_2+b_2-a_1))=f(a_2+b_2)=b_2$.

It is an $R$-module homomorphism:

  • $f((a_1+b_1)+(a_2+b_2))=f((a_1+a_2)+(b_1+b_2))=b_1+b_2=f(a_1+b_1)+f(a_2+b_2)$

  • Let $r\in R$, $f(r(a+b))=f(ra+rb)=rb=rf(a+b)$

$f$ is surjective, as $B\subset M$, so for all $b\in B$, we can find an a pre-image in $M$ (which is basically $b$ itself).

Since the kernal $K$ of $f$ is $A$, then by the first isomorphism theorem, we have $M/K=M/A\cong f(M)=B$. Is it correct? Thanks.

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  • $\begingroup$ Seems correct for me I would do the same $\endgroup$
    – Amer
    Commented Oct 19, 2022 at 23:27
  • $\begingroup$ Your proof of $a_1+b_1=a_2+b_2 \implies b_1=b_2$ is wrong, you used implicitly that $f$ is well-defined, which is what you want to prove. Note that you didn’t use the fact that $A \cap B=0$. $\endgroup$
    – azif00
    Commented Oct 19, 2022 at 23:35
  • $\begingroup$ @azif00 yeah I was worried that something is wrong as I didn’t use that $A \cap B=0$. For the well defined part, I was trying to prove that no element in $M$ has more than one $1$ image in $B$, so when $a_1+b_1=a_2+b_2$, we should get $f(a_1+b_1)=f(a_2+b_2)$. $\endgroup$
    – Dima
    Commented Oct 19, 2022 at 23:36
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    $\begingroup$ Hint: Assuming that $a_1+b_1=a_2+b_2$, in which submodule lives the element $a_1-a_2 = b_2-b_1$? $\endgroup$
    – azif00
    Commented Oct 19, 2022 at 23:38
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    $\begingroup$ The rest is perfectly fine :) $\endgroup$
    – azif00
    Commented Oct 20, 2022 at 0:23

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