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Let $B$ be a commutative ring. Let $A$ be a subring of $B$. If $M$ and $N$ are $\mathbb{Z}$-submodules of $B$, we denote by $MN$ the submodule of $B$ generated by the subset $\{ab\mid a \in M, b\in N\}$. If $M$ and $N$ are $A$-submodules of $B$, $MN$ is clearly an $A$-submodule of $B$.

If $M$ and $N$ are $A$-submodules of $B$, we denote by $(N : M)$ the set $\{x \in B\mid xM \subset N\}$. $(N : M)$ is clearly an $A$-submodule of $B$.

Is the following proposition correct? If yes, how do we prove it?

Proposition. Let $M$ be an $A$-submodule of $B$. Suppose there exists an $A$-submodule $N$ of $B$ such that $MN = A$. Then $N = (A : M)$.

Makoto Kato
  • 42,602

1 Answers1

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Yes, this is correct, and easy to prove.

$MN=A$ implies $N \subseteq (A:M)$ and $(A:M) = (A:M)MN \subseteq AN \subseteq N$, hence $(A:M)=N$.

Martin Brandenburg
  • 163,620