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I am self studying ring theory and modules from Rotman's Advanced Modern Algebra.
I would like some help on putting this thought to bed.

Let $A$ and $B$ be rings. Let $R=A\times B$. Is it possible for $R$-submodules of $R$ to be $A$-submodules of $A$ as well as $B$-submodules of $B$? If yes, I would like to see a proof or a guide to a proof. If not, I'd like to know why?
Thanks.

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    Oh. right...Thanks Rasmus2011-05-03

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An $R$-submodule of $R$ is exactly an ideal of $R$. So you ask about the ideals of $A \times B$.

If $I$ is an ideal of $A$ and $J$ is an ideal of $B$, then it is easy to see that $I \times J$ is an ideal of $A \times B$. Conversely, if $K$ is an ideal of $A \times B$, consider the projections to $A$ resp. $B$ to get ideals $I$ of $A$ resp. $J$ von $B$. Now show $K = I \times J$ (use the idempotents $(1,0)$ and $(0,1)$ of $A \times B$).

Thus there is a bijection between the ideals of $A \times B$ and pairs of ideals in $A$ resp. $B$.

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    Quick question. So ideals of $R$ are of the form $I \times J$. How then do we conclude that $R$-submodule of $R$ is exactly the same as an $A$-submodule of $A$. Thanks2011-05-03