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A lemma in my lecture notes claims, as in the title, given a surjective ring homoromorphism from R to S, ideals in S are the R submodules of S (in the first paragraph - picture attached), but I can't see why this is true. Note that we are assuming all rings are commutative. .lecture notes

Any help very much appreciated

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lkjhgfdsa
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    An ideal of $S$ is an $S$-module, hence becomes an $R$-module via the action $r\cdot s=\phi(r)s$. Conversely, if $I\subset S$ is an $R$-module, since $\phi$ is surjective we have $sI\subset I$ for all $s\in S$. – carmichael561 Mar 31 '16 at 16:22
  • Thank you for the help. I now understand how an Ideal of S is an R submodule of S, but I don't follow the other direction - would you mind elaborating :) – lkjhgfdsa Mar 31 '16 at 16:46
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    An $R$-submodule $I$ of $S$ is an abelian group such that $r\cdot I\subset I$ for all $r\in R$. Since $\phi$ is surjective, it follows that $sI\subset I$ for all $s\in S$, so $I$ is an ideal of $S$. – carmichael561 Mar 31 '16 at 16:48

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