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If $M$ is an $R$-module, and $I$ is an ideal of $R$, then what is $IM$? That is, does it consist of all products of some form or finite sums of some specific form?

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up vote 4 down vote accepted

$IM$ is the submodule of $M$ generated by all elements of the form $am$ with $a\in I$, $m\in M$. It is equal to the set of all finite sums of the form $$a_1m_1 + \cdots + a_km_k,\qquad a_i\in I,\ m_j\in M.$$

This is a submodule of $M$: it is an additive subgroup, since it contains $0$ (as $0m$ for any $m\in M$); and if $a_1m_1+\cdots + a_km_k$ and $b_1n_1+\cdots+b_{\ell}m_{\ell}\in IM$, with $a_i,b_j\in I$, then $a_i,-b_j\in I$ so $$(a_1m_1+\cdots+a_km_k)-(b_1n_1+\cdots+b_{\ell}m_{\ell}) \in IM.$$

And it is closed under $R$-multiplication, because $I$ is a (left) ideal: if $r\in R$, and $a_1m_1+\cdots+a_km_k\in IM$, then $ra_i\in I$ for all $i$, so $$r(a_1m_1+\cdots+a_km_k) = (ra_1)m_1 + \cdots + (ra_k)m_k\in IM.$$

For example, if $A$ is a $\mathbb{Z}$-module (an abelian group), and $I=n\mathbb{Z}$, then $IA = nA = \{na\mid a\in A\}$.

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Thanks! It's quite helpful to see this explanation explicitly. – Bongle Dec 10 '11 at 22:06

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