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I have a ring $R$, an ideal $I$ of $R$ and an $R$-module $M$. How does one define $IM$. Is it the set $\{im\ \vert\ i\in I, m\in M\}$ or is it the smallest module containing these elements?

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  • $\begingroup$ According to this website (first page), it is indeed the latter one (the module generated by these elements). $\endgroup$ – Kenny Lau Dec 24 '17 at 13:52
  • $\begingroup$ Is multiplication in $R$ commutative? If not then you need to be more careful about how things are defined (left/right/two-sided stuff). You want a submodule, and the mere set of products $im$ is usually not a submodule, but the set of finite sums $i_1m_1+\ldots+i_km_k$ is a submodule of $M$ if $R$ is commutative, so that set is defined to be $IM$. Note: if $I = (x)$ is a principal ideal then $IM = xM = \{xm : m \in M\}$. $\endgroup$ – KCd Dec 24 '17 at 14:08
  • $\begingroup$ Note the set $\{im \ | \ i \in I, m \in M\}$ is usually not closed under addition and hence is not a module. For example if you look at the polynomial ring $P = \mathbb R[w, x, y, z]$ and $I = (w, x)$ and $M = (y, z)$ (an ideal treated as a submodule of the free module $P$) then $wy$ and $xz$ are of the required form but $wy + xz$ is an irreducible polynomial so it can only be written as a product if one of the two elements is a unit, and $I$ and $M$ don't contain units so $wy + xz$ is not an element of the form $im$. $\endgroup$ – Jim Dec 28 '17 at 15:34
  • $\begingroup$ So the correct definition of $IM$ is the one given by Narcissus jewel in their answer. $\endgroup$ – Jim Dec 28 '17 at 15:36
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Assuming that $R$ is commutative, then we can proceed as in A&M page 19.

Let $I\subset R$ be an ideal, and let $M$ be an $R$-module. Then, $IM$ is the set of all finite sums $\sum r_ix_i$ where $r_i\in I$ and $x_i\in M$. This is of course a submodule of $M$.

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