I'm trying to get more of intuition for the cloud of ideas surrounding the abstract Cayley-Hamilton theorem, Nakayama's lemma, etc., so I'd like to see some concrete examples. The problem is that the main concrete applications I'm aware of are just showing that something is zero, so they don't provide great examples to build intuition.
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1The intuition is that isf something is very, very small it is zero. And smallness is measured in terms of the Jacobson radical. – Mariano Suárez-Álvarez Feb 16 '14 at 03:37
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Let $R=\mathbb{Z}$, let $I\subseteq R$ be any non-zero ideal, and let $M=\mathbb{Q}$. Then clearly $IM=M$.
More generally, if $R$ is an integral domain, $I\subseteq R$ is any non-zero ideal, and $M=\mathrm{Frac}(R)$, then $IM=M$.
A general class of examples: let $R$ be any ring, and let $f\in R$ be a non-zero-divisor, non-unit element. Then letting $I=(f)$ and $M=R_f$, the localization of $R$ at $f$, we have $IM=M$.
Zev Chonoles
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In general, for any $R$-module $M$ and $R$-ideal $I$, $M/IM \cong M \otimes_R R/I$, so asking when $M = IM$ is the same as asking when the tensor product $M \otimes_R R/I$ vanishes. If there are no conditions on $I$, then this can happen quite easily: if $M$ is finite, then $M \otimes_R R/I = 0$ iff $I$ and $\text{ann}(M)$ are comaximal, i.e. $I + \text{ann}(M) = R$. In particular, if $I$ and $J$ are any two comaximal ideals, then for $M = R/J$, $M = IM$.
As an aside, the previous statement actually implies Nakayama's lemma (although Nakayama is commonly used to prove it): if $I$ is contained in the Jacobson radical of $R$, then it cannot be comaximal to any proper ideal, so $M = IM \implies \text{ann}(M) = R$, i.e. $M = 0$. Nakayama's lemma can be viewed as recovering the intuition that for vector spaces, $V \otimes_k W = 0$, $W \ne 0 \implies V = 0$.
Another case in which $M \otimes_R R/I = 0$ is when $I$ acts as units on $M$ (the previous answer gave examples of this). But there are other cases: over $\mathbb{Z}$, if $A$ is divisible and $B$ is torsion, then $A \otimes_{\mathbb{Z}} B = 0$. As an interesting case, if $M = \mathbb{Q}/\mathbb{Z}$, $I = n\mathbb{Z} \ne 0$, then $M = IM$, even though $I$ annihilates nonzero elements of $M$.
zcn
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