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We know that if $M$ is a left $R$-module, then $(\{\text{submodules of }M\},\subseteq)$ is a modular lattice.

Taking $M\!=\!R$, we deduce that $(\{\text{ideals of }R\},\subseteq)$ is a modular lattice.

However, for $R=K[x,y,z]$, the ideals $\big\{0,\:\langle x\rangle,\:\langle y\rangle,\:\langle y,z\rangle,\:R\big\}$ form the lattice $N_5$, which contradicts modularity. Where did I make a mistake?

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    What I meant was that in $I(R)$, we have $\langle x\rangle\vee\langle y,z\rangle=\langle x,y,z\rangle\neq R$, so the operation $\vee$ is not induced from $I(R)$. Thank you for your answer.2012-05-24

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If you want to find an embedding of $N_5$ into $I(R)$, it does not suffice to merely find five ideals of $R$ that are ordered as $N_5$. Rather, the embedding must preserve $\vee$ and $\wedge$, so the image must be closed under these operations. In your example, it is not the case that $\langle x\rangle\vee\langle y\rangle=\langle x\rangle\vee\langle y,z\rangle$, therefore the subset you proposed is not an image of $N_5$.