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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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Perhaps this is not an embedding of $N_5$ into $I(R)$. For instance, is the join of $\langle x\rangle$ and $\langle y,z\rangle$ really $R$? –  Egbert May 24 '12 at 10:43
    
Ah, of course, $N_5$ must be a sublattice, not just a sub-poset. In our case, $\langle x\rangle\vee\langle y,z\rangle=\langle x,y,z\rangle\neq R$. Thanks. If you write it down, I'll accept. –  Leon May 24 '12 at 10:46
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my question should have read: is the $\langle x\rangle\vee\langle y\rangle$ really $\langle x\rangle\vee\langle y,z\rangle$? That it is not $R$ does not matter so much, because you could have just taken the smaller ideal instead. –  Egbert May 24 '12 at 10:53
    
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. –  Leon May 24 '12 at 11:08

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up vote 5 down vote accepted

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 $N_5$.

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