# Diamonds of ideals, part 3

I'd like to wrap up the line of questioning started first in this question and then continued in this question.

The only variant left to try is:

"How close can you get to the Diamond lattice with two-sided ideals of a ring?"

Naturally, the commutative example in the first post is an example with six ideals, the Diamond with one ideal on top.

I'm putting (what I think is) the solution below for review. If all is well then it contains an alternate proof of why the Diamond can never appear in a lattice of ideals with $R$ at the top, even for noncommutative rings. (The previous proof factored $R$ into local rings.)

This brings the line of questioning to closure to me, but maybe someone else has a good variant too!

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## 1 Answer

Suppose $R$ has three maximal two-sided ideals $M_1, M_2, M_3$, and that their pairwise intersection is an ideal $K$ so that $R, M_1, M_2, M_3, J$ forms the Diamond within the lattice of ideals of $R$.

Now $R/K$ obviously has three distinct maximal two-sided ideals which are the images of the three maximal ideals in $R/K$. But on the other hand, the Chinese Remainder Theorem says that $R/K\cong R/M_1\oplus R/M_2$, which is a product of two simple rings. However it is clear that the product of two simple rings has exactly two maximal ideals. Contradiction!

Reading back through this, I thought this was a little strange. The intersection of two different maximal ideals is never contained by a third distinct maximal ideal? Did I do something silly?

Edit: I suppose it also means that given $n$ distinct maximal ideals, $\cap_{i=1}^{n-1}P_i\not\subseteq P_n$. This is sounding a little more believable now, since if it were a containment, then $\prod_{i=1}^{n-1} P_i\subseteq P_n$, whence by primeness of $P_n$, one of the $P_i\subseteq P_n$, an absurdity.

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@JackSchmidt Not sure if you noticed, so I'm paging in case you're interested. –  rschwieb Jun 12 '12 at 0:37