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Assume we are given a set of ideals $I_1, \dots, I_s$ in a commutative polynomial ring. Let's define a subset indexed by $A\subseteq [s] = \{ 1,2,\dots, s\}$ as dependent if there exists an $a\in A$ such that $$I_a \subseteq \sum_{A \ni b\neq a} I_b.$$ A set is independent if it is not dependent. For example two ideals are dependent if and only if one is contained in the other. I would like to know if this notion of (in)dependence is meaningful and if there is literature about it.

For example, the ideals $(x), (y), (x+y)$ define the uniform matroid $U_{2,3}$ since any two of them are independent, but any of them is contained in the sum of the other two (which equals $(x,y)$). However, the resulting notion is not 'matroid' -- take any three ideals $I_1,I_2,I_3$ such that $I_1\subset I_2$, $I_1 \subset I_3$, but $I_2,I_3$ incomparable. Then $\{I_1\}$ and $\{I_2, I_3\}$ are maximal independent sets of different sizes.

This problem comes from an application in algebraic statistics. There each ideal comes from a conditional independence statement for certain random variables and a collection of statements corresponds to the sum of the ideals. One would like to say things like "a colletion of statements is (ir)redundant", but is there a well-studied notion?

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It seems like for every finite poset $P$, you could cook up a family of ideals with $P$ giving exactly the containment relations among your ideals and with no other dependencies. Is that right or am I misunderstanding your definition? Thanks. – Patricia Hersh Oct 16 '12 at 0:24
Yes that would be possible using ideals generated by variables, I think. It's probably unrelated, though. For simplicity we may assume that the original ideals are all pairwise containment incomparable. (that is not the case in the non-matroid example, but one could come up with another example). Then what is the poset you are thinking of? Is it the poset of all possible sums of those ideals? – Thomas Kahle Oct 16 '12 at 2:08
Indeed I was thinking of having a generator for each poset element and then associating to each poset element the ideal generated by its generator as well as those below it. I didn't think this was what you were actually trying to get at though and can see how the pair wise in comparability assumption might be handy. On the other hand, in the poset case I mention, your independent sets would be the poset anti chains, which certainly are something that some people study -- e.g. Trying to determine size of largest antichain. – Patricia Hersh Oct 16 '12 at 2:38
Silly comment: Prime Avoidance states that if the $I_i$s are prime ideals, then they form an independent set if and only if they are pairwise incomparable. I only mentioned it because Prime Avoidance is somewhat subtle and widely used in commutative algebra. – Hailong Dao Oct 16 '12 at 3:08
@Hailong: I think the sum which Thomas is using is not the same thing as "union". – Patricia Hersh Oct 16 '12 at 19:39

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