Intersection distributes over sum for monomial ideals.

Question

I want to show that for monomials ideals the intersection distributes over the sum in a basic expression like this: $$I \cap (J+K) = (I \cap J) + (I \cap K).$$

How can I prove this?

Answer 1

It is clear that $(I \cap J) + (I \cap K) \subseteq I \cap (J+K) $.

Then I express monomials ideals in the following manner:

$I = \langle \{x^{\alpha}:\alpha \in A\} \rangle$

Then one needs a couple of standard lemmas:

A polynomial is in a monomial ideal if and only if it is a K-combination of monomials in the ideal. A monomial is in the ideal if and only if it is divisible by some generator ($x^{\beta} = x^{\alpha + \gamma}, \alpha \in A, \gamma \in \mathbb{N}^n$

Therefore I can express $I = \{\sum k x^{\alpha + \gamma}:\gamma \in \mathbb{N}^n\}$.

Now I assume that I have one element of the left-hand side of the distributivity equation and I get:

$\sum kx^{i+ \gamma} = \sum kx^{j+\gamma}+\sum kx^{k+\gamma}$ where you can imagine that $x^i,x^j,x^k$ are the set of generators of each ideal.

Now, I take one $i$ and if there exists $j$ such that $i+\gamma \ge j$ I have that this monomial is in $J$ and clearly, it would be in the right-hand side. I do the same for each $k$. I therefore obtain the following expression:

$\sum_{i+\gamma \ge j} kx^{i+\gamma} + \sum_{i+\gamma \ge k} kx^{i+\gamma} + \sum_{i+\gamma < j,k} ?$

I claim that the third addend is empty. I think this is evident since in the previous equation I would get a term in the left-hand side that is not in the right-hand side.