# Intersection and Sum of Polynomial Ideals from different rings

It is well known that intersection and sum of polynomial ideals from the same ring are lattice operations. I wonder if this is still true for ideals from different rings (over the same field).

Specifically, let I be an ideal over the ring k[x,y], while J be an ideal over k[y,z]. (I have deliberately chosen the special case of three variables in order to keep notation as simple as possible). Then, the sum of the ideals is defined as:

$I + J = \{ f(x,y)+g(y,z) | f \in I \wedge g \in J \}$

The theorem that the basis of the sum is concatenation of the bases (which can be used as an alternative definition of the sum) generalizes verbatim.

Intersection of I and J is defined over the ring k[y]. First we eliminate variable x in I, and eliminate z in J, then intersect the elimination ideals over k[y].

Here are the lattice properties:

• $I + I = I$ (same as in classic case)
• $I + J = J + I$ (easy)
• $(I + J) + K = I + (J + K)$ (obvious in alternative definition)
• $I \cap I = I$ (same as in classic case)
• $I \cap J = J \cap I$ (easy, or redundant anyway)
• $(I \cap J) \cap K = I \cap (J \cap K)$ ?
• $I + (I \cap J) = I$ ??

Perhaps I can manage to prove associativity of ideal intersection if I establish simpler lemma that elimination of variables commutes with intersection. However, I'm entirely lost trying to prove absorption.

• Hint: How do you define intersections of ideals? – NasuSama Feb 11 '14 at 2:20
• Are you implying that $I \cap (I + J) = I$ easier to prove? ( containment $I \cap (I + J) \subset I$ is almost immediate). – Tegiri Nenashi Feb 11 '14 at 2:27
• Nope, I want you to look at the lattice properties you were unable to prove and determine how to approach such proofs. – NasuSama Feb 11 '14 at 2:28