0

Let $A$ and $B$ be arbitrary subsets of a ring. Then $V(A\cup B)=V(A)\cap V(B)$.

Here, $V(X)$ is the set of prime ideals containing $X$.

Let $W(X)$ be the set of ideals (any sort of ideals) containing the set $X$. Is $$W(A\cup B)=W(A)\cap W(B)$$ true? I think it is, but I'm not sure.

freebird
  • 69
  • 4

1 Answers1

0

Yes, $W(A \cup B) = W(A) \cap W(B)$:

Suppose $I \in W(A \cup B)$. Then $I$ is an ideal, and $I$ contains $A \cup B$. In particular $I$ is an ideal containing $A$ and $B$, i.e. $I \in W(A)$ and $I \in W(B)$. Thus $I \in W(A) \cap W(B)$.

Suppose $I \in W(A) \cap W(B)$. Then $I$ is an ideal, $I$ contains $A$ and $I$ contains $B$. So $I$ is an ideal containing $A \cup B$. Thus $I \in W(A \cup B)$.

So $W(A \cup B) = W(A) \cap W(B)$.

Matt Gallagher
  • 471