A problem in prime ideals in Noetherian Rings

Question

I'm trying to solve the following problem

Let $A$ be a commutative noetherian ring with identity $1 \neq 0$. Write $X (\mathfrak{a})$ for the set of prime ideals of $A$ containing a given ideal $\mathfrak{a}$. Suppose that $X(0) = X(\mathfrak{a}) \cup X(\mathfrak{b})$ and $X(\mathfrak{a}) \cap X(\mathfrak{b}) = \emptyset$ for two ideals $\mathfrak{a}$ and $\mathfrak{b}$. Prove the following facts

(a) $A = \mathfrak{a} + \mathfrak{b}$.

Clearly since $\mathfrak{a}, \mathfrak{b} \subset A$, we have $\mathfrak{a} + \mathfrak{b} \subset A$. However, the other direction is not clear

(b) $\mathfrak{a}\cap \mathfrak{b} =\mathfrak{a}\mathfrak{b}$

We have one direction $\mathfrak{a}\cap \mathfrak{b} \subset \mathfrak{a}\mathfrak{b}$, but the other direction is not clear

**I feel like in both cases Notherian and $X$ properties should be used somehow but I'm not sure how. **

Answer 1

Noetherianity and the assumption that $X(0) = X(\mathfrak{a}) \cup X(\mathfrak{b})$ are irrelevant. Both $(a)$ and $(b)$ follows from just the assumption that $X(\mathfrak{a}) \cap X(\mathfrak{b}) = \emptyset$. Also, for future reference, the notation $V(-)$ is much more standard than $X(-)$.

For $(a)$:

Check from the definition of $X$ that $X(\mathfrak{a} + \mathfrak{b}) \subseteq X(\mathfrak{a}) \cap X(\mathfrak{b})$ for any ideals $\mathfrak{a}, \mathfrak{b}$. By assumption, then $X(\mathfrak{a} + \mathfrak{b}) = \emptyset$. Thus $\mathfrak{a} + \mathfrak{b}$ is an ideal that is not contained in any prime. It is a very fundamental fact in commutative algebra that an ideal in a commutative ring does not contain the identity element if and only if it is contained in some prime ideal. This is a consequence of Zorn's lemma. See, e.g., here.

Thus for any ideal $\mathfrak{c}$, $X(\mathfrak{c}) = \emptyset$ iff $\mathfrak{c} = A$.

For $(b)$:

Note that for any ideals $\mathfrak{a}, \mathfrak{b}$, you have $(\mathfrak{a} + \mathfrak{b}) (\mathfrak{a} \cap \mathfrak{b}) \subseteq \mathfrak{a}\mathfrak{b}$. Therefore when $\mathfrak{a} + \mathfrak{b} = A$, as in part (a) you get $\mathfrak{a} \cap \mathfrak{b} \subseteq \mathfrak{a}\mathfrak{b}$.