Clarification on notation in Siegfried Bosch's Commutative Algebra book about primary decomposition of ideals.

123 Views Asked by At

I'm reading through Siegfried Bosch's Commutative Algebra book, and I'm confused on his notation in one his proofs. He uses this notation a lot, so I think I should I understand it. The notation first appears in his proof of Theorem 8:

Theorem 8. For a decomposable ideal $\mathfrak{a} \subset R$, let $\mathfrak{a} = \bigcap_{i=1}^r \mathfrak{q_i}$ be a minimal primary decomposition. Then the set of corresponding prime ideals $\mathfrak{p}_i = \text{rad}(\mathfrak{q}_i)$ coincides with the set of all prime ideals in $R$ that are of type $\text{rad}\big((\mathfrak{q}:x)\big)$ for $x$ varying over $R$.

In his proof, he writes: Let $x\in R$. The primary decomposition of $\mathfrak{a}$ yields $(\mathfrak{a}:x) = \bigcap_{i=1}^r(\mathfrak{q_i}:x)$ and therefore by Lemma 7, a primary decomposition

$ \text{rad}\big((\mathfrak{a}:x)\big) = \bigcap_{i=1}^r\text{rad}\big((\mathfrak{q}_i:x)\big) = \bigcap_{x\not\in \mathfrak{q}_i}\mathfrak{p}_i. $

What I'm confused about is that last equality, what's that mean? The indexing confuses me. If it's helpful, here's the part of Lemma 7 I think he's using.

Lemma 7. Consider a primary ideal $\mathfrak{q}\subset R$ and the corresponding prime ideal $\mathfrak{p} = \text{rad}(\mathfrak{q})$. Then, for $x\not\in \mathfrak{q}$, $(\mathfrak{q}:x)$ is $\mathfrak{p}$-primary.

Thank you.