1.3 Prime Avoidance and Variants
Before we get too far, we present the prime avoidance lemma, which will be used repeatedly during the rest of the blueprint.
(See BH 1.2.2) Let \(R\) be a ring, and \(\mathfrak {p}_{1}, \ldots , \mathfrak {p}_{m}\) be prime ideals. Let \(M\) be an \(R\)-module, and let \(x_{1}, \ldots , x_{n} \in M\). Let \(N\) be the submodule generated by \(x_{1}, \ldots , x_{n}\).
If \(N_{\mathfrak {p}_{j}} \not\subset \mathfrak {p}_{j}M_{\mathfrak {p}_{j}}\) for all \(j \in \{ 1, \ldots , m\} \), then there exist \(a_{2}, \ldots , a_{n} \in R\) such that \(x_{1} + \sum _{i=2}^{n} a_{i}x_{i} \not\in \mathfrak {p}_{j} M_{\mathfrak {p}_{j}} \) for all \(j \in \{ 1, \ldots , m\} \).
See BH 1.2.2
Let \(R\) be a ring, and \(\mathfrak {p}_{1}, \ldots , \mathfrak {p}_{m}\) be prime ideals. Let \(I\) be a finitely generated ideal of \(R\). Then if \(I \not\subset \mathfrak {p}_{j}\) for all \(j \in \{ 1, \ldots , m\} \), then there exists an \(x \in I\) such that \(x \not\in \mathfrak {p}_{j}\) for all \(j\).
Take \(M = R\) in the previous theorem. Using the correspondence theorem for localizations, the condition \(I_{\mathfrak {p}_{j}} \not\subset \mathfrak {p}_{j}R_{\mathfrak {p}_{j}}\) is the same as \(I \not\subset \mathfrak {p}_{j}\).
We also have a few variants (see wikipedia page for "Prime Avoidance")
Let \(E\) be an additive subgroup of \(R\) which is multiplicatively closed. Let \(I_{1}, \ldots , I_{n}\) be ideals such that \(I_{3}, \ldots , I_{n}\) are prime. Then if \(E\) is not contained in any one of the \(I_{j}\), then \(E\) is not contained in their union.
Let \(R\) be a ring and \(\mathfrak {p}_{1}, \ldots , \mathfrak {p}_{m}\) be prime ideals. Let \(x\) be some element of \(R\) and let \(J\) be an ideal. Define \(I := (x) + J\). If \(I \not\subset \mathfrak {p}_{j}\) for all \(j\), then there exists some \(y \in J\) such that \(x + y \not\in \mathfrak {p}_{j}\) for all \(j\)
Let \(\mathfrak {p}_{1}, \ldots , \mathfrak {p}_{n}\) be prime ideals. Let \(I\) be an ideal contained in in their union. Then \(I \subset \mathfrak {p}_{i}\) for some \(i\).
This is a weaker statement then both of the two above it.
Let \(I_{1}, \ldots , I_{n}\) be ideals, let \(\mathfrak {p}\) be a prime ideal containing the intersection of them all. Then \(I_{i} \subset \mathfrak {p}\) for some \(i\). If \(\mathfrak {p}\) is exactly the intersection, then \(I_{i} = \mathfrak {p}\) for some \(i\).