Ideal Contained in Finite Union of Prime Ideals
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Theorem
Let $A$ be a commutative ring with unity.
Let $\mathfrak p_1, \ldots, \mathfrak p_n$ be prime ideals.
Let $\mathfrak a \subseteq \ds \bigcup_{i \mathop = 1}^n \mathfrak p_i$ be an ideal contained in their union.
Then $\mathfrak a \subseteq \mathfrak p_i$ for some $i \in \{1, \ldots, n\}$.
Proof
The proof goes by induction on $n$.
For $n = 1$, the statement is trivial.
Let $n > 1$.
Suppose that the statement holds for $n - 1$ prime ideals.
It is to be shown that the statement holds for $n$ prime ideals.
Aiming for a contradiction, suppose suppose that $\mathfrak a \nsubseteq \mathfrak p_i$ for all $i \in \set {1, \ldots, n}$.
By the induction hypothesis, for all $i$ we have:
- $\mathfrak a \nsubseteq \ds \bigcup_{j \mathop \ne i} \mathfrak p_j$
For all $i$, let $x_i \in \mathfrak a \setminus \ds \bigcup_{j \mathop \ne i} \mathfrak p_j$.
Suppose there exists $i$ such that $x_i \notin \mathfrak p_i$.
Then:
- $x_i \in \mathfrak a \setminus \ds \bigcup_j \mathfrak p_j$
and we conclude that:
- $\mathfrak a \nsubseteq \ds \bigcup_j \mathfrak p_j$
which is a contradiction.
$\Box$
The other possibility is that for all $i$:
- $x_i \in \mathfrak p_i$
Let:
- $y = \ds \sum_{i \mathop = 1}^n \prod_{\substack{j \mathop = 1 \\ j \neq i}}^n x_j$
be defined as a summation of products.
All terms except the $i$th are in $\mathfrak p_i$.
Hence for all $i$:
- $y \notin \mathfrak p_i$
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Thus:
- $y \in \mathfrak a \setminus \bigcup_j \mathfrak p_j$
This also leads to a contradiction.
$\blacksquare$
Also see
Generalization
Sources
- 1969: M.F. Atiyah and I.G. MacDonald: Introduction to Commutative Algebra: Chapter $1$: Rings and Ideals: $\S$ Operations on Ideals: Proposition $1.11$