Ideal Contained in Finite Union of Prime Ideals

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 $i$:

$y \notin \mathfrak p_i$

Thus:

$y \in \mathfrak a \setminus \bigcup_j \mathfrak p_j$

This also leads to a contradiction.

$\blacksquare$

Also see

Generalization

• Subgroup Contained in Union of Two Subgroups

Sources

• 1969: M.F. Atiyah and I.G. MacDonald: Introduction to Commutative Algebra: Chapter $1$: Rings and Ideals: $\S$ Operations on Ideals: Proposition $1.11$