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 $\mathfrak a \subseteq \displaystyle\bigcup_{i \mathop = 1}^n \mathfrak p_i$ 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.

Aiming for a contradiction, suppose suppose that $\mathfrak a \nsubseteq \mathfrak p_i$ for all $i \in \{1, \ldots, n\}$.

By the induction hypothesis, for all $i$ we have $\mathfrak a \nsubseteq \displaystyle\bigcup_{j \neq i} \mathfrak p_j$.

For all $i$, let $x_i \in \mathfrak a \setminus \displaystyle\bigcup_{j \neq i} \mathfrak p_j$.

If there exists $i$ such that $x_i \notin \mathfrak p_i$, then $x_i \in \mathfrak a \setminus \bigcup_{j} \mathfrak p_j$ and we conclude that $\mathfrak a \nsubseteq \displaystyle\bigcup_{j} \mathfrak p_j$, a contradiction.

Suppose that for all $i$, $x_i \in \mathfrak p_i$.

Let:

$y = \displaystyle \sum_{i \mathop = 1}^n \prod_{\substack{j \mathop = 1 \\ j \neq i}}^n x_j$

be defined as a summation of products.

Then for all $i$, $y \notin \mathfrak p_i$, because all terms except the $i$th are in $\mathfrak p_i$.

Thus $y \in \mathfrak a \setminus \bigcup_{j} \mathfrak p_j$.

This is 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 $1.11$