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 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:

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

be defined as a summation of products.

All terms except the $i$th are in $\mathfrak p_i$.

Since Sum of an Element in an Ideal and an Element not in the Ideal is not in the Ideal:

$\forall i:\quad y \notin \mathfrak p_i$

Thus:

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

This also leads to a contradiction.

$\blacksquare$


Also see

Generalization


This page or section has statements made on it that ought to be extracted and proved in a Theorem page.
In particular: Group Contained in Union of Two Groups
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Let $G,H,K$ be groups.

If $G \subseteq H \cup K$, then $G \subseteq H$ or $G \subseteq K$.

Proof

Aiming for a contradiction, suppose that neither $G \subseteq H$ nor $G \subseteq K$.

Then there exists $x \in G \setminus H$.

By assumption $G \subseteq H \cup K$, then $x \in H \cup K$, so $x \in K$.

Similarly, there exists $y \in G \setminus K$.

By assumption $G \subseteq H \cup K$, then $y \in H \cup K$, so $y \in H$.

Since $x,y \in G$, we have $xy \in G$, so $xy \in H \cup K$.

Without loss of generality $xy \in H$.

Since $xy \in H, y \in H$, we have $x = (xy)y^{-1} \in H$, which is a contradiction.

Sources

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