# Ideal Contained in Finite Union of Prime Ideals

This article needs to be linked to other articles.You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{MissingLinks}}` from the code. |

This article needs to be tidied.In particular: Tighten it up and complete the sloppy work.Please fix formatting and $\LaTeX$ errors and inconsistencies. It may also need to be brought up to our standard house style.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Tidy}}` from the code. |

## 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 $i$:

- $y \notin \mathfrak p_i$

This article, or a section of it, needs explaining.In particular: this needs to cite some results about summationsYou can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by explaining it.To discuss this page in more detail, feel free to use the talk page.When this work has been completed, you may remove this instance of `{{Explain}}` from the code. |

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 $1.11$