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[parent] ideal included in union of prime ideals (Result)

In the following $ R$ is a commutative ring with unity.

Proposition 1   Let $ I$ be an ideal of the ring $ R$ and $ P_1, P_2, ... ,P_n$ be prime ideals of $ R$. If $ I\not\subseteq P_i$, for all $ i$, then $ I \not\subseteq \cup P_i$.
Proof. We will prove by induction on $ n$. For $ n=1$ the proof is trivial. Assume now that the result is true for $ n-1$. That implies the existence, for each $ i$, of an element $ s_i$ such that $ s_i\in I$ and $ s_i \not\in \bigcup_{j\ne i}P_j$. If for some $ i$, $ s_i\not\in P_i$ then we are done. Thus, we may consider only the case $ s_i \in P_i$, for all $ i$.
Let $ a_i = r_1...r_{i-1}r_{i+1}...r_n$. Since $ P_i$ is prime then $ a_i\not\in P_i$, for all $ i$. Moreover, for $ j \ne i$, the element $ a_i \in P_j$. Consider the element $ a = \sum a_j \in I$. Since $ a_i = a - \sum_{i\ne j} a_j$ and $ \sum_{i\ne j} a_j\;\in P_i$, it follows that $ a \not\in P_i$, otherwise $ a_i\in P_i$, contradiction. The existence of the element $ a$ proves the proposition. $ \qedsymbol$
Corollary 1   Let $ I$ be an ideal of the ring $ R$ and $ P_1,P_2,...,P_n$ be prime ideals of $ R$. If $ I \subseteq \cup P_i$, then $ I\subseteq P_i$, for some $ i$.



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Cross-references: proposition, contradiction, prime, implies, induction, prime ideals, ring, ideal, unity, commutative ring

This is version 4 of ideal included in union of prime ideals, born on 2007-04-01, modified 2007-04-02.
Object id is 9142, canonical name is IdealIncludedInUnionOfPrimeIdeals.
Accessed 454 times total.

Classification:
AMS MSC: 13C99 (Commutative rings and algebras :: Theory of modules and ideals :: Miscellaneous)
 16D99 (Associative rings and algebras :: Modules, bimodules and ideals :: Miscellaneous)
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