ideal included in union of prime idealsIdealIncludedInUnionOfPrimeIdeals2007-04-01 18:36:102009-10-14 16:10:08Resultprime ideal% this is the default PlanetMath preamble. as your knowledge
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In the following $R$ is a commutative ring with unity.
\begin{proposition} 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 $I \not\subseteq \cup P_i$.\end{proposition}\begin{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 $i$.\newline
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.\end{proof}
\begin{corollary} 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 $i$.\end{corollary}