# ideal included in union of prime ideals

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 $s_{i}\not\in\bigcup_{j\neq i}P_{j}$. 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 $j\neq i$, the element $a_{i}\in P_{j}$. Consider the element $a=\sum a_{j}\in I$. Since $a_{i}=a-\sum_{i\neq j}a_{j}$ and $\sum_{i\neq j}a_{j}\;\in P_{i}$, it follows that $a\not\in P_{i}$, otherwise $a_{i}\in P_{i}$, contradiction   . The existence of the element $a$ proves the proposition  .∎

###### 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 $i$.

Title ideal included in union of prime ideals IdealIncludedInUnionOfPrimeIdeals 2013-03-22 16:53:14 2013-03-22 16:53:14 polarbear (3475) polarbear (3475) 10 polarbear (3475) Result msc 16D99 msc 13C99 prime avoidance lemma IdealsContainedInAUnionOfIdeals