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 $P_{\mathrm{1}}\mathrm{,}{P}_{\mathrm{2}}\mathrm{,}\mathrm{\dots }\mathrm{,}{P}_n$ be prime ideals of $R$. If $I\mathrm{⊈}{P}_i$, for all $i$, then $I\mathrm{⊈}\mathrm{\cup }{P}_i$.

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 $s_i\notin {\bigcup }_{j\ne i}{P}_j$. If for some $i$, $s_i\notin P_i$ then we are done. Thus, we may consider only the case $s_i\in P_i$, for all $i$.
Let $a_i=r_1\mathrm{\dots }{r}_{i-1}{r}_{i+1}\mathrm{\dots }{r}_n$. Since $P_i$ is prime then $a_i\notin P_i$, 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 $a\notin P_i$, 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 $P_{\mathrm{1}}\mathrm{,}{P}_{\mathrm{2}}\mathrm{,}\mathrm{\dots }\mathrm{,}{P}_n$ be prime ideals of $R$. If $I\mathrm{\subseteq }\mathrm{\cup }{P}_i$, then $I\mathrm{\subseteq }{P}_i$, for some $i$.