If an ideal contained in union of finite number of ideal then ideal is contained in some of them.

Question

Let $R$ be a ring which contains an infinite field as a subring and let $ I, J_1, J_2, ..... J_n$ are ideals of $ R$ such that $I \subseteq \cup J_i $ . Prove that $I\subseteq J_i $ for some $ i$ with $ 1 \leq j \leq n$ .

I have proved it when $ n=2$ . For $n\geq 3$ I wanted to use induction. But unable to do the problem.

Edit : Here $R$ is commutative but $ J_i$ need not be prime ideal

Answer 1

Let $k$ be the infinite field contained in $R$. Then $R$ is a $k$-vector space, and the ideals $I, J_1,J_2,\dots, J_n$ are subspaces of $R$. It is a consequence of this lemma for vector spaces:

Let $k$ be a field,$V$ a vector space which is the union of $n$ proper subspaces $W_1,W_2,\dots, W_n$. Then $\;|k|\le n-1$. (for a proof, you can look at my answer to this question

So, over an infinite field, a vector space cannot be the union of a finite number of proper subspaces.

Now, if $I$ is not contained in any $J_k$, this means that $\;I\cap J_k\varsubsetneq I$ for each $k$, so the vector space $I$ is the union of the $n$ proper subspaces $I\cap J_k$ $(k=1, \dots ,n)$n which contradicts the infiniteness of $k$.