Taken from: [1, Exercise 3.63].
Exercise 3.63
Let $R$ be a commutative ring which contains an infinite field $K$ as a subring. Let $I$ and $J_1, \ldots, J_n$, where $n \geq 2$, be ideals of $R$ such that
\[I \subseteq \bigcup_{i=1}^n J_i.\]Prove that $I \subseteq J_j$ for some $j$ with $1 \leq j \leq n$.
Proof
It is easy to verify that $ R $ is a $ K $-vector space with the same addition and multiplication, and in turn, each of $ I, J_1, \ldots, J_n $ is a subspace of $ R $.
Define $ U_i = I \cap J_i $ for each $ i \in {1, \ldots, n} $. Each $ U_i $ is a subspace of $ R $ since it is the intersection of subspaces; in particular, it is also a subspace of $ I $. We have
\[I = I \cap \bigcup_{i=1}^{n} J_i = \bigcup_{i=1}^{n} (I \cap J_i) = \bigcup_{i=1}^{n} U_i.\]If $ I = U_j = I \cap J_j $ for some $ j $, there is not much left to prove, as this directly implies that $ I \subseteq J_j $.
By contradiction, suppose that all $ U_i $ are proper subspaces of $ I $. Consider a nonzero element $ x \in U_1 $ and $ y \in I \setminus U_1 $. Notice that there are infinitely many elements of the form $ x + \alpha y $ with $ \alpha \in K $ nonzero. Moreover, observe that $ x + \alpha y $ never belongs to $ U_1 $. By the pigeonhole principle, there must exist some $ U_j $ that contains two elements $ x + \alpha_1 y $ and $ x + \alpha_2 y $ with $ \alpha_1 \neq \alpha_2 $.
It follows that
\[(x + \alpha_1 y) - (x + \alpha_2 y) = (\alpha_1 - \alpha_2)y \in U_j.\]Since $ \alpha_1 - \alpha_2 \neq 0 $, we conclude that $ y \in U_j $. Now, since
\[(x + \alpha_1 y) - \alpha_1 y = x \in U_j,\]and $ x $ was chosen arbitrarily, we deduce that
\[U_1 \subseteq \bigcup_{i=2}^{n} U_i.\]Thus, we obtain $ I = \bigcup_{i=2}^{n} U_i $. Repeating this argument iteratively, we get
\[I = \bigcup_{i=3}^{n} U_i, \quad I = \bigcup_{i=4}^{n} U_i, \quad \dots, \quad I = U_n,\]which is a clear contradiction1.
Therefore, $ I \subseteq J_j $ for some $ j $. $\blacksquare$
References
- Sharp, Rodney Y. (2001). Steps in Commutative Algebra (2nd ed.). Cambridge University Press. http://dx.doi.org/10.1017/CBO9780511623684
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In fact, we have proven that no vector space over an infinite field can be written as a finite union of proper subspaces. ↩