Counterexample to noncommutative prime avoidance theorem

Question

For a commutative ring $R$, if $I,P_1,\ldots,P_n$ are ideals with the $P_i$ prime and $I\subseteq \bigcup P_i$, then there is an index $j$ such that $I\subseteq P_j$ (the prime avoidance theorem). Now, the Atiyah-MacDonald proof of this fact is not directly transferable to the noncommutative setting (since to use that $P$ is prime we have to check any condition equivalent to $aRb\subseteq P$, which includes an undesirable degree of freedom).

I suspect that the result is false for noncommutative rings. Can you help me find a counterexample (or prove the result)? Note that a counterexample cannot be an algebra over an infinite field $K$, since a vector space over $K$ cannot be the union of a finite number of proper subspaces.

Answer 1

This has been investigated in the article The Prime Avoidance Lemma Revisited , where it is proved for non-commutative rings.