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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.

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

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  • $\begingroup$ Now I see! The author does not elaborate much the fundamental point, because it is easy enough: If we pick $x\in P_2\setminus P_1$ and $y\in P_3\setminus P_1$, then $xRy\subseteq P_2P_3$ and cannot be contained in $P_1$, for otherwise $x$ or $y$ is in $P_1$. Therefore there is an element $z\in P_2P_3\setminus P_1$, and this is enough (pick $w\in P_1\setminus(P_2\cup P_3)$ and consider $w+z$). $\endgroup$
    – Jose Brox
    Commented Aug 29, 2017 at 18:23

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