Let $\{\mathfrak{p}_i\}_{i\in I}$ ($I$ is an infinite set) be a family of minimal prime ideals in a commutative reduced ring $R$ with identity, and let $a, b \in R$. If the ideal $\langle a, b\rangle$ generated by $a$ and $b$ is contained in the union $\bigcup_{i\in I}\mathfrak{p}_i$, can we deduce that $\langle a, b\rangle \subseteq \mathfrak{p}_j $ for some $j\in I$?
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$\begingroup$ I previously made this incorrect answer, I move it to a comment for the record: Let $k$ be a field, $\{q_i : i \in I\}$ the (infinite) set of irreducible polynomials of the polynomial ring $k[a,b]$ contained in the ideal $(a,b)$. Let $R=k[a,b,z_i : i \in I]_{(a,b,z_i: i \in I)}/(q_{i}z_{i} : i \in I)$. Then $P_i := (q_i,z_j:j \neq i)$ are minimal primes of $R$ providing a counterexample. Sorry, this is wrong: $q_1 + z_{1}q_2 \notin \cup P_i.$ $\endgroup$– David LampertCommented Jul 11, 2016 at 12:11
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$\begingroup$ Context: the prime avoidance lemma en.wikipedia.org/wiki/Prime_avoidance_lemma says that an ideal contained in a finite union of prime ideals is contained in one of those prime ideals. In particular, since a noetherian commutative ring has finitely many minimal prime ideals, the answer is positive then, so possible counterexamples are to be looked for outside the noetherian world. $\endgroup$– YCorCommented Jul 11, 2016 at 12:29
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$\begingroup$ Another remark: while Boolean algebras are a source of counterexamples when we allow arbitrary ideals (in place of $\langle a,b\rangle$), they cannot provide counterexamples here because any finitely generated ideal in a Boolean algebra is generated by a single element, and the prime avoidance lemma is trivial in the case of such an ideal. $\endgroup$– YCorCommented Jul 11, 2016 at 12:57
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This is an interesting question. Quentel's Example provides an example. Let $R$ be a reduced ring, $Min(R)$ its space of minimal prime ideals, and $q(R)$ its classical quotient ring.
Theorem The following are equivalent.
$q(R)$ is von Neumann regular.
$R$ satisfies Property $A$ and $Min(R)$ is compact.
$Min (R)$ is compact and whenever a f.g. ideal is contained in the union of minimal prime ideals, then it is contained in one them.
Consequently, Quentel's Example is a reduced ring $R$ with $Min(R)$ compact but $q(R)=R$ not von Neumann regular. Therefore, it must possess a f.g. ideal which is contained in a union of minimal prime ideals but contained in any single one.
Glaz, Sarah Controlling the zero divisors of a commutative ring. Commutative ring theory and applications (Fez, 2001), 191–212, Lecture Notes in Pure and Appl. Math., 231, Dekker, New York, 2003.
J. A. Huckaba, Commutative Rings With Zero Divisors, Marcel Dekker, Inc. (1988)
