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Recall that a right ideal $P$ of a prime Goldie ring $R$ is called completely prime right ideal if $aP\subseteq P$ and $ab\in P$ implies that $a\in P$ or $b\in P$ for all $a, b\in R$. Now let $P_1$, $P_2$ and $P_3$ be completely prime right ideals which are not comparable, Is $P_1\cap P_2\cap P_3$ proper subset of $P_1\cap P_2$ ?

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For a counterexample, consider the ring $R = \mathbb{M}_2(F)$ for a field $F$. View this as a ring of linear transformations acting from the right on the space $F^2$ of row vectors.

Because $R$ is an artinian ring, then it follows from Corollaries 2.7 and 2.10 of this paper that every completely prime right ideal of $R$ is maximal. Any maximal right ideal of $R$ is of the form $eR$ where $e$ is an idempotent element that projects onto a 1-dimensional subspace of $F^2$.

Let $L_1, L_2, L_3$ respectively denote the lines in $F^2$ spanned by $(1,0)$, $(0,1)$, and $(1,1)$. For $i=1,2,3$, let $e_i \in R$ denote the projection with kernel $L_i$ and image $L_{i+1}$, indices taken mod 3. Consider the distinct (hence incomparable) maximal right ideals $P_i = e_i R$. Then for $i \neq j$, any element of $P_i \cap P_j = e_i R \cap e_j R$ has kernel containing $L_i + L_j = F^2$. Thus $P_i \cap P_j = 0$ for any $i \neq j$. In particular $P_1 \cap P_2 \cap P_3 = P_1 \cap P_2$.

(This example can be generalized to $\mathbb{M}_n(F)$ for any $n \geq 2$, by fixing an idempotent $f \in \mathbb{M}_n(F)$ with rank $n - 2$ and constructing three idempotents $e_1,e_2,e_3$ with rank $n-1$ such that $e_iR \cap e_jR = fR$ for any $i \neq j$.)

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