Let $A$ be a commutative unitary ring, $P_1,\dots , P_r$ prime ideals, $I$ an ideal, and $x$ an element of $A$. If $xA+I\not \subset P_1\cup \cdots \cup P_r$, then there is $y\in I$ such that $x+y\not \in P_1\cup \cdots \cup P_r$.

I have tried several methods but nothing works. For example, I tried to apply the following theorem:

Let $\mathfrak{a}$ be an ideal, $P_1,\dots ,P_r$ prime ideals. If $\mathfrak{a}\subset P_1\cup \cdots \cup P_r$, there exists $1\leq k \leq r$ such that $\mathfrak{a}\subset P_k$.

And of course, I tried to show contraposition but failed. If you know how to solve, please help me!!


1 Answer 1


Matsumura attributes this to E. Davis, and the only paper of Davis in Matsumura's References section is "Ideals of the principal class, $R$-sequences and a certain monoidal transformation" (link). The following argument is from the proof of the Lemma in the Appendix.

We may assume that there are no containments between the $P_{i}$, i.e. $P_{i_{1}} \not\subseteq P_{i_{2}}$ if $i_{1} \ne i_{2}$.

By assumption, there exist $a \in A$ and $u \in I$ such that $xa+u \not\in P_{i}$ for all $i$. This implies that if $x \in P_{i}$ then $u \not\in P_{i}$. Set $S := \{i \in \{1,\dotsc,r\} \;:\; x \in P_{i}\}$ and $S' := \{1,\dotsc,r\} \setminus S$.

If $S'$ is empty, then we take $y := u$. Suppose $S'$ is nonempty; then by prime avoidance, the set $\bigcap_{i \in S'} P_{i} \setminus \bigcup_{i \in S} P_{i}$ is nonempty; let $b$ be an element of this set; we take $y := bu$.

If $i \in S$, then $x \in P_{i}$ implies $u \not\in P_{i}$; since $b \not\in P_{i}$, we have $x+bu \not\in P_{i}$.

If $i \in S'$, then $x \not\in P_{i}$ but $b \in P_{i}$ so $x+bu \not\in P_{i}$.

  • $\begingroup$ See also math.stackexchange.com/q/3311495. $\endgroup$ Nov 21, 2019 at 11:12
  • $\begingroup$ What if $S$ is empty and $\bigcap_{i \in S;}P_{i}=0$? $\endgroup$
    – user124697
    May 16, 2021 at 19:54
  • $\begingroup$ @user124697 yes, then we can take $b = 0$ so that $y = 0$. $\endgroup$ May 16, 2021 at 21:49

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