All the ideals mentioned below are in a commutative ring $A$ with identity. Let me start with a special case. Note that the ideals in Lemma 1 below do not need to be prime.

Lemma 1. If $\mathfrak a\subset \mathfrak b\cup\mathfrak c$, then $\mathfrak a\subset \mathfrak b$ or $\mathfrak a\subset \mathfrak c$.

Here is my proof: If $\mathfrak a\subset \mathfrak b\cup\mathfrak c$, then it’s not hard to see that $\mathfrak a=(\mathfrak a\cap\mathfrak b)\cup(\mathfrak a\cap\mathfrak c)$. The right hand side union is an ideal thus one of them is contained in another, i.e. $\mathfrak a= \mathfrak a\cap\mathfrak b$ or $\mathfrak a= \mathfrak a\cap\mathfrak c$.

The above result is not true in general for three or more unions. Now I am trying to give a proof without involving argument about elements for Prime Avoidance Lemma:

If an ideal contained in a finite union of prime ideals, say $\mathfrak p_1,...,\mathfrak p_n$, then it is contained in at least one $\mathfrak p_i$ for some $i$.

Attempt for a proof:

Lemma 2. An ideal $\mathfrak p$ is prime if and only if $\mathfrak{ab}\subset\mathfrak p$ implies that $\mathfrak a\subset \mathfrak p$ or $\mathfrak b\subset\mathfrak p$.

Proof is omitted for Lemma 2. Let $\mathfrak a$ be an ideal contained in $\bigcup\limits_{i=1}^n\mathfrak p_i\subset \mathfrak p_n\cup\sum\limits_{i=1}^{n-1} \mathfrak p_i$. By applying Lemma $1$ we can reduce the question to the following:

Prove that if $\mathfrak a\subset \sum\limits_{i=1}^{n-1} \mathfrak p_i$, then $\mathfrak a\subset \mathfrak p_i$ for some $i$.

By the induction hypothesis if $\mathfrak a\subset \bigcup\limits_{i=1}^{n-1}\mathfrak p_i$ then we are done. Thus by applying the minimality of the sum of ideals we have restricted our attention to the following two cases:

(1). $\mathfrak a=\sum\limits_{i=1}^{n-1}\mathfrak p_i$.

(2). $\mathfrak a$ is not contained in $\bigcup\limits_{i=1}^{n-1}\mathfrak p_i$ and is properly contained in $\sum\limits_{i=1}^{n-1}\mathfrak p_i$.

How do I finish the proof? Any other different proof is also welcomed. Thank you.


1 Answer 1


Note first that for a ring with three distinct maximal ideals $\mathfrak{a},\mathfrak{b},\mathfrak{c}$, then $\mathfrak{a} \subset \mathfrak{b}+\mathfrak{c}$.

So we use indeed induction: assume that $\mathfrak{a} \subset \bigcup_{i=1}^n{\mathfrak{p}_i}$.

If any smaller reunion of the $\mathfrak{p}_i$ contains $\mathfrak{a}$, we are done by induction hypothesis. So in particular, we have some $x_i \in \mathfrak{p}_i \backslash \mathfrak{p}_n$. We also have some $a \in \mathfrak{a}$ not in any $\mathfrak{p}_k$ (induction hypothesis) with $k <n$, and some $b \in \mathfrak{a} \backslash \mathfrak{p}_n$.

If $a \notin \mathfrak{p}_n$, we are done. Otherwise, consider $a+bx_1 \ldots x_{n-1}$.

  • $\begingroup$ I think you have not honored the post’s request (trying to give a proof without involving the elements) and you’re just parroted a proof given on this site and in every book on the subject... $\endgroup$
    – rschwieb
    Oct 27, 2019 at 22:07

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