# Alternate proof of prime avoidance lemma.

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.

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 $$k , and some $$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}$$.