# A question in the proof of Prime Avoidance Lemma

I was unable to prove Prime Avoidance Lemma for union of n prime ideals (did it for 2 prime ideals). So, looked on internet for help.

I found a proof here in question where proof is written by OP but I have a question in his proof and he is not seen since 6 years. I also saw the proof of wikipedia but I was really confused in understanding that. This proof appears better understood too me.

So, I am asking question by posting his proof here.

Proof:"In an homework of mine, I gave the following proof for the prime avoidance lemma, i.e the lemma saying if $$R$$ is a commutative ring and $$I$$ an ideal of $$R$$ and $$I \subseteq \bigcup_{i=1}^{n} P_i$$ for a finite collection a finite collection of prime ideals $$\{P_1,...,P_n \}$$, then $$I \subseteq P_i$$ for some $$i$$. I got full points on the proof, but afterwards I discovered what seem to be a mistake. My conclusion in the proof is that $$I$$ is contained in $$T_k$$ for some $$k$$ (see below), but this is not true! Since $$T_k$$ does not include $$0$$. Is there a mistake in my proof? I really cant find it and neither could my teacher, but the conclusion is obvivously false!

This is the proof I gave:

We proceed by induction on the number of prime ideals. If $$n=1$$, the result is trivial. Now, suppose the result is true for $$n-1$$ primes. Now, let $$I$$ be an ideal such that $$I \subseteq P_1 \cup...\cup P_n$$. For each $$k=1,...,n$$ define the set $$T_k := (\bigcup_{i=1}^{n} P_i) \setminus P_k.$$ Now, assume $$I$$ is not contained in $$T_k$$ for any $$k$$, otherwise removing $$P_k$$ from the union would let us apply the induction hypothesis. Furthermore, pick an element $$x_k \in I \setminus T_k$$ for each $$k$$ (so that $$x_k$$ is contained in $$P_k$$ but no other $$P_i$$)

Let $$a =x_1 + \prod_{j=2}^{n} x_j$$, since $$x_k \in I$$ it follows that $$a \in I$$. Moreover, $$a \notin P_1$$ because if $$a \in P_1$$ then so is $$a-x_1$$ so that one factor in $$\prod_{j=2}^{n} x_j$$ is in $$P_1$$ which is a contradiction. I claim that $$a \notin P_k$$ for $$k \geq 2$$. Suppose $$a \in P_i$$ for some $$i$$, then $$-x_1 = \prod_{j=2}^{n} x_j-a \in P_i.$$ So, $$a \notin P_k$$ for any $$k$$, but this is a contradiction since $$a \in I \subseteq P_1 \cup...\cup P_n$$. Hence the assumption is not true, so we can remove one ideal from the union and apply the induction hypothesis.

My question: In third para of the proof which is gave how does $$x_k \in I$$ implies that $$a\in I$$ since it is not known that $$x_1$$ always belongs to I. Kindly help.

• Didn't you take the $x_i$ in $I \setminus T_i$ ? So then $x_1$ belongs to $I$. Jan 16, 2022 at 10:28

The proof looks correct to me, answering the last part, we have $$x_1\in I$$ because we have $$x_k\in I$$ for all $$k$$.
Also note the part where primality is used is to conclude that $$\prod\limits_{j=2}^n x_j$$ does not belong to $$P_1$$. This is because it is a product of elements that do not belong to $$P_1$$.
• Can you also tell how In the last para how does $-x_1 = \prod_{j=2}^{n} x_j-a \in P_i$ implies $a\nin P_k$ for any k?
• That part seems roundabout, but $a\not \in P_k$ for $k\neq 1$ is easy because $x_1\not \in P_k$ while $\prod\limits_{j=2}^n x_j\in P_k$, so it is the sum of an element in $P_k$ and an element not in $P_k$, and hence is not in $P_k$. Jan 22, 2022 at 14:06
• For $k=1$ you have that $x_1\in P_1$ and you have $\prod\limits_{j=2}^n x_j \not \in P_1$. Jan 22, 2022 at 14:17