1
$\begingroup$

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.

Please help me to sort out my confusion."

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.

$\endgroup$
1
  • $\begingroup$ Didn't you take the $x_i$ in $I \setminus T_i$ ? So then $x_1$ belongs to $I$. $\endgroup$
    – Asinomás
    Jan 16, 2022 at 10:28

1 Answer 1

0
$\begingroup$

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 $P_1$.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$
    – user775699
    Jan 22, 2022 at 13:49
  • 1
    $\begingroup$ 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 $P_k$. $\endgroup$
    – Asinomás
    Jan 22, 2022 at 14:06
  • $\begingroup$ Wyhy didn't you considered the case k=1? $\endgroup$
    – user775699
    Jan 22, 2022 at 14:15
  • 1
    $\begingroup$ For $k=1$ you have that $x_1\in P_1$ and you have $\prod\limits_{j=2}^n x_j \not \in P_1$. $\endgroup$
    – Asinomás
    Jan 22, 2022 at 14:17

You must log in to answer this question.