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.
Best Answer
If you define $T_k$ to be $\bigcup_{i\neq k}\,P_i$, then everything is good. If you want to know why your definition of $T_k$ doesn't work, then look at "Now, assume $I$ is not contained in $T_k$ [...]." You can't derive a contradiction from here since $I$ is never contained in any $T_k$ (in your definition).