Union of Prime Ideals Contains an Ideal

[Bashyboy]

Homework Statement

Let $R$ be a commutative ring with identity and suppose that $A$ is an ideal in $R$ contained in the finite union of prime ideals $P_1 \cup \cdots \cup P_n$. Show that $A \subseteq P_i$ for some $i$.

Homework Equations

The Attempt at a Solution

The base case ($n=1$) is trivial. I will prove the contrapositive of the induction step. That is, assume that the following conditional holds:

If $I$ is an ideal such that $I \not\subseteq Q_i$ for all $i \in \{1,...,n-1\}$, where $Q_i$ is a prime ideal, then $I \not\subseteq \bigcup_{i=1}^{n-1} Q_i$.

And suppose that $A \not\subseteq P_i$ for $i=1,2,...,n$, where $P_i$ is a prime ideal. If $A$ does not intersect any of the $P_i$, the conclusion follows trivially. If it intersects at most $n-1$ of them, then we can apply the induction hypothesis to arrive at the conclusion. Hence, suppose that $A$ intersects each one of them.Let $j \in \{1,...,n\}$ be arbitrary. If $A \cap P_j \subseteq P_i$ for some $i$, then we can remove $P_j$ from the collection of prime ideals and apply the induction hypothesis. Hence assume that $A \cap P_j \not\subseteq P_i$ for all $i$. Since $A \cap P_j$ is an ideal, there exists an $a_j \in A \cap P_j - \bigcup_{i \neq j} P_i$ for every $j$. Now clearly $a_1 + a_2...a_n \in A$. Suppose that $a_1 + a_2...a_n \in \bigcup_{i=1}^n P_i$ for contradiction. Then $a_1 + a_2...a_n \in P_i$ for some $i$ or $a_1 + a_i(a_2...a_n) \in P_i$. But $a_i(a_2...a_n) \in P_i$, so $a_1 \in P_i$, but this is a contradiction unless $i=1$, but $i$ isn't equal to $1$, so we get a contradiction anyways.

How does this sound? My only worry is, where did I use the fact that the $P_i$ are prime?

 

[fresh_42]

To be honest? I'm confused. I cannot see how the intersections come into play, e.g. in

If it intersects at most $n-1$ of them, then we can apply the induction hypothesis to arrive at the conclusion.

I don't arrive at the conclusion at all, because we have a statement about "not contained in" and you use it for "non empty intersections of".
So either I'm too stupid to see, or it's wrong. In any case I'm too lazy to figure it out. The proof I know also uses induction on $n$, but doesn't use intersections. If they play such an important role, you should define them. Also the switches between $I$ and $A$ on one hand and $P_i$ and $Q_i$ on the other hand is confusing. I know why you do it, but then you shouldn't stop there and write your proof completely in this formal logic: $\forall_{1 \leq i < n} A^{(1)}_i (Q_i;I) \Longrightarrow A^{(2)}(Q_1, \ldots , Q_{n-1};I)$ and the special cases with $A^{(1)}_i (P_i;A)$ etc. A mixture of these two ways is a bit confusing and unnecessary.

With the elements $a_j$, which do exist, only that I don't see how you arrive there, you should build $a := \sum_j \prod_{i \neq j} a_i$. To see why $a \notin \cup_i P_i$ you then use the prime property.

 

[Bashyboy]

I cannot see how the intersections come into play, e.g. in

I'm not really worried about this part, as I have already had it verified; I just included the whole proof for completeness. The intersections plays a role because, I argue, each $A \cap P_j$, is nonempty and is such that $A \cap P_j \not\subseteq P_i$ for all $i \neq j$, which is $n-1$ prime ideals; since $A \cap P_j$ is itself a ideal, by the induction hypothesis there exists an $a_j \in A \cap P_j - \bigcup_{i \neq j} P_i$; and this is how I build an element contained in $A$ but not in $\bigcup_{i=1}^n P_i$. At least this is what I attempted to argue do at the beginning of my proof. Let me try arguing it again.

If $A$ does not intersect any $P_i$, then $A \not\subseteq \bigcup_{i=1}^n P_i$ follows trivially. If $A$ intersects some but not all sets, say it misses $P_n$, then $A \not\subseteq P_i$ for $i=1,...,n-1$ and so the inductive hypothesis says $A \not\subseteq \bigcup_{i=1}^{n-1} P_i$; but since $A$ completely misses $P_n$, we can further conclude $A \not\subseteq \bigcup_{i=1}^n P_i$, which is want to prove. Hence just assume that $A$ intersects each $P_i$ nontrivially, and from here I build $a_1 + a_2 a_3 \dots a_n$.

What I am really concerned with is, where the primeness of the $P_i$ is used? I only seemed to use the fact that it is an ideal, nothing more.

 

[fresh_42]

I don't see where you excluded $i=1$. This is the case where you use primeness. Suppose $i=1$. Then $a_2\ldots a_n \in P_1$. But this means, because $P_1$ is prime, that at least one of the $a_i$ has to be in $P_1$. And this is the contradiction in this case.

 

[Bashyboy]

I read here https://math.stackexchange.com/questions/1027428/prime-avoidance-lemma?noredirect=1&lq=1 that the assumption that the ideals involved in union be prime is unnecessary for $n=2$. I think I have a proof, but something about it seems a bit 'flimsy', for lack of a better term.

Suppose that $A \subseteq I \cup J$, where $I,J$ are ideals in $R$, and that $x \in A-J$, which means that $x \in I$. Let $a \in A$ be arbitrary. Then $a \in I \cup J$ and either $a \in I$ or $a \in J$. If $a \in I$, we are finished. Hence, suppose that $a \in J$. Note that $a+x \in I \cup J$ also. If $a+x \in J$, then $x \in J$, which is a contradiction. Hence, $a+x \in I$. But since $x \in I$, we get $a \in J$. Thus, in all cases, we get $a \in I$.

 

[fresh_42]

Yes, this is correct, except the last $J$ should be $I$. It would be a bit easier to read if you wrote: Note that $a+x \in A \subseteq I \cup J$ because without it, it is not clear why it has to be the case, as $I \cup J$ isn't an ideal and you need this property from $A$. Without it, it looks like an error at first glance.

As said, the general proof can be done with $a := \sum_j \prod_{i \neq j} a_i \; (a_i \notin P_j \; , \; i \neq j)$ which shows, that the case $n=2$ doesn't need multiplications and therefore not the prime ideal property.