Dummit and Foote problem 11 in section 7.4 I am trying to solve problem 11 in Dummit and Foote section 7.4. The problem is the following:
Assume $R$ is commutative. Let $I$ and $J$ be ideals of $R$ and assume $P$ is a prime ideal of $R$ that contains $IJ$. Prove either $I$ or $J$ is contained in $P$.
I came up with the following proof and I just want to check if it is right. It's as follows:
We know $IJ \subset P$. Then if we consider $i \in I$ and $j \in J$. Then, we know $ij \in IJ \subset P$. By primality of $P$ and since $ij \in P$, we know that either $i \in P$ or $j \in P$. Thus, we must have that $I \subset P$ or $J \subset P$, as desired.
I would appreciate any suggestion or comments on this proof. Thanks!
 A: I want to more directly point out the problem in your proof. You have proved the following statement: $$\forall i \in I\ \forall j \in J \ [i \in P \lor j \in P].$$ However, you need to prove the following statement: $$[(\forall i \in I\ i \in P) \lor (\forall j \in J\ j \in P)].$$ Can you see the difference? In words, your proof only shows that given some $ij \in IJ$, you can determine that one of those two must be in $P$, but not which one. In particular, you have no reason to conclude that every $i$ should be in $P$, or that every $j$ should be in $P$, as you need to prove.
Robert Lewis gives an excellent answer for the correct proof, but to complete my own, here is a hint: Suppose that $IJ \subseteq P$, while $I$ is not contained in $P$. This gives you some $i \in I$ such that $i \notin P$. What can we say now, given our hypotheses?
A: Well, let's see . . . 
Suggestion for a proof:
We are given that
$IJ \subset P; \tag 1$
suppose then that
$I \not \subset P; \tag 2$
then there must be some $i \in I$ such that
$i \notin P; \tag 3$
now,
$iJ = \{ij \mid j \in J\} \subset IJ, \tag 4$
and thus
$iJ \subset IJ \subset P; \tag 5$
this says that
$\forall j \in J, \; ij \in P; \tag 6$
so with
$i \notin P, \; \text{a prime ideal}, \tag 7$
we must have
$j \in P, \; \forall j \in J, \tag 8$
which shows that
$J \subset P. \tag 9$
Comments on the OP's proof: it seems OK to me through the assertion, 
"either $i \in P$ or $j \in P$"; but it doesn't follow that "$I \subset P$ or $J \subset P$", since we need to show for example that $j \in P$ for every $j \in J$, which is why we need, in my proof above, that $iJ \subset P$ with $i \notin I$.
