5
$\begingroup$

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!

$\endgroup$
6
  • 3
    $\begingroup$ No this doesn't work, since it might be $i$ sometimes and $j$ sometimes. Instead, assume there exists $i$ not in $P$ and $j$ not in $P$; can you reach a contradiction? $\endgroup$
    – Steve D
    Commented Aug 16, 2018 at 1:39
  • $\begingroup$ I see, so we want to show that it is always the case that only $I$ or only $J$ is always contained in $P$, right? Not that we sometimes have $I \subset P$ and $J \subset P$?? Thanks for the comments! $\endgroup$ Commented Aug 16, 2018 at 1:42
  • $\begingroup$ No, your proof only shows that given $ij\in IJ$, we have $i\in P$ or $j\in P$. What you need is for every $i\in I$, $i\in P$ (or analogously for $J$). $\endgroup$
    – Clayton
    Commented Aug 16, 2018 at 1:46
  • $\begingroup$ But then since $i$ and $j$ are arbitrary, isn't the proof saying this works for every $i\in I$, since we can just take $j \in J$ to be the identity in $J$? $\endgroup$ Commented Aug 16, 2018 at 2:03
  • 1
    $\begingroup$ I added the "ring-theory" tag to your post. Cheers! $\endgroup$ Commented Aug 16, 2018 at 3:47

2 Answers 2

4
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Thank @rwbogl! That made it clear what was wrong with my proof! $\endgroup$ Commented Aug 19, 2018 at 20:04
4
$\begingroup$

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 $i \notin I$.

$\endgroup$
2
  • 2
    $\begingroup$ Thanks @Robert This makes it very clear where my confusion was. $\endgroup$ Commented Aug 19, 2018 at 20:05
  • $\begingroup$ @BOlivianoperuano84: glad to help out! $\endgroup$ Commented Aug 19, 2018 at 21:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .