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!

$\endgroup$
  • 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 Aug 16 '18 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! $\endgroup$ – BOlivianoperuano84 Aug 16 '18 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 Aug 16 '18 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 $J$? $\endgroup$ – BOlivianoperuano84 Aug 16 '18 at 2:03
  • 1
    $\begingroup$ I added the "ring-theory" tag to your post. Cheers! $\endgroup$ – Robert Lewis Aug 16 '18 at 3:47
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?

$\endgroup$
  • $\begingroup$ Thank @rwbogl! That made it clear what was wrong with my proof! $\endgroup$ – BOlivianoperuano84 Aug 19 '18 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 $i \notin I$.

$\endgroup$
  • 1
    $\begingroup$ Thanks @Robert This makes it very clear where my confusion was. $\endgroup$ – BOlivianoperuano84 Aug 19 '18 at 20:05
  • $\begingroup$ @BOlivianoperuano84: glad to help out! $\endgroup$ – Robert Lewis Aug 19 '18 at 21:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.