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Let $R$ be a ring with unity and $0\neq I,J\lhd R.$ Can it be that $IJ=0?$

It is possible in rings without unity. Let $A$ be a nontrivial abelian group made a ring by defining a zero multiplication on it. Then any subgroup of $S$ of $A$ is an ideal, because for any $a\in A,$ $s\in S,$ we have $$sa=as=0\in S.$$ Then if $S,T$ are nontrivial subgroups of $A,$ we have $ST=0.$

A non-trivial ring with unity cannot have zero multiplication, so this example doesn't work in this case. So perhaps there is no example? I believe there should be, but I can't find one. If there isn't one, is it possible in non-unital rings with non-zero multiplication?

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up vote 7 down vote accepted

Even simpler: Take any direct product of two nonzero rings $R\oplus S$. Then $R\oplus 0$ and $0\oplus S$ are nonzero ideals multiplying to zero.

One definition for prime ring is a ring in which this does not happen, so any non-prime ring will suffice.

Also, if you believe in nonzero nilpotent ideals, then you would easily find an example: If $I^n=\{0\}$ with $n$ minimal, then $I I^{n-1}=\{0\}$.

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Eh, I seem to be asking sillier and sillier questions. I knew about all of these things! :( Thank you very much! – user23211 May 9 '12 at 13:09
Don't worry about it: after this happens a few times, it spurs the growth of the part of your brain that looks for obvious answers first. – rschwieb May 9 '12 at 13:34

Take $n=ab \in \mathbb Z$ and the ideals $(a)$ and $(b)$ in $\mathbb Z/(n)$.

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Thank you very much for this. Very simple! – user23211 May 9 '12 at 13:07

Let $R=K[x]/\langle x^2\rangle $ for some field $K$, and let $I=J=\langle x\rangle $. Then these ideals are non-zero, but the product of any pair of elements from $I=J$ has a factor of $x^2$, so is zero in $R$. So $I^2=IJ=J^2=0$.

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Thank you. I'm happy to see such an example because I've had to do some work in $K[x]/\langle x^n\rangle$ lately. – user23211 May 9 '12 at 13:10

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