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Let $I$ and $J$ be two ideals of a ring $R$, the usual definition of the product of $I$ and $J$ is $$IJ = \left\{\left.\sum_{i=0}^n a_ib_i\ \right|\ a_i \in I, b_i \in J \text{ and } n \ge 0\right\}$$ because the subset of $R$ $$\big\{a_ib_i\ \big|\ a_i \in I \text{ and } b_i \in J \big\}$$ isn’t an ideal in general.

I would like to find an example where the last set isn’t an ideal. But I’m unable to finish all the sketch of examples I find on forums. Do you have examples ? The easier the better ;)

Thanks ! :D

Arturo Magidin
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MarcAntony
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  • I don't know, pick like seven pairs of non-principal ideals and do the first one that works. –  Oct 28 '21 at 07:05
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    Maybe you can explain what you can and can't do in those proofs, so that we understand what the issue is. – Captain Lama Oct 28 '21 at 07:27
  • @CaptainLama I misspoke my question. I was looking for an example of $R$, $I$ and $J$ where the last set isn't ideal, i.e it isn't stable for addition. Nightflight did it :) – MarcAntony Oct 28 '21 at 14:02
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    This question has been asked and answeed multiple times. For instance https://math.stackexchange.com/questions/290229/explaining-the-product-of-two-ideals/290230 – Aitor Iribar Lopez Oct 28 '21 at 14:43
  • @AitorIribarLopez Nice find: I could only find lesser duplicates in my first pass... – rschwieb Oct 28 '21 at 14:46

1 Answers1

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The last set is not an additive monoid, in general!

I'll give a specific example.

Let $I,J\lhd \mathbb Z[X]$ as $I=(2,X), J=(3,X)$, $S=\{ij\mid i\in I, j\in J\}$.

Then $2+X, 4+X\in I, 3+X, 6+X\in J$.

$(2+X)(6+X), (3+X)(4+X)\in S$ is to trivial.

But $(2+X)(6+X)+(3+X)(4+X)=24+15X+2X^2$ is irreducible polynomial over $\mathbb Z$

So there exists no $i,j$ such that $i\in I, j\in J, ij=24+15X+2X^2$.

So $24+15X+2X^2\notin S$.