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"One would like the product ideal to be $IJ=\{ij : i\in I,j\in J\}$ but we can easily see that there is a problem. It must be closed under addition, so $ij+i'j'$ must be in $IJ$. Can you find $i''\in I$, $j''\in J$ such that $ij+i'j'=i''j''$ so that it's in $IJ$ as defined above? Not in general, no. The natural way to allow for additive closure is to define $IJ$ as you did, including arbitrary finite sums of products."

I don't understand how closed under addition is the problem with $IJ=\{ij : i\in I,j\in J\}$ being an ideal. What is wrong with the following reasoning?

$(I,+),(J,+)$ are both normal subgroups of $R$, hence if $IJ=\{ij : i\in I,j\in J\}$ then $(IJ,+)$ is a subgroup of $R$ and therefore closed under addition.

Mike L
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  • You say that $(IJ,+)$ is a subgroup, but that is not true. What principle are you using to say that it is a subgroup? – Captain Lama Apr 01 '23 at 20:43

1 Answers1

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I think you are confusing addition and multiplication. If $I$ and $J$ were subgroups (of $R^\times$) for multiplication, then $IJ$ would be a subgroup for multiplication.

$I$ and $J$ are subgroups for addition, so $I+J$ is a subgroup for addition.

But there is no reason for $IJ$ to be a subgroup for addition (and it is not).

Captain Lama
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