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Let $R$ be a commutative ring with identity and let $I$, $J$ be ideals. Define $IJ$ to be all elements of $R$ of the form $ a_1b_1 + a_2b_2 + . . . + a_nb_n $ where $n ≥ 1$ and $a_1, a_2, . . . , a_n $ are in I and $b_1, b_2, . . . , b_n $ are in J.

Prove that $ IJ \subset I \cap J $

This seems really surprising at first glance until I wrote out a couple of examples and found that ideals of the same ring have a lot of things in common. In fact they had everything in common in the cases I found. I'm not really sure how to formally write out this argument and would like to know: why they are not simply equal?

rschwieb
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Faust
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1 Answers1

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$IJ\subseteq IR\subseteq I$

$IJ\subseteq RJ\subseteq J$

Therefore $IJ\subseteq I\cap J$.

An example to show they don't have to be equal:

Consider the ideal generated by $x$ in $\mathbb R[x]/(x^2)$, and use that ideal for both $I$ and $J$.

rschwieb
  • 153,510