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Let $R$ be a non-trivial commutative unital ring and let $I,J$ be ideals of $R$. The ideal product is defined as $IJ:=\{ i_1j_1+\dots+i_lj_l:l\ge 1, i_n\in I, j_n \in J, n\in\mathbb{N} \}$. Prove that $IJ$ is an ideal of $R$.

My approach was to prove that $IJ=I\cap J$, which implies that $IJ$ is an ideal.

Let $i\in I\cap J$, then $i\in I, J$, $=i_kj_k$ for some $i_k, j_k\in J$. But $i_kj_k\in IJ$, thus $i\in IJ$. Thus $I\cap J \subset IJ$.

For the other direction, I know that it is true, so I will skip the proof thereof. However, I'm not so sure that my proof above is correct since then $P^2=PP=P\cap P = P$, which is not true in general.

Can someone please point out what my error could possibly be? Maybe I didn't understand the definition of the product of ideal the way it was to be meant?

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    $\begingroup$ Look at the case $I=J$. $\endgroup$
    – Bob Jones
    Mar 17, 2017 at 23:26
  • $\begingroup$ I don't get how you justify that $i=i_kj_k$ $\endgroup$
    – Exodd
    Mar 17, 2017 at 23:27
  • $\begingroup$ @Exodd Since $i\in I\cap J$, which is an ideal, there exist some elements in $i$ and $j$ whose product is $i$. $\endgroup$
    – sequence
    Mar 17, 2017 at 23:42
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    $\begingroup$ @sequence Try $I=J=(2)$ in $\mathbb{Z}$. This is not necessarily the case. $\endgroup$
    – Bob Jones
    Mar 17, 2017 at 23:43

1 Answer 1

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Let $i\in I\cap J$, then $i\in I, J$, $=i_kj_k$ for some $i_k, j_k\in J$. But $i_kj_k\in IJ$, thus $i\in IJ$. Thus $I\cap J \subset IJ$.

The problem with your proof is that $i\in I\cap J$ doesn't mean necessarily that $i=i_kj_k$ for some $i_k\in I$, $j_k\in J$. To see that this is not always true we can take the counterexample given by Bob Jones. Indeed, if $I=J=(2)$, then if your idea were right, we could write $2=(2a)(2b)$ for $2a, 2b\in (2)$, and this would imply that $4\mid 2$, which is clearly false.

In general the result you want to prove is only true if $I$ and $J$ are comaximal ideals, i.e., $I+J=R$. In general, to prove that $IJ$ is an ideal of $R$ we need to show:

i) For every $a,b\in IJ$, $a+b\in IJ$. To show this let's write $a=\sum_{i=1}^n x_iy_i$ and $b=\sum_{j=1}^m x_j'y_j'$, then $$a+b=\sum_{k=1}^{n+m}z_kw_k,$$ where $z_k=x_i$ and $w_k=y_i$ for $1\le k\le n$, and $z_k=x_j'$ and $w_k=y_j'$ for $n+1\le k\le n+m$. As $z_k\in I$ and $w_k\in J$ we have that $a+b\in IJ$.

ii) For every $r\in R$ and $a\in IJ$, $ra\in IJ$. Again, let's write $a=\sum_{i=1}^n x_iy_i$, so $$ra=r(\sum_{i=1}^n x_iy_i)=\sum_{i=1}^n(rx_i)y_i.$$

Since $rx_i\in I$ and $y_i\in J$ we deduce that $ra\in IJ$. Hence, $IJ$ is an ideal of $R$.

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  • $\begingroup$ Thanks. I understand that by using the definition one can prove this, I was just trying to find a shorter, "fancier", way. $\endgroup$
    – sequence
    Mar 18, 2017 at 0:39
  • $\begingroup$ @sequence yeah I understand, but you have to be careful about the results you want to prove. $\endgroup$
    – Xam
    Mar 18, 2017 at 2:51
  • $\begingroup$ I still do not quite understand what exactly is incorrect in my proof that $I\cap J \subset IJ$ @Xam $\endgroup$
    – sequence
    Mar 18, 2017 at 2:55
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    $\begingroup$ @sequence I edited my answer. Hope it helps you. $\endgroup$
    – Xam
    Mar 18, 2017 at 3:42
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    $\begingroup$ I'm pretty sure in general $IJ \subset I \cap J $ not the other way around. $\endgroup$
    – Andres Mejia
    Sep 12, 2018 at 18:02

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