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Let I and J be ideals of a ring R (not necessarily unitary). I need to prove that IJ is an ideal of R contained in $I\cap J$. I can prove all the other properties of ideal. However, I cannot prove that IJ is closed under multiplication. I think I need this because I have to show that IJ is a subring of R and IJ is a subring of R if and only if IJ is closed under multiplication.

user10024395
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  • You need more than that, don't you? You need that it's stable under multiplication by any element of $R$, not just the ones in $IJ$. – Hoot Jan 16 '15 at 02:39
  • I know but that is the only thing that I can't prove. The rest I know how to prove. – user10024395 Jan 16 '15 at 02:48
  • @user2675516 Hoot's point is this: One of the things you need to show is that $R \cdot IJ \subset IJ$. If you're saying that you know how to prove this, then what you want follows immediately: $IJ \cdot IJ \subset R \cdot IJ \subset IJ$. – Andrew Dudzik Jan 16 '15 at 04:16

2 Answers2

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There seems to be some confusion here. To show $IJ$ is an ideal you need to show: (1) $IJ$ is a subgroup; and (2) If $r \in R$ and $\alpha \in IJ$ then $r\alpha$ and $\alpha r$ are both in $IJ$. As mentioned in the comments this is stronger then just being closed under multiplication (where you only consider $r$ from the subset $IJ$).

So let $r \in R$ and let $\alpha = i_1 j_1 + \ldots + i_1 j_1 \in IJ$. Then $$ r \alpha = (r i_1) j_1 + \ldots + (r i_n) j_n \in IJ$$ since each $r i_k$ is in $I$, since $I$ is an ideal (again using that same property of ideals).

You should argue similarly that $\alpha r \in IJ$.

Morgan Sherman
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Let $\sum_{i=1}^na_ib_i,\sum_{i=1}^nc_id_i\in IJ$

then $(a_1b_1+a_2b_2+...+a_nb_n)(c_1d_1+c_2d_2+...c_nd_n)=(a_1b_1c_1)d_1+(a_1b_1c_2)d_2+...+(a_ib_ic_j)d_j+..\in IJ$

use the distributive property and the fact that both $I$ and $J$ are ideals

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