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So I basically have to prove what is on the title.

Given $R$ a commutative rng (a ring that might not contain a $1$), with the property that $I+J=R$, (where $I$ and $J$ are ideals) we have to prove that $IJ=I\cap J$.

One inclusion is easy. If $x\in IJ$, then $x=\sum a_ib_i$ where $a_i\in I$ and $b_i\in J$. Thus for any fixed $i$, we have that since $a_i\in I$, we have that $a_ib_i\in I$, and the same argument shows that $a_ib_i\in J$, thus $\sum a_ib_i\in I$ and $\sum a_ib_i\in J$, this means that $x=\sum a_ib_i\in I\cap J$, and thus $IJ\subset I\cap J$.

I am having troubles proving the other inclusion. Any comments?

Thanks

user26857
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Daniel Montealegre
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    Wait, ins't this false for general commutative rngs? Consider $\mathbb Z$ with the normal addition and multiplication $\cdot$ defined by $a\cdot b=0,\forall a,b\in\mathbb Z$. Then $(2)$ and $(3)$ are comaximal ideals (since we can write $x$ as $2a+3b$ in integers with regular multiplication so as $2 + \cdots + 2 + 3 + \cdots + 3$ in our rng), $(2)\cap (3)=(6)\neq (0)$ yet clearly $(2)(3)=0$. – Alex Becker Mar 19 '12 at 06:03
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    Im thinking its false as well with $I=4\mathbb{Z}$ and $J=6\mathbb{Z}$ and $R=2\mathbb{Z}$ – Daniel Montealegre Mar 19 '12 at 06:05
  • what happens if the ring R is not Commutative?? – User8976 Feb 25 '15 at 02:23
  • @user8795 : see https://math.stackexchange.com/questions/1222474/assume-r-is-commutative-if-ij-r-prove-that-ij-i-cap-j-provide-a-count?noredirect=1&lq=1 – Watson Dec 22 '16 at 17:45

1 Answers1

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The statement you are trying to prove is only necessarily true for commutative rings with $1$. In this case, you can argue that $$I\cap J\subseteq (I\cap J)R=(I\cap J)(I+J)=I(I\cap J)+ J(I\cap J)\subseteq IJ+ IJ=IJ$$

A counterexample to the statement for general rngs is given by endowing the group $\mathbb Z$ with the zero product, that is defining $a\cdot b=0,\forall a,b\in\mathbb Z$. The ideals $(2)$ and $(3)$ are still comaximal, as for any $x\in\mathbb Z$ we can write $x=a\times 2+b\times 3$ for some $a,b\in\mathbb Z$, where $\times$ denotes regular multiplication, and $a\times 2=2+\cdots+2\in(2)$ where addition is performed $a$ times, and similarly $b\times 3\in (3)$. But $(2)\cap (3)=(6)\neq (0)$, yet clearly $(2)(3)=(0)$.

user26857
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Alex Becker
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  • Thank you so much. Perhaps a little easier example its just $R$ be the even numbers, $I$ the multiples of $4$ and $J$ the multiples of $6$. – Daniel Montealegre Mar 19 '12 at 07:30
  • I apologize for the confusion whether or not it had a $1$. I got it out from Dummit and Foote, and they usually have rings without a $1$, but in the beggining of the section they said to consider only unital rings. – Daniel Montealegre Mar 19 '12 at 07:31