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If $R$ is a commutative ring and $I$ and $J$ are ideals s.t. $I+J=R$ then show that $IJ=I\cap J$.

I've already shown that $IJ \subset I\cap J$, now I need to show the reverse inclusion.

I'm a bit lost, so far i'm just figuring out what pieces I have to work with.

Such as:

$\forall r\in R$ $\exists i\in I ,j\in J$ s.t. $i+j=r$

$\forall ij\in IJ$, $ij=i_1$ and $ij=j_1$ for some $i_1\in I$, $j_1\in J$.

Also, if I let $x\in I\cap J$, then $x=i_2=j_2=i+j$ for some $i_2\in I$, $j_2\in J$

Anyone, having problem getting to the conclusion here, thanks in advance

Math is hard
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2 Answers2

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There are $i\in I, j\in J$ such that $i+j=1$. Then, for all $a\in I\cap J$, $$a=a1=a(i+j)=ai+aj\in JI+IJ=IJ.$$

Rafael
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    This proof need not to be true in general. Because there are commutative rings without unity; for example, $R=2 \mathbb Z$. –  Aug 01 '20 at 09:22
  • @user808894 But in that case, the statement itself need not be true. For e.g. let $R=2\mathbb Z$, $I=4\mathbb Z$ and $J=6\mathbb Z$. $I+J=R$, $I\cap J=12\mathbb Z$ and $IJ=24\mathbb Z$. – Nothing special Oct 27 '24 at 19:05
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$I \cap J = (I \cap J) \cdot R \\ = (I \cap J) \cdot (I + J) \\ = (I \cap J) \cdot I + (I \cap J) \cdot J \\ \subseteq IJ + IJ \\ = IJ.$

QU Binggang
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    what does the dot operator mean here? –  Dec 09 '19 at 23:05
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    I think it is just Ideal multiplication. i.e. if $I$ and $J$ are Ideals in the same ring $R$, then $I \cdot J = {a_1 \cdot b_1 + ... + a_n \cdot b_n | a_i \in I, b_i \in J, n \in Z^+ }$. This is all the finite sums of products of elements of $I$ and $J$. And yes, that is also an ideal and it distributes over the sum of ideals. – Diego Andres Gomez Polo Dec 10 '19 at 04:20
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    There is obviously nothing wrong with this proof, but I feel like a student who needs to see a proof that $I+J=R \Rightarrow I\cap J=IJ$ would also need to see an element-wise proof of the various equalities contained in this proof. – Bobby Grizzard May 13 '20 at 15:31