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So far I have that if I + J = R, then 1 ∈ I and/or 1 ∈ J. Then I = R and/or J = R. If both I = J = R, then IJ = I ∩ J = R must be true because we already know that multiplying every element in I or J by every element in R will end up giving us R again. So IJ = I ∩ J = R.

If I = R, but J ≠ R, then IJ = J = I ∩ J because we already know that when all elements of J are multiplied by all elements of R, the result is J again. Therefore, IJ = I ∩ J = J.

Therefore, it is true that if I + J = R, then IJ = I ∩ J.

Does this proof work or am I not able to say that 1 ∈ I or 1 ∈ J?

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    $\begingroup$ "if I + J = R, then 1 ∈ I and/or 1 ∈ J" This is not true. You need a completely different argument. $\endgroup$ Dec 10 '18 at 23:51
  • $\begingroup$ How does $I+J = R$ imply that $1 \in I$ or $1 \in J$? This is false: consider for instance the ideals $2 \mathbb{Z}$ and $3 \mathbb{Z}$ in $\mathbb{Z}$. Then $1 = -2 + 3 \in 2 \mathbb{Z} + 3 \mathbb{Z}$, but $1 \notin 2\mathbb{Z}$ and $1 \notin 3 \mathbb{Z}$. $\endgroup$ Dec 10 '18 at 23:52
  • $\begingroup$ Try $I \cap J(I+J)$ $\endgroup$ Dec 10 '18 at 23:54
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Hint:

$IJ\subset I\cap J$ is trivial. For the reverse inclusion:

By hypothesis, there exist $u\in I$, $v\in J$ such that $u+v =1$. Now let $x\in I\cap J$. Use that $x=1\cdot x$.

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  • $\begingroup$ Why do you consider the first line to be trivial? Am I missing something obvious about multiplying ideals? $\endgroup$
    – Jon D.
    Dec 11 '18 at 0:11
  • $\begingroup$ It is indeed obvious. Check the definition of a product of ideals. $\endgroup$
    – Bernard
    Dec 11 '18 at 0:27
  • $\begingroup$ I've been trying to find a good one, my professor never taught us about adding and multiplying ideals, so this question blindsided me a bit. $\endgroup$
    – Jon D.
    Dec 11 '18 at 0:45
  • $\begingroup$ Ok, correct me if I'm wrong, but to get the product, you essentially multiply every element of I by every element of J and then also see what else you can make by adding and subtracting those products you already made? $\endgroup$
    – Jon D.
    Dec 11 '18 at 0:54
  • $\begingroup$ Yes. Isn't each of these products in $I\cap J$? $\endgroup$
    – Bernard
    Dec 11 '18 at 1:05

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