# Let R be a commutative ring, and I, J denote two ideals in R such that I + J = R. Is it true that IJ = I ∩ J? [duplicate]

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?

• "if I + J = R, then 1 ∈ I and/or 1 ∈ J" This is not true. You need a completely different argument. Dec 10 '18 at 23:51
• 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 $1 \notin 3 \mathbb{Z}$. Dec 10 '18 at 23:52
• Try $I \cap J(I+J)$ Dec 10 '18 at 23:54

$$IJ\subset I\cap J$$ 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 $$x=1\cdot x$$.
• Yes. Isn't each of these products in $I\cap J$? Dec 11 '18 at 1:05