# Form of coprime ideals in commutative rings

I was trying to prove that there exists a bijection between the idempotent elements in a commutative ring $R$, and possible factorisations of $R$ into two other commutative rings $R_{1} \times R_{2}$. With not too much difficulty, it can be shown that for every idempotent element $i$ there is a factorisation, the converse is pretty hard though. I've reduced the problem to the question if for every pair of coprime ideals $I, J$ in $R$ there exists an idempotent $i \in R$ such that $I = iR$ and $J = (1-i)R$. Is this statement true? Or should I start my proof over again? If it is true, could I get a good hint on how to prove that statement?

• No, it is not true: $I=2\mathbb Z$ and $J=3\mathbb Z$ are coprime ideals in $R=\mathbb Z$. – user26857 Sep 24 '16 at 18:00

The product $R_1 \times R_2$ has two obvious idempotents, namely $(1,0)$ and $(0,1)$, and indeed, $R_1 \times R_2 = (R_1 \times R_2)(1,0) \times (R_1 \times R_2) (0,1)$. Letting $\phi: R_1 \times R_2 \overset{\sim}{\to} R$ be the given isomorphism, we simply push this factorization forward by $\phi$.
Let $e = \phi(1,0)$. Note that $\phi(1,1) = 1$ (since $(1,1)$ is the multiplicative identity in the product $R_1 \times R_2$), so $$\phi(0,1) = \phi((1,1) - (1,0)) = \phi(1,1) - \phi(1,0) = 1 - e$$ as desired.
I would simply, for a factorisation $R=R_1\times R_2$, consider $e=p_{1}(1)$ and show that $e$ is an idempotent of $R$, $1-e$ is the orthogonal idempotent, and that $R_1=Re$, $R_2=R(1-e)$.
• I'm sorry but I don't really get what we're doing when we consider $e = p_1$, and why can we even consider that $R_1 = Re$ if we don't even necessarily know that $R_1$ is a principal ideal? What am I missing? – pokemonfan Sep 22 '16 at 23:10
• Sorry for the typo (fixed). I meant $e=p_1(1)$ (and $1-e=p_2(1)$). Yes, $R_1$ is a principal ideal of $R$ (but considered for itself, it is a ring which is not necessarily principal). – Bernard Sep 22 '16 at 23:14
• I'm still kind of confused, my problem is that I don't see how we can even consider $R_1$ to be a principal ideal when it could be an ideal of any form really, although I understand it atleast must be an ideal which I've already verified. Also I don' understand what you mean with the notation $e = p_{1}(1)$ – pokemonfan Sep 22 '16 at 23:27
• I denote $e$ (the standard name for an idempotent) the projection of the unit $1$ of $R$ onto the 1st factor $R_1$. – Bernard Sep 22 '16 at 23:34