Two comaximal radical ideals, with product the nilradical, are radicals of principal ideals generated by complementary idempotents

Question

Let $\mathfrak a,\mathfrak b\subset R$ be two comaximal radical ideals of the ring $R$, with $\mathfrak{ab}=0$. Then $\mathfrak a=Ra$ and $\mathfrak b=Rb$ for complementary idempotents $a,b\in R$.

The map $\rho:R\to R/\mathfrak a\times R/\mathfrak b$ defined as the projection in each coordinate is an isomorphism, according to the CRT.

Set $a:=\rho^{-1}(0,1)$ and $b:=\rho^{-1}(1,0)$. The kernel of the projection $R\to R/\mathfrak a$ is $\mathfrak a$, but it is also $Ra$ (for the kernel of the projection $R/\mathfrak a\times R/\mathfrak b\to R/\mathfrak a$ is the ideal generated by $(0,1)$). Hence $\mathfrak a=Ra$ and analogously $\mathfrak b=Rb$. Plus $a,b$ are complementary idempotents since $(0,1),(1,0)$ are.

Let $\mathfrak a,\mathfrak b\subset R$ be two comaximal radical ideals of the ring $R$, with $\mathfrak{ab}=\operatorname{nil}R$. Then $\mathfrak a=\sqrt{Ra}$ and $\mathfrak b=\sqrt{Rb}$ for complementary idempotents $a,b\in R$.

Set $R':=R/\operatorname{nil}R$, and let $\frak A,B$ be the images of $\frak a,b$ in the quotient. The first italic fact implies that there are complementary idempotents $a',b'\in R'$ such that $\mathfrak A=R'a'$ and $\mathfrak B=R'b'$. Now let $a,b\in R$ be lifting a of $a',b'$: one has $Ra+\operatorname{nil}R=\mathfrak a$ and $Rb+\operatorname{nil}R=\mathfrak b$, hence $\mathfrak a=\sqrt{Ra}$ and $\mathfrak b=\sqrt{Rb}$. But $a,b$ are not complementary idempotents, although $(a+b-1)^n=0$, $(ab)^m=0$.

At this point I don't see how to find two complementary idempotents using $a,b$; would you please give an idea? Thanks in advance

Answer 1

Suppose that $I,J$ are comaximal ideals and that $IJ = \sqrt{0}$ (no need in the following to assume that $I, J$ are radical).

Find $x \in I, y \in J, n \in \mathbb{N}$ such that $x + y = 1$ and $(xy)^n = 0$. Then $1 = (x+ y)^n = x^n + y^n + r$ for some $r$ divisible by $xy$, which is therefore nilpotent. Hence $1 -r$ is a unit with inverse $u$, and we can write $x^n u + y^n u = 1$, where $x^n y^n = 0$.

Can you find your complementary idempotents from here? Hint below

In general, check that $a^{2n} b = a^n$ implies that $a^nb$ is an idempotent.