Question about an "idempotent" in the ring

Question

Let A denote a commutative ring and let e denote an element of A such that $e^2 = e$. How to prove that $eA \times (1 - e)A \simeq A$? I thought that $\phi: A \mapsto eA \times (1 - e)A, \ \phi(a) = (ea, (1-e)a)$ is an isomorphism but I don't know how to prove that $\phi$ is a bijection.

Answer 1

Hint:

$\phi$ is $A$-linear, so all you have to prove is:

• $\ker\phi=\{0\}$.
• $\phi$ is surjective, i.e. given $\bigl(ea,(1-e)b\bigr)\in eA\times (1-e)A$, you can solve the system of equations \begin{cases} ex=ea,\\(1-e)x=(1-e)b. \end{cases} You'll need to check that $1-e\;$ is another idempotent, and that $e$ and $1-e$ are orthogonal idempotents, i.e. $\;e(1-e)=0$.

Answer 2

Your map is good, you just have to check (it is not clear to me from the question if you've already managed to prove that $\phi$ is a ring homomorphism):

• $\phi(ab)=\phi(a)\phi(b)$, how would you define the multiplication (this is a ring isomorphism)? For a natural multiplication, you'll need to find both $e^2$ and $(1-e)^2$.
• $\phi(a+b)=\phi(a)+\phi(b)$
• $\ker(\phi)=0$
• Show that $\phi$ is surjective, that is, find $c\in A$ such that $\phi(c)=(a,b)\in eA\oplus(1-e)A$. More details for this step:

  Starting from $(a,b)\in eA\oplus (1-e)A$ you want to find some $c\in A$ that maps to $(a,b)$. Since $a\in eA$ we may write $a=me$, similarly we may write $b=n(1-e)$. What does $a+b$ map to? What is $e(1-e)$?

Alt. the two last steps may be skipped if you construct an inverse.