# Complete set of orthogonal idempotents of a direct product of modules

Let $$R$$ be a ring and let $$A_1, \ldots, A_n$$ be $$R$$-modules.

I have to show that there are $$R$$-homomorphisms $$e_i : \prod\limits_{i =1}^n A_i \rightarrow \prod\limits_{i =1}^n A_i$$ for $$i = 1, \ldots, n$$ satisfying the following conditions :

(I) $$\quad$$ $$e_i \circ e_i = e_i$$ for all $$i = 1, \ldots ,n$$

(II) $$\ \ \ \ e_i \circ e_j = 0$$ for all $$1 \leq i, j \leq n$$ and $$i \neq j$$

(III) $$\ \ \ id_A = e_1 + \cdots + e_n$$

My problem is that I have no idea how to construct such a map $$e_i$$.

Furthermore, I have to prove that $$\varphi : A \rightarrow \prod\limits_{i =1}^n A_i$$ is an $$R$$-isomorphism, where $$A$$ is an $$R$$-module, $$e_1, \ldots, e_n \in Hom_R(A, A)$$ a complete set of orthogonal idempotents of $$M$$ and $$A_i = e_i(A)$$.

It is clear to me how to prove that a map between modules is an $$R$$-isomorphism, but in this case I may need some hints how $$\varphi$$ is defined.

• My guess is projection maps $e_i : (x_1, x_2, \cdots, x_i, \cdots, x_n)\to (0, 0, \cdots, x_i,\cdots,0)$ Nov 22, 2017 at 17:22
• The obvious candidate (which Bumblebee points out) works, and the verification of $I-III$ is all routine after that. It's also trivial to show that $\phi = \sum e_i$ is the map you seek in the second part. Nov 22, 2017 at 17:31
• @Bumblebee and rschwieb. Thank you for your helpful hints :). Finally, I could solve the task. Nov 23, 2017 at 16:01
• @Crystal: That is great. I am happy to up see your answer. You can post the answer for your own question which would help for many students using this site. Nov 24, 2017 at 3:39
• @Bumblebee : I apologize for the late answer. Of course I can post my solution. In fact, the first part is trivially verified if one just considers your comment. I am sure that every student at my level will succeed in proving the first part. So I only post an answer to the second part. Dec 1, 2017 at 22:42

By the universal property of $\prod\limits_{i = 1}^n A_i$, there exists a unique homomorphism $\phi : A \rightarrow \prod\limits_{i = 1}^n A_i$ such that $\pi_i \circ \phi = e_i$ for all $i \in \{1, \ldots ,n\}$. By condition III we have $(e_1 + \cdots + e_n) \circ \phi = id_A$.
It remains to show that $\phi$ is a bijection.
Let $x \in ker(\phi)$. Then $(\pi_1 + \cdots + \pi_n) \circ \phi(x) = x$ implies that $x = 0$. Hence, $\phi$ is injective.
We have $A_i = e_i(A)$. Since $\pi_i$ is surjective, $\pi_i(\prod\limits_{i = 1}^n A_i) = A_i$. Moreover, $(\pi_i \circ \phi)(A) = e_i(A)$. It follows that $\pi_i(\prod\limits_{i = 1}^n A_i) = \pi_i(\phi(A))$ for all $i = 1, \ldots,n$.
Finally, $\prod\limits_{i = 1}^n A_i = \phi(A) \Rightarrow \phi$ is surjective.