1
$\begingroup$

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.

Thanks for your help.

$\endgroup$
5
  • 1
    $\begingroup$ My guess is projection maps $e_i : (x_1, x_2, \cdots, x_i, \cdots, x_n)\to (0, 0, \cdots, x_i,\cdots,0)$ $\endgroup$
    – Bumblebee
    Nov 22, 2017 at 17:22
  • 1
    $\begingroup$ 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. $\endgroup$
    – rschwieb
    Nov 22, 2017 at 17:31
  • $\begingroup$ @Bumblebee and rschwieb. Thank you for your helpful hints :). Finally, I could solve the task. $\endgroup$
    – Crystal
    Nov 23, 2017 at 16:01
  • $\begingroup$ @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. $\endgroup$
    – Bumblebee
    Nov 24, 2017 at 3:39
  • 1
    $\begingroup$ @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. $\endgroup$
    – Crystal
    Dec 1, 2017 at 22:42

1 Answer 1

1
$\begingroup$

Since the first part is obvious, I answer the second part of the question.

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .