Let $R=R_1\oplus R_2\oplus\cdots\oplus R_n$ be the external direct sum of a finite number of rings $R_i$ with identity; $i=1,2,\ldots,n$. For fixed $i$, I could prove that $\pi_i:R\to R_i$ given by $\pi((a_1,a_2,\ldots,a_n))=a_i$ is an onto homomorphism. Now it is required to prove that an ideal $I$ of $R$ is of the form $I=I_1\oplus I_2\oplus...\oplus I_n,$ with $I_i$ an ideal of $R_i$. The following is my attempt.

Let $I$ be an ideal of $R$. Put $I_i=\pi_i(I)$. Then $I_i$ is an ideal of $R_i$ as the map $\pi_i$ is an onto homomorphism. Now for any $a_i\in I_i$, we can choose $(a_1,\ldots,a_i,\ldots,a_n)\in I$. Now since $(0,\ldots,1,\ldots,0)\in R$ and $I$ is an ideal of $R$ we have $$(a_1,\ldots,a_i,\ldots,a_n)(0,\ldots,1,\ldots,0)=(0,\ldots,a_i,\ldots,0)\in I.$$

Put $J_i=\{0,\ldots,a_i,\ldots,0):a_i\in I_i\}$. Then $J_i$ is an ideal of $I$ such that $J_i\cong I_i$. Moreover $I=J_1\dotplus J_2\dotplus\cdots\dotplus J_n$, where $\dotplus$ denotes internal direct sum. Hence $I=I_1\oplus I_2\oplus\cdots\oplus I_n$.

Is this argument alright? Thanks.


Your argument is fine. You could simplify the notation by proving it in the case where $R = R_1\oplus R_2$ and then extending by induction.

  • $\begingroup$ Ok. Thank you very much. $\endgroup$ – Janitha357 Jul 20 '17 at 18:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.