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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.

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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.

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

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