- #1

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## Homework Statement

I am reading Paul E. Bland's book: Rings and Their Modules and am currently focused on Section 2.1 Direct Products and Direct Sums ... ...

I need someone to check my solution to Problem 3(a) of Problem Set 2.1 ...

Problem 3(a) of Problem Set 2.1 reads as follows:

## Homework Equations

## The Attempt at a Solution

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My attempt at a solution follows:We claim that

__right ideal of the ring ##R_1 \times R_2 \times \ ... \ ... \ \times R_n## is of the form ##A_1 \times A_2 \times \ ... \ ... \ \times A_n## ...Proof:__

*every*Suppose ##A## is a right ideal of ##R_1 \times R_2 \times \ ... \ ... \ \times R_n## ...

Let ##a \in A## and put ##A_1 = \pi_1 (A) , A_2 = \pi_2 (A) , \ ... \ ... \ , A_n = \pi_n (A)##

Now ##a \in A## ...##\Longrightarrow \pi_1(a) = a_1, \pi_2(a) = a_2, \ ... \ ... \ , \pi_n(a) = a_n##

for some ##a_1 \in A_1, a_2 \in A_2, \ ... \ ... \ , a_n \in A_n##Hence ...

##a = ( i_1 \pi_1 + i_2 \pi_2 + \ ... \ ... \ + i_n \pi_n ) (a)####= (a_1, 0, 0, \ ... \ ... \ , 0) + (0, a_2, 0, \ ... \ ... \ , 0) + \ ... \ ... \ + ( 0, 0, \ ... \ ... \ , a_n ) ####= ( a_1, a_2, \ ... \ ... \ , a_n)##Hence ... ##A \subseteq A_1 \times A_2 \times \ ... \ ... \ \times A_n## ... ... ... ... ... (1)

*Conversely ...*

Let ##a_1 \in A_1, a_2 \in A_2, \ ... \ ... \ , a_n \in A_n##Note that again ... ##A## is a right ideal of ##R_1 \times R_2 \times \ ... \ ... \ \times R_n## ...

... and ##a \in A## and put ##A_1 = \pi_1 (A) , A_2 = \pi_2 (A) , \ ... \ ... \ , A_n = \pi_n (A)##Then there are ##b_1, b_2, \ ... \ ... \ , b_n \in A## such that ...

##\pi_1 (b_1) = a_1, \pi_2 (b_2) = a_2, \ ... \ ... \ , \pi_n (b_n) = a_n## ... Hence ...

##b_1 ( 1,0,0, \ ... \ ... \ , 0 ) + b_2 ( 0, 1,0, \ ... \ ... \ , 0 ) + \ ... \ ... \ + b_n ( 0, 0,0, \ ... \ ... \ , 1 )####= ( i_1 \pi_1 + i_2 \pi_2 + \ ... \ ... \ + i_n \pi_n ) ( b_1 ( 1,0,0, \ ... \ ... \ , 0 ) + b_2 ( 0, 1,0, \ ... \ ... \ , 0 ) + \ ... \ ... \ + b_n ( 0, 0,0, \ ... \ ... \ , 1 ) )####= ( a_1, a_2, \ ... \ ... \ , a_n)##So ... ##A_1 \times A_2 \times \ ... \ ... \ \times A_n \subseteq A## ... ... ... ... ... (2)Now ... ##(1), (2) \Longrightarrow A = A_1 \times A_2 \times \ ... \ ... \ \times A_n##

Can someone please critique my proof ... and either confirm it is correct or point out the errors and shortcomings ... ...

**Problem/Issue**... there is part of the above proof I do not fully understand ... I will relate the issue to text solution for ##n = 2## ...In Bland's text on the problem we read the following:

"... ... Hence ##a(1,0) + b(0,1) = ( i_1 \pi_1 + i_2 \pi_2 ) ( a(1,0) + b(0,1) ) = (a_1, a_2)##, so ##A_1 \times A_2 \subseteq A.## ... ... "I have two questions regarding the above quote:(1) Exactly why/how is the equation ##a(1,0) + b(0,1) = ( i_1 \pi_1 + i_2 \pi_2 ) ( a(1,0) + b(0,1) ) = (a_1, a_2)## ... true?

Can someone please explain in detail why/how this is true ...(2) Exactly why/how does the above equation being true imply that ##A_1 \times A_2 \subseteq A## ... ?

Help with the above will be much appreciated ...

Peter