MHB Ring Direct Products .... Bland Problem 3(a), Problem Set 2.1, Page 49 ....

[Math Amateur]

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:https://www.physicsforums.com/attachments/8062

My attempt at a solution follows:We claim that every 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:

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

 

[steenis]

Bland uses that $A$ is a right ideal of $R_1 \times R_2$.
So, if $a=(x,y) \in A$ and $u=(r,s) \in R_1 \times R_2$ then $au=(x,y)(r,s)$ exists and belongs to $A$.
Also $a(1,0)$ and $b(0,1)$, where $a,b \in A$, both exist and belong to $A$.
He then proves that $(a_1,a_2)=a(1,0)+b(0,1) \in A$
Now, try it again.

 

[Math Amateur]

Bland uses that $A$ is a right ideal of $R_1 \times R_2$.
So, if $a=(x,y) \in A$ and $u=(r,s) \in R_1 \times R_2$ then $au=(x,y)(r,s)$ exists and belongs to $A$.
Also $a(1,0)$ and $b(0,1)$, where $a,b \in A$, both exist and belong to $A$.
He then proves that $(a_1,a_2)=a(1,0)+b(0,1) \in A$
Now, try it again.

Hi Steenis ... sorry to be slow in responding ...

But Hugo (one of my friends large standard poodles - Hugo and Pierre...) chewed up my glasses ...

Will be back in touch soon ...

Peter

 

[steenis]

No worries, I fill my time with studying.

(Hugo and Pierre ?)

 

[Math Amateur]

No worries, I fill my time with studying.

(Hugo and Pierre ?)

Hugo and Pierre are the names of my friends large standard poodles ...

Peter

 

[steenis]

The point is: Hugo is my first name and Pierre = Peter is your first name ....

 

[Math Amateur]

The point is: Hugo is my first name and Pierre = Peter is your first name ....

Quite a coincidence ... :) ...

Peter

 

[steenis]

Does this make us antipodal poodles ?

 

[Math Amateur]

Does this make us antipodal poodles ?

Well ... the two standard poodles are intelligent... but quite eccentric ... :) ...

Peter