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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 2(b) of Problem Set 2.1 ...
Problem 2(b) of Problem Set 2.1 reads as follows:
View attachment 8060
My attempt at a solution follows:We claim that $$\bigoplus_\Delta A_\alpha$$ is a right ideal of $$\prod_\Delta R_\alpha$$ Proof ... Let $$(x_\alpha ) , (y_\alpha ) \in \bigoplus_\Delta A_\alpha$$ and let $$(r_\alpha ) \in \prod_\Delta R_\alpha$$Then $$ (x_\alpha ) + (y_\alpha ) = (x_\alpha + y_\alpha )$$ ...
... further ... if $$(x_\alpha )$$ has $$m$$ non-zero components and $$(y_\alpha $$) has $$n$$ non-zero components then $$ (x_\alpha + y_\alpha )$$ has at most $$(m+n)$$ non-zero components ... that is $$ (x_\alpha + y_\alpha )$$ has only a finite number of non-zero components ...
... so ... since each $$x_\alpha + y_\alpha \in A_\alpha$$ we have that $$(x_\alpha + y_\alpha ) \in \bigoplus_\Delta A_\alpha$$ ...
Hence ... $$(x_\alpha ) + (y_\alpha ) \in \bigoplus_\Delta A_\alpha$$ ... ... ... ... ... (1)
Also ... we have ...
$$(x_\alpha ) (r_\alpha ) = (x_\alpha r_\alpha)$$
... and assuming $$x_\alpha$$ has $$m$$ non-zero components, then $$(x_\alpha r_\alpha$$) has at most $$m$$ non-zero components ...... and ...$$x_\alpha r_\alpha \in A_\alpha$$ since $$A_\alpha$$ is a right ideal of $$R_\alpha$$so $$(x_\alpha r_\alpha) \in \bigoplus_\Delta A_\alpha$$and it follows that $$(x_\alpha) ( r_\alpha) \in \bigoplus_\Delta A_\alpha$$ ... ... ... ... ... (2)$$(1) (2) \Longrightarrow \bigoplus_\Delta A_\alpha$$ is a right ideal of $$\prod_\Delta R_\alpha $$
Can someone please critique my proof either by confirming it to be correct and/or pointing out errors and shortcomings ...
Peter
I need someone to check my solution to Problem 2(b) of Problem Set 2.1 ...
Problem 2(b) of Problem Set 2.1 reads as follows:
View attachment 8060
My attempt at a solution follows:We claim that $$\bigoplus_\Delta A_\alpha$$ is a right ideal of $$\prod_\Delta R_\alpha$$ Proof ... Let $$(x_\alpha ) , (y_\alpha ) \in \bigoplus_\Delta A_\alpha$$ and let $$(r_\alpha ) \in \prod_\Delta R_\alpha$$Then $$ (x_\alpha ) + (y_\alpha ) = (x_\alpha + y_\alpha )$$ ...
... further ... if $$(x_\alpha )$$ has $$m$$ non-zero components and $$(y_\alpha $$) has $$n$$ non-zero components then $$ (x_\alpha + y_\alpha )$$ has at most $$(m+n)$$ non-zero components ... that is $$ (x_\alpha + y_\alpha )$$ has only a finite number of non-zero components ...
... so ... since each $$x_\alpha + y_\alpha \in A_\alpha$$ we have that $$(x_\alpha + y_\alpha ) \in \bigoplus_\Delta A_\alpha$$ ...
Hence ... $$(x_\alpha ) + (y_\alpha ) \in \bigoplus_\Delta A_\alpha$$ ... ... ... ... ... (1)
Also ... we have ...
$$(x_\alpha ) (r_\alpha ) = (x_\alpha r_\alpha)$$
... and assuming $$x_\alpha$$ has $$m$$ non-zero components, then $$(x_\alpha r_\alpha$$) has at most $$m$$ non-zero components ...... and ...$$x_\alpha r_\alpha \in A_\alpha$$ since $$A_\alpha$$ is a right ideal of $$R_\alpha$$so $$(x_\alpha r_\alpha) \in \bigoplus_\Delta A_\alpha$$and it follows that $$(x_\alpha) ( r_\alpha) \in \bigoplus_\Delta A_\alpha$$ ... ... ... ... ... (2)$$(1) (2) \Longrightarrow \bigoplus_\Delta A_\alpha$$ is a right ideal of $$\prod_\Delta R_\alpha $$
Can someone please critique my proof either by confirming it to be correct and/or pointing out errors and shortcomings ...
Peter