Direct Products .... Bland Probem 2(b), Problem Set 2.1 ....

[Math Amateur]

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 2(b) of Problem Set 2.1 ...

Problem 2(b) of Problem Set 2.1 reads as follows:

Homework Equations

The Attempt at a Solution

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

 

[andrewkirk]

The proof looks sound.

I like to state the justification for each move in a proof, except where it's staggeringly obvious.

So where you say $x_\alpha+y_\alpha\in A_\alpha$ I would add 'because ideals are closed under addition'.

 

[Math Amateur]

The proof looks sound.

I like to state the justification for each move in a proof, except where it's staggeringly obvious.

So where you say $x_\alpha+y_\alpha\in A_\alpha$ I would add 'because ideals are closed under addition'.

Thanks Andrew ...

Yes, understand your point regarding proofs ...

Peter