Direct Products of Rings and Ideals .... Bland Problem 2(c)

[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(c) of Problem Set 2.1 ...

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

Homework Equations

The Attempt at a Solution

My attempt at a solution follows:We claim that $\bigoplus_\Delta R_\alpha$ is a right ideal of $\prod_\Delta R_\alpha$Proof ...Let $(x_\alpha ) , (y_\alpha ) \in \bigoplus_\Delta R_\alpha$ and let $(r_\alpha ) \in \prod_\Delta R_\alpha$Then $(x_\alpha ) + (y_\alpha ) = (x_\alpha + y_\alpha )$... by the rule of addition in direct products ...Now ... $x_\alpha + y_\alpha \in R_\alpha$ for all $\alpha \in \Delta$ ... by closure of addition in rings ...Thus $(x_\alpha + y_\alpha ) \in \prod_\Delta R_\alpha$ ...... but also ... since $(x_\alpha )$ and $(y_\alpha )$ each have only a finite number of non-zero components ...

... we have that $(x_\alpha + y_\alpha )$ has only a finite number of non-zero components ...

... so ...$(x_\alpha + y_\alpha ) \in \bigoplus_\Delta R_\alpha$..

Hence $(x_\alpha ) + (y_\alpha ) \in \bigoplus_\Delta R_\alpha$ ... ... ... ... ... (1)
Now we also have that ...$(x_\alpha ) (r_\alpha ) = (x_\alpha r_\alpha)$ ... ... rule of multiplication in a direct product ...

Now ... $x_\alpha r_\alpha \in R_\alpha$ for all $\alpha \in \Delta$ ... since a ring is closed under multiplication ...

and ...

$(x_\alpha r_\alpha)$ has only a finite number of non-zero components since $(x_\alpha )$] has only a finite number of non-zero components ...

So ... $(x_\alpha r_\alpha) \in \bigoplus_\Delta R_\alpha$

$\Longrightarrow (x_\alpha) (r_\alpha) \in \bigoplus_\Delta R_\alpha$ ... ... ... ... ... (2)
$(1) (2) \Longrightarrow \bigoplus_\Delta R_\alpha$ is a right ideal of $\prod_\Delta R_\alpha$
Can someone please critique my proof ... ... and either confirm its correctness or point out the errors and shortcomings ...

Such help will be much appreciated ...

Peter

 

[andrewkirk]

The proof looks sound. My only comment is that in (2) you have used the fact that in a ring, 0.r = 0 for any element r. That is a theorem, not a ring axiom, so should be stated rather than implicitly assumed. It would be over the top to prove it every time one uses it, but everybody should prove it once in their life, or read a proof of it.

 

[Math Amateur]

Hi Andrew ... thanks for the confirmation...

Also thanks for the point regarding 0.r = 0 ... I definitely missed that ...

Appreciate your help ...

Peter

 

[fresh_42]

Hi Andrew ... thanks for the confirmation...

Also thanks for the point regarding 0.r = 0 ... I definitely missed that ...

Appreciate your help ...

Peter

Here are the steps for this, which @andrewkirk mentioned: https://de.wikipedia.org/wiki/Ring_(Algebra)#Folgerungen
I suggest to try it on your own first and then look up the solution. It is the wrong language, but the words in this paragraph are so similar, that it doesn't matter.