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January 17th, 2012, 05:41 PM  #1 
Newbie Joined: Jan 2012 From: Tasmania, Australia Posts: 14 Thanks: 0  Ring Theory  0a = a0 = 0
In Dummit and Foote Chapter 7 Introduction to Rings, Proposition 1 Part (1) reads as follows: ============================ Let R be a ring. 0a = a0 = 0 for all a in R ============================ The 'proof' given is as follows: ============================================== The proposition follows from 0a = (0 + 0)a = 0a + 0a ============================================== I do not completely follow the proof. I can see that from the definition of the additive identity that a + 0 = 0 + a = a that if we put a = 0 we have 0 + 0 = 0 Thus using this equation and distributivity we can write 0a = (0 + 0)a = 0a + 0a BUT ... how/why does 0a  0a + 0a imply 0a = 0 ? Is it because the additive identity 0 is the only element in the ring for which x = x + x ? If so, how do we know this? Peter 
January 17th, 2012, 06:13 PM  #2 
Senior Member Joined: Aug 2010 Posts: 195 Thanks: 5  Re: Ring Theory  0a = a0 = 0
In a ring, every element has an additive inverse. So when we have the expression for some in our ring, we can add to both sides of the equation, yielding . What is the result? 
January 17th, 2012, 06:41 PM  #3 
Newbie Joined: Jan 2012 From: Tasmania, Australia Posts: 14 Thanks: 0  Re: Ring Theory  0a = a0 = 0
Thanks Turgui!! The result is 0 = x so if x + x = x then we know x is the additive identity! Thanks again peter 
January 19th, 2012, 02:30 PM  #4 
Senior Member Joined: Dec 2011 From: Argentina Posts: 216 Thanks: 0  Re: Ring Theory  0a = a0 = 0
We know that Let be the inverse additive of 
January 20th, 2012, 01:31 AM  #5 
Newbie Joined: Jan 2012 From: Tasmania, Australia Posts: 14 Thanks: 0  Re: Ring Theory  0a = a0 = 0
Thanks Watson! That shows a x 0 = 0! I suppose to show 0 x a = 0 we simply start from 0 x a = (0 + 0) x a = (0 x a) + (0 x a) and proceed similarly so that both a x 0 and 0 x a equal 0 and hence also each other! Peter 

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