My Math Forum Ring Theory - 0a = a0 = 0

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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 $x= x + x$ for some $x$ in our ring, we can add $-x$ to both sides of the equation, yielding $x + (-x)= x + x + (-x)$. 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 $a\times0= a\times0 + a\times0$ Let $w$ be the inverse additive of $a\times0$ 
 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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