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January 17th, 2012, 05:41 PM   #1
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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
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January 17th, 2012, 06:13 PM   #2
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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?
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January 17th, 2012, 06:41 PM   #3
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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
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January 19th, 2012, 02:30 PM   #4
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Re: Ring Theory - 0a = a0 = 0

We know that



Let be the inverse additive of

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January 20th, 2012, 01:31 AM   #5
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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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