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1=0 only in trivial {0} ring. 
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#1
Feb811, 02:40 PM

P: 8

Hi, I found a couple of proofs proving that 1=0 only in the trivial ring {0}. They say
Suppose 1 = 0. Let a be any element in R; then a = a ⋅ 1 = a ⋅ 0 = 0. But what I don't understand is that they say a = a ⋅ 1. But that is only true if a ring has unity (x*1=1*x=x), and it is possible to have a ring without unity, so why is it okay to say a = a ⋅ 1? 


#2
Feb811, 02:49 PM

Sci Advisor
P: 905

What could 1 possibly mean in a ring without unity?



#3
Feb811, 02:57 PM

P: 8

Yeahh okay that's what I was thinking. We know 1 is in R.... and there is no other way for the number one to behave... 1*x = x always. And so since 1 is in R, we must have unity. Thanks!



#4
Feb811, 03:03 PM

Sci Advisor
P: 905

1=0 only in trivial {0} ring.
So the correct statement should be: Let R be a ring with 1. If 1=0, then R={0}. The first sentence is essential, because otherwise the second sentence does not make any sense. 


#5
Feb811, 04:19 PM

P: 8

Okay.. but if we are talking about a ring where 1=0, don't we already know 1 is in the ring?



#6
Feb811, 04:25 PM

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P: 905




#7
Feb811, 04:34 PM

P: 8

I'm sorry, I'm just trying to understand... my initial question is: we want to prove that the only time 1=0 is in the trivial ring {0}. And in the proof, it is said a=a*1. And so I am trying to clarify... we can use the property a=a*1, because we are talking about a ring where 1=0, we know the ring contains the identity element 1 since 1=0 in our ring? http://en.wikipedia.org/wiki/Proofs_...ing_properties



#8
Feb811, 04:42 PM

Sci Advisor
P: 905




#9
Feb811, 04:50 PM

P: 8

So if our ring does not contain 1... then our ring does not have unity (there is no element such that a*1=a). Then 1=0 would mean... I'm not sure.. that the only element must be 0 because 1's not in there?



#10
Feb911, 07:28 AM

P: 836

I think this sounds like a contradiction. First you say 1 is not in R. Then you say 1=0 leads to R={0}?
What do you think is confusing about Landau's statement? 


#11
Feb911, 08:45 AM

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P: 1,741




#12
Feb1011, 05:45 PM

P: 209

Proof:
If all "a" in R(Ring) such that ab = b = ba then on one hand "b" is a zero. Then consider if ab = a = ba, then all "b" is identity. But then all elements are identity and zero, so this set is trivial, only one element is acting on itself. Equivalently, the first equation says all "a" is identity" and the second equation says all "a" is "identity", but then there must only be one element because everything is zero and identity.... There 1=0 and set is trivial. QED 


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