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Confusing ring problem

StefanM

New member
Jan 30, 2012
28
Let R be a ring in which 1_R = 0_R .Show that R has only one element.
I'm assuming the idea behind the problem is to prove that the additive identity and multiplicative identity are the same.This can only happen if either 1 or 0 or both are part of the Ring.
If R={1},then all the axioms that define a ring are satisfied.
If R={0},then again all the axioms that define a ring are also satisfied.....and if If R={0,1} it is the same thing so in order for 1_r=0_r the Ring must have one element which is either 0 or 1...is this correct?
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,503
Let R be a ring in which 1_R = 0_R .Show that R has only one element.
I'm assuming the idea behind the problem is to prove that the additive identity and multiplicative identity are the same.
What I find somewhat troubling about your post is the apparent confusion about the concept of a proof. The meaning of the phrase "If A, then B" and the ways of proving it, besides being familiar from everyday language, are studied in secondary school (e.g., using geometric proofs) or, at most, in the very first university courses. Certainly by the time students encounter rings and other topics of abstract algebra, they should know very well what a good proof is and how to construct one.

The problem can be reformulated as follows: "If a ring R has $0_R=1_R$, then R has a single element." To prove this statement, one assumes that $0_R=1_R$; one does not need to prove it.

This can only happen if either 1 or 0 or both are part of the Ring.
Of course, by using $1_R$ the problem implicitly assumes that this is a ring with a multiplicative unity. Additive unity always belongs to a ring.

If R={1},then all the axioms that define a ring are satisfied.
If R={0},then again all the axioms that define a ring are also satisfied.....and if If R={0,1} it is the same thing so in order for 1_r=0_r the Ring must have one element which is either 0 or 1...is this correct?
Checking that one-element set is a ring is the converse of what the problem is asking. The problem is to show that if some arbitrary ring, which a priori may be assumed to have many elements, has 0 = 1, then it in fact has only one element.

You need to know that $x\cdot 0 = 0$ in a ring (this is derivable from axioms). Then assume that x is an arbitrary ring element and prove that x = 0 using the fact that 1 = 0.
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
Let R be a ring in which 1_R = 0_R .Show that R has only one element.
I'm assuming the idea behind the problem is to prove that the additive identity and multiplicative identity are the same.
As Evgeny.Makarov said, you have this backward. You are given that the additive identity and multiplicative identity are the same. You are to prove that R contains no other elements.

This can only happen if either 1 or 0 or both are part of the Ring.
Yes, by definition every ring contains 0 (additive identity). Since you are told that 0_R= 1_R it must also contain 1 (multiplicative identity).

If R= {1}, then all the axioms that define a ring are also satisfied.....and if If R={0,1} it is the same thing so in order for 1_r=0_r the Ring must have one element which is either 0 or 1...is this correct?
You have done two cases- R= {1} and R= {0, 1} which are, of course, the same thing. Now what about the possible cases that R contains some element other than 0 or 1. Suppose the does in fact exist m in R not equal to either 0 or 1. What can you say about 0*m? What can you say about m*1? How does that give a contradiction?
 
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StefanM

New member
Jan 30, 2012
28
A multiplicative identity gives the following a*1_r=1_r*a=a...since 1_r=0_r then this implies a*0_r=a*1_r which is not true unless a=0...or a=1 which preserves the given identity equality.
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,503
A multiplicative identity gives the following a*1_r=1_r*a=a...since 1_r=0_r then this implies a*0_r=a*1_r which is not true unless a=0...or a=1 which preserves the given identity equality.
You are saying that $a\cdot0=a\cdot1$ implies $a=0$. This is correct, but you need to show this from the ring axioms.
 

StefanM

New member
Jan 30, 2012
28
a=1 is also an option,correct?
You are saying that a⋅0=a⋅1 implies a=0. This is correct, but you need to show this from the ring axioms.
I'm using the identity axioms....but apparently I'm doing something wrong
 

Evgeny.Makarov

Well-known member
MHB Math Scholar
Jan 30, 2012
2,503
a=1 is also an option,correct?
If a = 0, then also a = 1 because by assumption 0 = 1.

I'm using the identity axioms....but apparently I'm doing something wrong
As I said in post #2, you need to show that $a\cdot 0 = 0$ for all $a$. Start by representing 0 as 0 + 0.
 

StefanM

New member
Jan 30, 2012
28
Well if 0=0+0=>a*0=a*(0+0)=a*0+a*0=0.
Also a*0=0 should hold for all powers of a which can be proved by induction:
since a^2*0=a*a*0=a*0=0. ...and assuming this hold for n then a^(n+1)*0=a^n*a*0=0.
 

Evgeny.Makarov

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MHB Math Scholar
Jan 30, 2012
2,503
Well if 0=0+0=>a*0=a*(0+0)=a*0+a*0=0.
I agree until you get $$a\cdot0=a\cdot0+a\cdot0\hspace{3cm}(\mbox{*})$$ By how do you justify that $a\cdot0+a\cdot0=0$? I don't know how to show this immediately, but I know that $a\cdot0=a\cdot0+a\cdot0$ implies that $a\cdot0=0$. Indeed, since a ring is a group with respect to addition, there exists an additive inverse of $a\cdot0$. Add it to both sides of (*) and simplify both sides (which involves applying other axioms about addition).

Also a*0=0 should hold for all powers of a which can be proved by induction:
since a^2*0=a*a*0=a*0=0. ...and assuming this hold for n then a^(n+1)*0=a^n*a*0=0.
I agree, but I am not sure this is relevant to the problem.
 

StefanM

New member
Jan 30, 2012
28
0=0+0=>a*0=a*(0+0)=a*0+a*0=a*1_r+a*1_r=2a=>a*0=2a=>a=0....is this correct?
 

Evgeny.Makarov

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MHB Math Scholar
Jan 30, 2012
2,503
First, note that 2 is a mere notation for 1 + 1. For all we know, the ring in question may consists of dinasaurs rather than natural numbers, as long as someone gives names 0 and 1 to some dinos (in this case, to a single one) and defines two operations. The definition of the ring does not speak of 2, so the best way to define 2 is just 1 + 1. Also, note that multiplication in rings is not necessarily commutative, so we have no reasons to conclude that a * 1 + a * 1 = 2a. What we know by distributivity is that a * 1 + a * 1 = a(1 + 1) = a * 2 (the latter equality follows by the definition of 2).

Second, I still don't see how 0 * a = 2a implies a = 0 right away. My outline of the proof that a = 0 is in post #9.

You need to resist the temptation to use the facts you know about integers or real numbers. Integers is just one example of a ring, and not everything that is true for integers holds in all rings. Instead, you have to scrupulously apply the ring axioms one step at a time. Can you say which axiom is used in each particular equality? If not, this equality is unjustified and may be wrong in some rings. There is a tradeoff, which is common in mathematics. On the one hand, you can use only a limited number of very basic laws, which makes deriving things harder. On the other hand, your results apply not only to numbers, but to a huge collection of other objects: matrices, hyperreal and superreal numbers, permutations, etc.
 
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StefanM

New member
Jan 30, 2012
28
Very true.Ok let me try again:
a*0 is always 0 and a*1 is always a ....but in this context 0 = 1 therefore 0=1>a*0=a*1=a>0=a
 

Evgeny.Makarov

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MHB Math Scholar
Jan 30, 2012
2,503
a*0 is always 0 and a*1 is always a ....but in this context 0 = 1 therefore 0=1>a*0=a*1=a>0=a
Yes, but you have to explain why a * 1 = a and a * 0 = 0.
 

StefanM

New member
Jan 30, 2012
28
because 0 is the additive identity,the multiplicative zero so by definiton a*0=0...and 1 is the multiplicative identity so by definition a*1=a...
 

Evgeny.Makarov

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MHB Math Scholar
Jan 30, 2012
2,503
and 1 is the multiplicative identity so by definition a*1=a...
Yes. This is axiom #8 in MathWorld.

because 0 is the additive identity,the multiplicative zero so by definiton a*0=0...
The term "multiplicative zero" is undefined, at least, it does not occur in the ring definition. Unless you cite an axiom number from the ring definition, you don't have a proof. To repeat,
you have to scrupulously apply the ring axioms one step at a time. Can you say which axiom is used in each particular equality? If not, this equality is unjustified and may be wrong in some rings.
 

StefanM

New member
Jan 30, 2012
28
Ok ,by Axiom number 3.. a+0=a which can be written as a+0=a=a*1 which is true because of Axiom number 8 ..but 0=1 therefore a*1=a*0>a=0
 

Evgeny.Makarov

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MHB Math Scholar
Jan 30, 2012
2,503
Ok ,by Axiom number 3.. a+0=a which can be written as a+0=a=a*1 which is true because of Axiom number 8
Agree so far: we have a + 0 = a * 1.

but 0=1 therefore a*1=a*0>a=0
Yes, a * 1 = a * 0, but how does a = 0 follow?

Edit: It is better to use => rather than > as an abbreviation for "therefore" since > can be confused with "greater than."
 

StefanM

New member
Jan 30, 2012
28
Well...Since a*1=a by the axiom and a*0=0 because I'm multiplying with the additive identity or 0 then a=0...I guess..I'm walking in circles
 

Evgeny.Makarov

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MHB Math Scholar
Jan 30, 2012
2,503
Well...Since a*1=a by the axiom and a*0=0 because I'm multiplying with the additive identity or 0 then a=0
How do you know that a * 0 = 0?

You need to know that $x\cdot 0 = 0$ in a ring (this is derivable from axioms).
As I said in post #2, you need to show that $a\cdot 0 = 0$ for all $a$. Start by representing 0 as 0 + 0.
Yes, but you have to explain why ... and a * 0 = 0.
How many times do I need to suggest proving a * 0 = 0 by carefully applying ring axioms one at a time? And the way to do this was outlined in post #9:
I agree until you get $$a\cdot0=a\cdot0+a\cdot0\hspace{3cm}(\mbox{*})$$ By how do you justify that $a\cdot0+a\cdot0=0$? I don't know how to show this immediately, but I know that $a\cdot0=a\cdot0+a\cdot0$ implies that $a\cdot0=0$. Indeed, since a ring is a group with respect to addition, there exists an additive inverse of $a\cdot0$. Add it to both sides of (*) and simplify both sides (which involves applying other axioms about addition).
 

StefanM

New member
Jan 30, 2012
28
Yes after adding the additive inverse on both sides I get 0 on the left and 'a' on the right (since a*1=a*0=a. )
 

Evgeny.Makarov

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MHB Math Scholar
Jan 30, 2012
2,503
Bingo!