# Ring problem

1. Mar 12, 2012

### Stephen88

1. The problem statement, all variables and given/known data
Let R be a ring in which 1_R = 0_R .Show that R has only one element.

2. Relevant equations

3. The attempt at a solution
I'm trying to show that a*0_r=a*1_r implies a*0_r=0_r.
if 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? If not
Is there something I should do?

2. Mar 12, 2012

### tiny-tim

Hi Stephen88!
No, you should be trying to prove that a = 0.

Hint: what, by definition, is a*0 ?

what, by definition, is a*1 ?

3. Mar 12, 2012

### Stephen88

yes a = 0 is something I need to prove that was just I way that I thought it would be possible.
For instance a*0=a*(0+0)=a*0+a*0=a*1+a*1=2a=>a*0=2a= >a=0..otherwise 2=0 which is a contradiction.Is this correct?

4. Mar 12, 2012

### tiny-tim

Sorry, but this is fantasy …

there is no "2", is there?

All you're allowed to play with is a 0 and 1.

what, by definition, is a*0 ?

what, by definition, is a*1 ?​

5. Mar 12, 2012

### Stephen88

Uh sorry about that...I;m used to integers and real numbers....hmm..given the context both represent the identity of the ring so a*0=a*1=a

6. Mar 12, 2012

### tiny-tim

what, by definition, is a*0 ?

7. Mar 12, 2012

### Stephen88

a product of a and the multiplicative identity?

8. Mar 12, 2012

### tiny-tim

(the multiplicative zero)

9. Mar 12, 2012

### Stephen88

Yes normally 0 is the additive identity and 1 the multp. one...but here they are both the same.
So if 1=0->1*a=0*a=a=>a=1=0?
If not I'm confused

10. Mar 12, 2012

### tiny-tim

I can't tell whether you've got it or not … you need to make it clear.

Let's spell it out.

a*0 by definition is always 0

a*1 by definition is always a

but 0 = 1

so a*0 = a*1

so 0 = a