Proving that b is 1 or -1 in an Integral Domain

[silvermane]

Homework Statement

Suppose R is an integral domain. Suppose c in R has c^2 = 1.
Prove that b must equal either 1 or -1.
Also, in Z/12Z, find an element b not equal to -1 or 1, with b^2 = 1

The Attempt at a Solution

So for the first part, I'm guessing it would be a good idea to factor (b^2 - 1) and notice that the roots are -1, 1. Is that all I have to show? I'm kinda worrying that I don't have enough here. I think I need to incorporate the fact that this is an integral domain. :/

For the second part (Z/12Z) I know that such an element,7^2 \equiv 49mod12 \equiv1mod12

 

[fzero]

The Attempt at a Solution

So for the first part, I'm guessing it would be a good idea to factor (b^2 - 1) and notice that the roots are -1, 1. Is that all I have to show? I'm kinda worrying that I don't have enough here. I think I need to incorporate the fact that this is an integral domain. :/

Well how does (b-1)(b+1)=0 contradict that R is an integral domain?

 

[silvermane]

Well how does (b-1)(b+1)=0 contradict that R is an integral domain?

Well, in an integral domain, the multiplicative identity is not equal to the additive identity. Thus wouldn't one of the statements (b-1) or (b+1) have to be zero to satisfy this statement?

I'm very new to the idea of an integral domain. Please let me know if I've misinterpreted what it means to be an integral domain. Thank you for your help too. :)

 

[fzero]

Well, in an integral domain, the multiplicative identity is not equal to the additive identity. Thus wouldn't one of the statements (b-1) or (b+1) have to be zero to satisfy this statement?

I'm very new to the idea of an integral domain. Please let me know if I've misinterpreted what it means to be an integral domain. Thank you for your help too. :)

You're correct. Therefore think in terms of a proof by contradiction of the stated lemma.

Edit: Actually, I was too quick. There's one more condition for a ring to be an integral domain. Quote the definition you're using in class (or text) if you need help applying it.

 

[Eynstone]

The Attempt at a Solution

So for the first part, I'm guessing it would be a good idea to factor (b^2 - 1) and notice that the roots are -1, 1. Is that all I have to show? I'm kinda worrying that I don't have enough here. I think I need to incorporate the fact that this is an integral domain. :/

For the second part (Z/12Z) I know that such an element,7^2 \equiv 49mod12 \equiv1mod12

Z/12Z is not an integral domain- 3*4=0.
You can factor b^2 - 1 = (b-1)(b+1) =0 (even if the domain is not commutative). If b-1 is not zero, it has an inverse.Multiply on the left by the inverse & we get b+1=0.

 

[silvermane]

Z/12Z is not an integral domain- 3*4=0.
You can factor b^2 - 1 = (b-1)(b+1) =0 (even if the domain is not commutative). If b-1 is not zero, it has an inverse.Multiply on the left by the inverse & we get b+1=0.

Thanks everyone for their help!

(pertaining to the second part, he just wanted an example of an element in Z/12Z that when squared is equivalent to 1.
Such an element is 7. When squared, we obtain 1mod12. It didn't really relate with the integral domain concept, but he wanted us understand what it meant when squaring an element gives 1.

 

[silvermane]

You're correct. Therefore think in terms of a proof by contradiction of the stated lemma.

Edit: Actually, I was too quick. There's one more condition for a ring to be an integral domain. Quote the definition you're using in class (or text) if you need help applying it.

An integral domain is a commutative ring with 1 ≠ 0 (the multiplicative identity is not equal to the additive identity) that has no zero divisors.

That's the definition we're using in class. Would I use contradiction for the fact that it should have no zero divisors?

 

[fzero]

An integral domain is a commutative ring with 1 ≠ 0 (the multiplicative identity is not equal to the additive identity) that has no zero divisors.

That's the definition we're using in class. Would I use contradiction for the fact that it should have no zero divisors?

Yes, compare (b-1)(b+1)=0 with the definition of zero divisor.