Answer: Properties of A Ring: I, II, & III True?

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Homework Statement


If s is a ring with the property that s=s^{2} for each
s\in S, which of the following must be true?

I. s + s = 0 for each s in S.

II. (s+t)^{2}=s^{2}+t^{2} for each s,t in S.

III. S is commutative

Homework Equations



none

The Attempt at a Solution



The answer is I, II, and III. I understand why III is true, but not the other two. How can s + s = 0 for all s?!? In fact, I don't see how this can be a ring at all, since there don't appear to be any additive inverses in the set.

For II, I tried this:
(s+t)^{2}=(s^{2}+t^{2})^{2}, which is only equal to s^{2}+t^{2} when both s and t are the additive identity element.
 
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s is its own additive inverse if s + s = 0. Do any of those properties imply each other (two of them together imply a third one.)?
 
deluks917 said:
s is its own additive inverse if s + s = 0. Do any of those properties imply each other (two of them together imply a third one.)?

I don't see how one would figure that s + s = 0, though. As far as my understanding goes, only III is true.
 
darkchild said:
I don't see how one would figure that s + s = 0, though. As far as my understanding goes, only III is true.

(-s)^2=s^2.
 
I hate to say it, but II is immediately obvious from the form of the equation and the identity satisfied by elements of the ring.
 
Dick said:
(-s)^2=s^2.

Yes, but that doesn't imply that s=-s, right?
 
darkchild said:
Yes, but that doesn't imply that s=-s, right?
What identity does this ring satisfy?
 
darkchild said:

Homework Statement


If s is a ring with the property that s=s^{2} for each
s\in S, which of the following must be true?

I. s + s = 0 for each s in S.

II. (s+t)^{2}=s^{2}+t^{2} for each s,t in S.

III. S is commutative

I:(s+s)^{2}=s + s = 4s^2 = 4s

II: (s+t)^{2}=s+t
 
lavinia said:
I:(s+s)^{2}=s + s = 4s^2 = 4s

II: (s+t)^{2}=s+t

Oh, I see how II works now.
 
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Hurkyl said:
What identity does this ring satisfy?

Ok, I see. -s must be in S by definition of a ring, and
-s=(-s)^{2}=s^{2}=s. Thanks
 
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