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

[darkchild]

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.

 

[deluks917]

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.)?

 

[darkchild]

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.

 

[Dick]

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.

 

[Hurkyl]

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.

 

[darkchild]

(-s)^2=s^2.

Yes, but that doesn't imply that s=-s, right?

 

[Hurkyl]

Yes, but that doesn't imply that s=-s, right?

What identity does this ring satisfy?

 

[lavinia]

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

 

[darkchild]

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

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

Oh, I see how II works now.

 

[darkchild]

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