## Sunday, May 23, 2010

### GR9768.60: Ring

60. 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.
(A)  III only.
(B)  I and II only.
(C)  I and III only.
(D)  II and III only.
(E)  I, II, and III.

Solution:
Consider the ring $\mathbb{Z}_2 = \{0,1\}$. And observe that $0 = 0^2$ and $1 = 1^2$.
Then
$0+0 = 0$ and $1 + 1 = 0$ so I is true.
$(0+1)^2 = 1^2 = 1$ and $0^2 + 1^2 = 1$ so II is true.
$0 = 0 \cdot 1 = 1 \cdot 0 = 0$ so III is true.
-sg-

1. What about a general proof? This shows that these three properties are true for one example, not that they MUST be true. I am confused about property I in particular.

2. I found a proof here:
http://www.mathematicsgre.com/viewtopic.php?f=1&t=87

1. (s + s)^2 = s^2 + 2ss + s^2 = s + 2s + s = s + s, so 2s = 0

2. (s + t)^2 = s + t = s^2 + t^2

3. All rings are commutative with respect to addition, so here we want to show that S is commutative with respect to multiplication. (s + t)^2 = s^2 + st + ts + t^2 = s + t = s + st + ts + t
so st + ts = 0
We already know that s + s = 0, so each element is its own additive inverse. This means that ts = st.