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.

The answer is E.

-sg-

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.

ReplyDeleteI found a proof here:

ReplyDeletehttp://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.

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ReplyDelete