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.
The answer is E.
-sg-






3 comments:

  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.

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  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.

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  3. Could you review my website, freestudytools.com and possibly suggest it as a way to access free study material for the GRE? The website is totally free, but members just need to create and rate questions and flash cards in order to keep their account active.

    ReplyDelete