A ring element 'a' is called an idempotent if a^2=a. Prove that the only idempotents in an integral domain are 0 and 1.

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  awesome1563 answered this

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  Suppose a is an idempotent in an integral domain. Then a^2=a.
  
  So a^2 - a = 0.
  
  a(a-1) = 0
  
  So either a = 0 or a - 1 = 0 or a and a - 1 are zero divisors. But there are no nonzero zero divisors since it is an integral domain. So either a = 0 or a = 1.

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  SeriousPawn2640 answered this

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  i use x=a
  (i) For any idempotents x,y in R, (xy)^2
  = (xy)(xy)
  = x(yx)y
  = x(xy)y, since R is commutative
  = x^2 y^2
  = xy, since x^2 = x and y^2 = y.
  
  Hence, xy is idempotent.
  -----------------
  (ii) Assuming you mean Z6 x Z12:
  
  The idempotents in Z6 are 0, 1, 3, 4, and the idempotents in Z12 are 0, 1, 4, 9.
  (square these, and you'll get the same element back).
  
  So, there are 4 * 4 = 16 idempotents in Z6 x Z12 of the form
  (a, b) where a = 0, 1, 3, or 4, and b = 0, 1, 4, or 9.

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