How Does kR Equal Zero in Ring Theory?

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Z = field of integers

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If R is a ring and k is an element of Z, write kR = {kr | r is an element of R}. It is not too difficult to verify that {k is an element of Z | kR = 0} is an additive subgroup of Z.

I am confused on how kR would equal 0? Wouldn't that mean that k would have to equal zero? How could all of the elements of R multiplied by some integer k ever equal 0?

The book goes on to say that kR = 0 if and only if k1 = 0, where 1 = the unity of R.

Anyone want to help me understand what's going on here?
 
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Take Zk, the ring of integers modulo k, where k is a positive integer. Clearly, kr (that is, r+r+...+r, k times) equals 0 in Zk, for any r in Zk, right?
 
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Oh, thank you! So the complex plane, real numbers, integers all have char R = 0?
 
PsychonautQQ said:
Oh, thank you! So the complex plane, real numbers, integers all have char R = 0?
Yes.
 
Quick comment and a nitpick ; nitpick , re post #1 ,is that the integers are not a field, they are a ring. As a general comment notice that you may have non-trivial torsion in infinite algebraic structures.
 
The ring ℤ of integer numbers isn’t a field. Skipped all after such introduction.
 
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