How Does kR Equal Zero in Ring Theory?

[PsychonautQQ]

Z = field of integers

.

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?

 

[Erland]

Take Z_k, the ring of integers modulo k, where k is a positive integer. Clearly, kr (that is, r+r+...+r, k times) equals 0 in Z_k, for any r in Z_k, right?

 

[PsychonautQQ]

Oh, thank you! So the complex plane, real numbers, integers all have char R = 0?

 

[Erland]

Oh, thank you! So the complex plane, real numbers, integers all have char R = 0?

Yes.

 

[WWGD]

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.

 

[Incnis Mrsi]

The ring ℤ of integer numbers isn’t a field. Skipped all after such introduction.