Re: Is every subring of Z is of the form mZ ?
- From: hagman <google@xxxxxxxxxxxxx>
- Date: Sun, 10 Jun 2007 10:30:28 -0700
On 10 Jun., 16:54, student <n...@xxxxxx> wrote:
Can somebody help me with this? It is easily understood that every
subset of Z of the form mZ is a subring of Z, but, why the opposite?
In fact, every subgroup H of the (additive) group Z is of the from mZ.
The trivial case is {0} = 0Z.
If H is not trivial and a in H\{0} then -a in H, hence H contains
at least one positive integer.
Let m be the smallest positive element of H.
Then clearly mZ subset H.
Assume h in H. Then h=q*m+r for q in Z, 0<=r<m (division with
remainder).
Thus r=h-q*m is in H.
By choice of m, we have r=0.
Therefore H=mZ.
.
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