Subring Math Problem: Find Number of Elements in \mathbb{Z}_{2000} Subring

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Homework Statement



Let R be a ring and a be an element of R. Let [tex]S= \left\{ x \in R: ax=0_R \right\}[/tex]. S is a subring of R.

Let [tex]R= \mathbb{Z}_{2000}[/tex] and [tex]a=850[/tex]. Determine the elements of the subring S as defined previously. How many elements are in S?


The Attempt at a Solution



The elements of the subring S will be elements x from [tex]\mathbb{Z}_{2000}[/tex] such that [tex]850.x=0_R[/tex].

And I think since 850x=0-5000n, [tex]x= \frac{2000}{850} n = \frac{40}{21} n[/tex] then

n=k.21

But what I do I need to do to find the number of elements in S? Is there a quick way of finding this?
 
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What's the largest multiple of 21 in Z/2000? What k does this correspond to? The other elements of S follow from this.
 


fzero said:
What's the largest multiple of 21 in Z/2000? What k does this correspond to? The other elements of S follow from this.

The largest multiple of 21 [tex]\mathbb{Z}_{2000}[/tex] is 1995. It ocrresponds to k=95, since 21x95=1995. What do you mean "other elements of S follow from this"? How do I need to figure out how many elements are in S?
 


roam said:
The largest multiple of 21 [tex]\mathbb{Z}_{2000}[/tex] is 1995. It ocrresponds to k=95, since 21x95=1995. What do you mean "other elements of S follow from this"? How do I need to figure out how many elements are in S?

Sorry, I thought your method actually determined the elements of S. I would look at the prime factorizations 2000 = 24 53, 850 = 2 52 17. Now, by comparing the prime factorizations, what is the smallest x (call it xg) such that a xg = 0R? Now note that all multiples m xg are also in S.
 


fzero said:
Sorry, I thought your method actually determined the elements of S. I would look at the prime factorizations 2000 = 24 53, 850 = 2 52 17. Now, by comparing the prime factorizations, what is the smallest x (call it xg) such that a xg = 0R? Now note that all multiples m xg are also in S.

Firstly, how do you determine this xg from the prime factorization? Also, how does it help to determine the number of elements in S?
 


roam said:
Firstly, how do you determine this xg from the prime factorization? Also, how does it help to determine the number of elements in S?

The condition [tex]a x =0_R[/tex] requires that [tex]a x = 2^4 5^3 k[/tex] for some k. Looking at the prime factors in a leads us to conclude that [tex]x \in S[/tex] have the form [tex]x_{k'}=2^m 5^n k'[/tex]. [tex]m,n[/tex] are easily determined, while the [tex]k'=1,\ldots k'_{\text{max}}[/tex] are constrained by the condition that [tex]x\in R[/tex].
 


fzero said:
The condition [tex]a x =0_R[/tex] requires that [tex]a x = 2^4 5^3 k[/tex] for some k. Looking at the prime factors in a leads us to conclude that [tex]x \in S[/tex] have the form [tex]x_{k'}=2^m 5^n k'[/tex]. [tex]m,n[/tex] are easily determined, while the [tex]k'=1,\ldots k'_{\text{max}}[/tex] are constrained by the condition that [tex]x\in R[/tex].

Could you please explain a bit more and maybe give some examples? Because I'm very confused... I think all of the elements which will be zero must be factors of 2000=2.52.17. I mean 850 x (something x k)=2000k' Do I need to try these:

400x 24.5 x k
200 x 23.52 k
1000 x 24.53 k

for different k's.
 


Since [tex]850 = (2) (5^2) ( 17)[/tex], we compute

[tex]a x_{k'} = 2^{m+1} 5^{n+2} 17 k' .[/tex]

This is [tex]0 (\mod 2000)[/tex] if [tex]m=3,n=1[/tex], so

[tex]x_{k'} = 40 k'.[/tex]

Note that (850)(40)=(17)(2000), so your intuition is correct. Since 17 is prime, 40 is the smallest integer for which this works. Now [tex]x_{50} = 2000 = 0_R[/tex] gives us [tex]k'_{\text{max}}[/tex].