Let r, s integers such that r/ ks+1. For some k belogs to [1,r)
Prove that the subset {r,2r,...,(s-1)r} of Z/rsZ es a ring with identity ks+1.
I tried to do it but I couldīt. Please help
Ok, I thought that {r,2r,...} is a sibset of a ring, so if I want to see if is a ring or not, is only necessary to prove is a subring of Z/rsZ. So,
I consider 2 different elements in {r,2r,...}, lets say $\displaystyle $s_1r$$ and $\displaystyle $s_2r$$. Then:
$\displaystyle $s_1r - s_2r$$= $\displaystyle $(s_1-s_2)r$$ But, is this element is my subset?
$\displaystyle $(s_1r)(s_2r)$$ = ??
And I really dont know what to do with the identity element given....
well if s2 < s1 < s, then certainly (s1-s2)r is in {r,2r,3r,.....,(s-1)r}. but hopefully you just forgot to include 0, because otherwise s1 = s2 leads to certain problems, eh?
as for your second question, (s1r)(s2r) = (s1s2r)r, but remember, we are working mod rs, so we wind up with (s1s2r (mod rs))r.
my bad i had a typo. it should have been (s1s2r)r.
after all s1,s2 and r are integers, correct? so (s1r)(s2r) is still SOME multiple of r (which may be some other multiple when s1s2r is reduced mod rs, but will still be some multiple of r).
proving the set {0,r,2r,...,(s-1)r} is a subring of Z/(rs)Z isn't the interesting part.
showing it has a multiplicative identity is.
Ok, for the multiplicative identity I have done the following:
First we know that r divides ks+1 then : ar= ks+1 , some interger a.
Now , to prove ks+1 is the identity I have to check that if mr belongs tu my subset then mr (ks+1) = mr.
So: mr (ks+1) = mr (ar) = (mra)r = (amr)r, but arm is congruent with 0(mod mr ) son arm = mr, so mr (ks+1) = mr (ar) = (mra)r = (amr)r = (mr)r Now I dont know what to do....
well, look at it this way:
ks+1 = ar, so ks+1 is in our set of multiples of r (mod rs) (it might look different when reduced).
(ks+1)(mr) = (ks)(mr) + (mr) = (km)(rs) + (mr), and because we are working mod (rs)......?
sometimes it helps to look at a specific example.
suppose r = 4, and s = 13. then r divides 1 + 3(13), so we have the following set:
{0,4,8,12,16,20,24,28,32,36,40,44,48} in Z/52Z.
according to our theorem, 1+ks = 40 should be a multiplicative identity.
for example (28)(40) = 1,120 = (52)(21) + 28, so (28)(40) = 28 (mod 52).
and in fact, 40 = (3)(13) + 1, so
(4m)(40) = (4m)((3)(13) + 1) = (3m)(52) + 4m = 4m (mod 52). see how that works?