Ring theory- zero divisors and integral domains

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Homework Statement



Consider the ring Z/mZ, show that S = {[0], [a], [2a], · · · , [m − a]} forms a (possibly
nonunitary) subring of Z/mZ when a divides m. (i.e. show that (S,+, ·) is closed
the usual addition and multiplication. (We are not require to find a multiplicative identity).



The Attempt at a Solution



Since a divides m then m=ab so I tried subbing in ab for m and got [m-a]=[ab-a]=[a(1-b)]... but not too sure where to go from here. From looking at the set S it does not seem to be closed under addition or multiplication? Just a hint at how to go about/start/ approach this question would much appreciated! THank you!
 
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Well, let's check closure under addition. Two general elements of ##S## look like ##[ra]## and ##[sa]##, where ##r## and ##s## are integers. So what is ##[ra] + [sa]##? Evaluate it in the ring ##\mathbb{Z}/m\mathbb{Z}##, and see if the answer is in ##S##.
 
OK, I see how to prove it now- the addition of two elements of S will always give a number which is a multiple of a, therefore will be an element of S. (Similarly for multiplication)...?

(Thank you)
 
Yes, similarly for multiplication.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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