- #1

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I know that both R and Z/pZ have the same number of elements (up to equivalence) and that R isomorphic to Z/pZ implies R must be cyclic (I think) but am otherwise lost on where to start. Help much appreciated!

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- Thread starter rachellcb
- Start date

- #1

- 10

- 0

I know that both R and Z/pZ have the same number of elements (up to equivalence) and that R isomorphic to Z/pZ implies R must be cyclic (I think) but am otherwise lost on where to start. Help much appreciated!

- #2

jgens

Gold Member

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Let [itex]r \in R[/itex] be an element generating the additive group of [itex]R[/itex]. If [itex]a,b \in R[/itex] are non-zero, then it follows that [itex]a = r + \cdots + r[/itex] (n times) and [itex]b = r + \cdots + r[/itex] (m times) where neither [itex]n,m[/itex] divides [itex]p[/itex]; and in particular it follows that [itex]ab = r^2 + \cdots + r^2 \; (mn \; \mathrm{times}) = ba[/itex]. This proves two things:

- If [itex]r^2 = 0[/itex], then [itex](R,+,\cdot)[/itex] would be a trivial ring. This means that [itex]r^2 \neq 0[/itex].
- Since [itex]p[/itex] is prime and [itex]r^2 \neq 0[/itex], this means that [itex]ab \neq 0[/itex]. This shows that [itex]R[/itex] contains no zero-divisors.

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