To show a ring of order p (prime) is isomorphic to the integers mod p.

  • Thread starter rachellcb
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If R is a finite ring of of order p where p is prime, show that either R is isomorphic to Z/pZ or that xy=0 for all x,y in R


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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jgens
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Suppose that [itex](R,+,\cdot)[/itex] is not a trivial ring. Notice that [itex](R,+) \cong (\mathbb{Z}/p\mathbb{Z},+)[/itex] by the Lagrange Theorem.

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:
  1. 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].
  2. 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.
Using the fact that [itex]R[/itex] contains no zero-divisors, you should pretty easily be able to show that [itex]R[/itex] is a unital ring; and therefore, [itex]R[/itex] is an integral domain. From there, it should be pretty simply to construct the desired isomorphism.
 

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