1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Mar 18, 2012 #1
    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!
     
  2. jcsd
  3. Mar 18, 2012 #2

    jgens

    User Avatar
    Gold Member

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




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