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Abstract algebra: proving an ideal is maximal, Constructing quotient rings

  1. May 8, 2012 #1
    1. The problem statement, all variables and given/known data
    M = {(pa,b) | a, b are integers and p is prime}
    Prove that M is a maximal ideal in Z x Z

    2. Relevant equations



    3. The attempt at a solution

    I know that there are two ways to prove an ideal is maximal:

    You can show that, in the ring R, whenever J is an ideal such that M is contained by J, then M=J or J=R.

    Or you can show that the quotient ring R/M is a field.

    I think it will be much easier to show that R/M is a field, but I'm not familiar with how to construct it from the given information. My understanding is that it is the set of all cosets of M (congruence classes modulo M).

    Can anyone point me in the right direction? Thanks.
     
  2. jcsd
  3. May 9, 2012 #2
    I assume that p is fixed in M?

    Do you know what Z/pZ is isomorphic to, for a given prime p?

    Do you know how to show that ZxZ / Zx{0} is isomorphic to Z?
     
  4. May 9, 2012 #3
    P is fixed.

    I know that Z/pZ is a field, for a given prime p. Is that what you mean?

    To show that ZxZ /Zx{0} is isomorphic to Z, I'd need to create a bijective function from ZxZ / Zx{0} where f(a + b) = f(a) + f(b) and f(ab)= f(a)*f(b).

    Are you hinting that since p is a prime integer, Z/M must be a field?
     
  5. May 9, 2012 #4
    Take a look at what M is, when written in the form RxS for the sets R and S. You will see that both R and S are well known sets and that should help you figure out what ZxZ/M should look like.
     
  6. May 9, 2012 #5
    To clarify what I mean, consider that {0}xZ = {(a,b) | a = 0 and b is an integer}.
     
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