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Homework Help: A subring in the same domain?

  1. Mar 30, 2007 #1
    1. The problem statement, all variables and given/known data
    If A is a subring of B and B is in a particular domain like a UFD than does it imply A is also in that domain B lives in hence also a UFD?


    3. The attempt at a solution
    Z is a ring but not a field. However Z mod p is a field. And Z mod p is a subring of Z. Although Z mod p is a ring also so a subring may live inside a more 'specialised' domain but can always be denoted the name of the domain it is a subring of. So the answer is yes to my above question.
     
    Last edited: Mar 30, 2007
  2. jcsd
  3. Mar 30, 2007 #2
    Actually it is possible to construct a subring of the rationals which is a field such that this subring is not a field. So the answer is no. i.e the subring could be all fractions in Q such that the denominator does not contain a factor integer.
     
  4. Mar 30, 2007 #3

    matt grime

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    Z mod p is definitely not a subring of the integers.

    Also, fields are a silly thing to talk about as UFDs since every element is a unit. And why is your second example (the p-locals) not a UFD? Suppose we choose

    Z_{(p)} = { a/b in Q such that hcf(b,p)=1}

    then the only non-units are powers of p, and those factorize uniquely into irredicuibles (up to units). More mathematically it is a PID - the ideals are those of the form (p^r).
     
    Last edited: Mar 30, 2007
  5. Mar 30, 2007 #4
    Why 'Z mod p is definitely not a subring of the integers'?

    Is it because for A to be a subring of B, A must satisfy being a ring and also the properties of B? i.e Z is not a field but Z mod p is so the latter can't be a subring of Z? But 'properties' is a vague word.

    I took UFD just as an example.

    So the answer to " If A is a subring of B and B is in a particular domain D then A must also be in that domain D" true? What about the converse?
     
    Last edited: Mar 30, 2007
  6. Mar 30, 2007 #5

    Hurkyl

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    The only subring of Z is Z.

    Z mod p and Z disagree arithmetically -- for example, the value of (p-1) + 1.

    And technically, Z mod p isn't even a subset of Z. (Although we might pretend, for convenience)


    By "in", I will assume you meant "is a subset of", "is a subring of", "is a subdomain of", or something like that.

    If A is a subring of D then A is a subring of B and B is a subring of D is clearly false. For example, pick A = Q, D = Q, and B = Z.

    I'm not sure which converse you meant, but I suspect it's not true. And I bet the domain Z[x] and its subdomains Z[x^2] and Z[x^3] will yield a counterexample.
     
    Last edited: Mar 30, 2007
  7. Mar 31, 2007 #6

    matt grime

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    It isn't because it isn't. The map sending 1 in Z/pZ to 1 in Z does not define a ring homomorphism, and is nothing to do with one being a field, and the other not. Z/mZ for m composite is not a field, and not a subring of Z. Q is a field, and it is isomorphic to a subring of M_2(Q) [2x2 matrices over Q], which is not a field.


    As Hurkyl points out, Z/pZ isn't even a subset of Z, but it is also not isomorphic to a subring of Z either.
     
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