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Simple counterexample for claim about integral domains

  1. Jun 2, 2008 #1
    So i'm looking for an example of an infinite integral domain with finite characterestic. That is a infinite integral domain such that there is a prime p such that p copies of any element added together is the additive identity.

    I'm just looking for a simple counterexample. I'm working through Herstein's (Topics..) Ring chapter and this is the problem set after chapter 3.2.
     
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  3. Jun 3, 2008 #2

    morphism

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    Think about, for example, polynomial rings.
     
  4. Jun 3, 2008 #3
    take the finite ring J_17 (addition mod 17 and multiplication mod 17). Consider the set of polynomials in one variable X (which, say, runs through the real line) with coefficients from J_17. This set of polynomials is a ring: all the ring axioms can be verified using the fact that J_17 is a ring and that the reals form a ring as well. This polynomial ring is clearly finite. Take any polynomial and add 17 copies of it together and we'll be able to collect terms with similar coefficients, since for any element,a, in J_17, 17a=0 we get that adding 17 copies of any such polynomial gives us the zero polynomial.

    The idea of a polynomial ring was not introduced before this problem set. What does the variable X traditionally vary over? Does it matter?
     
  5. Jun 3, 2008 #4

    matt grime

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    Don't think of it as a variable. It is commonly called an 'indeterminate'. It isn't varying over anything.

    How about an algebraic closure of F_p? By the way, why are you looking for a *counter*-example?
     
  6. Jun 3, 2008 #5
    i was going to state the false claim: integral domains of finite characteristic are finite.

    An indeterminate, ok.

    I presume F_p is the field of 0,1,...,p-1 where addition is modp and similarly for multiplication. If you could elaborate as to 1. what the algebraic closure of F_p is, 2. how it is infinite and 3. how it has finite characteristic.

    I haven't been formally introduced to algebraic closures or polynomial rings as yet in Herstein yet i'm expected to come up with this example. There must be something simpler...or maybe not.

    Thanks for the help.
     
  7. Jun 3, 2008 #6

    matt grime

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    An algebraic closure F of F_p is a minimal field containing F_p, so it necessarily has finite characteristic p, and such that every polynomial with coefficients in F has a root in F.

    It is easy to show that no field F_q where q=p^r is algebraically closed (you should try this - hint the non zero elements a a group of order p^r - 1).

    Thus if such a thing as an algebraic closure exists, it is infinite.

    Anyway, that is probably beyond what you were expected to know.

    Since we don't necessarily know what is in Herstein, we cannot say what you're supposed to know or not, but the polynomial ring idea should have been something you can come up with. Just because it is not formally introduced yet doesn't mean you weren't supposed to come up with something like that. After all, you knew what polynomials are before starting book, right?
     
  8. Jun 3, 2008 #7
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