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Field theory

  1. May 13, 2007 #1
    1. The problem statement, all variables and given/known data

    I'm trying to find a commutative ring, not a field, who's only ideals are {0} and itself.

    2. Relevant equations

    Definition: A subset of a ring R is an ideal if it is a subring of R and is closed under multiplication by elements of R.

    3. The attempt at a solution

    I claim [itex]\mathbb{Z}_4[/itex] has the desired properties.
    [itex]\mathbb{Z}_4[/itex] is a commutative ring.
    [itex]\mathbb{Z}_4[/itex] is not a field since it has zero divisors.
    The only proper nontrivial subring of [itex]\mathbb{Z}_4[/itex] is {0,2} and {0,2} is not an ideal (not closed under multiplication by 2).
  2. jcsd
  3. May 13, 2007 #2
    Was this actually the question? Or was it something more of the form "does there exist a commutative ring whose only ideals are {0} and itself that is not a field?"

    This definition isn't quite right. An ideal, I, is a subring such that rI and Ir are subsets of I for every r in R. (for a two-sided ideal).

    But if you multiply the elements in {0,2} by 2 you get {0} which is contained in {0,2}.

    If the condition were that rI=I for every r in the ring than there would only be one ideal {0}, because 0 must be in the ideal since it is an additive group and multiplying by 0 will only give zero then 0*I would be {0}.
  4. May 13, 2007 #3
    You're correct and in fact [itex]Z_4[/itex] does not satisfy the required properties since {0,2} is not an ideal.
  5. May 13, 2007 #4

    matt grime

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    {0,2} is an ideal of Z_4: there is at least one too many negatives in your last sentence.

    Look, let's just write down what we know:

    suppose we have a ring R and the only ideals in R are 0 and R.

    Let's suppose that x is a non-identity element in R and consider the ideal xR, since 1R=R, and this tells us nothing about R.

    xR is either 0 or R.

    Since rings normally assumed to have identities you can now proceed to deduce some things.
  6. May 13, 2007 #5
    My bad again. I meant to say that Z4 IS an ideal. Thanks.

    Am I missing something or does the trivial ring R = {0} not count??

    R is a commutative ring.
    R is not a field since it does not have a nonzero identity.
    The only subrings of R are {0} and R.
  7. May 13, 2007 #6

    matt grime

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    Most people require a ring to have a multiplicative identity. Why is it weird that there are (degenerate) counter examples. If you follow my suggestion you will find all of the counter examples.
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