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Difference between a field and a ring

  1. Aug 1, 2004 #1
    I think the title describes my question fairly well. Could someone please explain to me the difference between a field and a ring? While you're at it, feel free to explain the concept of "mod." I see these all the time when I'm reading, but I've never had anyone to tell me what they actually are. I have a rough idea, but I think it would serve me better if someone who knew what they were talking about shared their knowledge with me. Thanks.
     
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  3. Aug 1, 2004 #2
  4. Aug 1, 2004 #3

    Hurkyl

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    A field is a ring that is commutative and has division.

    An example of a field is Q, the rational numbers.
    An example of a ring that is not a field is Z, the integers.

    However, Z is something called a domain, which means that if ab = 0, then a = 0 or b = 0.

    For any commutative domain, one can define its fraction field. As its name suggests, it consists of all fractions whose numerator and denominator are in the domain. The fraction field of Z is, of course Q.

    Another example is R[x], the ring of real polynomials in x. It is a domain, and it has a fraction field, R(x), the field of all real rational functions in x.


    I'm going to assume that you understand modulo arithmetic of integers...

    Let R be the the ring of integers modulo 8. This ring is not a domain because, for instance, 2 * 4 = 0. Thus, R does not have a fraction field. However, we can make some fractions. For example, we could take S = {1, 3, 5, 7} as the set of allowable denominators. This is called the localization of R at S, denoted RS. The same thing is also described in the opposite way; the localization of R at T where T = {0, 2, 4, 6}, also denoted as RT


    moduli, in general, are a way of making a new ring from an old ring by specifying a set of elements that should be equal to zero, called an ideal. For example, we make Z mod 8 by declaring the set {..., -16, -8, 0, 8, 16, ...} to be equal to zero. If R is the ring, and I is the ideal, then we call the ring formed in this way the quotient ring of R over I, written R/I.

    To satisfy this intuition, an ideal has special properties. I is an ideal of a ring, R, iff:
    for any r in R and x in I: rx and xr are in I. (because multiplying by zero yields zero)
    for any x, y in I: x - y are in I. (because zero - zero = zero)


    (However, Z mod p, where p is a prime, is a field!!!)


    An example of this is the ring S = R / (i^2 + 1)... (where R is the real numbers)

    R is the ring of all real polynomials in the variable i.
    i^2 + 1 is an element of R, and we use the notation (i^2 + 1) to mean the smallest ideal containing i^2 + 1.

    The resulting quoteint ring consists of things that look like: {x + yi | x, y in R}. To see that this set is closed under multiplication:

    (a + bi) (c + di) = ac + (ad + bc)i + (bd)i^2
    = ac + (ad + bc)i + (bd)(i^2 + 1 - 1)
    = ac + (ad + bc)i + (bd)(-1) (because, in S, i^2 + 1 = 0!!!!)
    = (ac - bd) + (ad + bc)i

    which is, again, of the form x + yi.

    In fact, you'll recognize that S is just the ring of complex numbers!


    In general, if F is a field, and f is an irreducible polynomial in F[x], then it turns out that F[x]/(f), that is F[x] mod f, is always a field.
     
  5. Aug 1, 2004 #4

    Gza

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    Just out of curiosity(or naiivete, take your pick), what is the rationale behind naming the concept a "ring"? Is there some geometric basis for this?
     
  6. Aug 1, 2004 #5

    Hurkyl

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    Ask Fraenkel. :smile: According to my text, the term "Zahlring" appeared in algebraic number theory; I'm not sure what it meant.
     
  7. Aug 1, 2004 #6

    AKG

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    Gza

    I'll pick laziness ;). From the article on Rings linked above:

    The word ring is short for the German word 'Zahlring' (number ring). The French word for a ring is anneau, and the modern German word is Ring, both meaning (not so surprisingly) "ring." The term was introduced by Hilbert to describe rings like
    [tex]\mathbb{Z}[\root 3\of 2] = {a + b\root 3\of 2 + \root 3\of 4 \mbox{ such that } a,b,c \in \mathbb{Z}}[/tex]​
    By successively multiplying the new element [itex]\root 3\of 2[/itex], it eventually loops around to become something already generated...
     
  8. Aug 1, 2004 #7

    Janitor

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    Thanks for the information. I myself have wondered about the reason for that word being used.
     
  9. Aug 2, 2004 #8
    Every field is a ring but not every ring is a field. A field is a ring with a few more properties like commutivity and that fields have multiplicative inverses are well defined for each non-additive-zero element.

    For example, Z_4={0,1,2,3} is not a field because there is no well-defined multiplicative inverse of 2. 2*0=0, 2*1=2, 2*2=0, 2*3=2. None of these is 1.
    The other nonzero elements of Z_4 have inverses: 1*1=1 and 3*3=1. Note that 2 does not have an inverse becuase it is a zero divisor: 2*2=0. I think Hurkyl talked about zero divisors.

    Z_n is a field whenever n is a prime and if n is not a prime then Z_n is not a field (I think). These finite fields are important in cryptography unless I've completely lost my mind.
     
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