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Why is every finite Integral Domain a field?

  1. Apr 2, 2006 #1
    Any explanation would be great. Also, are there any examples of finite Integral domains that aren't fields?
     
  2. jcsd
  3. Apr 2, 2006 #2

    Hurkyl

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    Counting!

    How many nonzero multiples does a nonzero element r have? How many nonzero elements are there in your domain?
     
  4. Apr 3, 2006 #3

    shmoe

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    Nope.

    Maybe you wanted an example of an integral domain with an infinite number of elements that isn't a field? What infinite integral domains do you know?!?
     
  5. Apr 12, 2006 #4

    mathwonk

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    hurkyls approach is to take a non zero element x and consider the multiplication map from the domain to itself taking y to xy. since it is a domain, the map is injective and thus since it is finite any injective map is surjective, so it has 1 in its oimage, so there is some y suich that xy = 1, hence x is a invertible and we have a field.

    a different proof is the following in the case of Z/(p): if n is non zero in Z/(p) then n and po are relativelt porime, so there are integers r,s, such that nr + ps = 1, so r is the inverse of n mod p.

    this proof generalizes to quotient rings of form k[X]/(f), where k is a field, and f is irreducible in k[X], even though these rings are not always finite.

    i.e. such a quotient is a domain iff it is a field.
     
  6. Apr 13, 2006 #5

    matt grime

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    I believe there is a stronger and harder result that relies on the fact that finite division rings are commutative, thus any finite domain (which is a division ring) is field without assuming comutativity.
     
  7. Apr 15, 2006 #6
    Here is proof. I had to edit somewhat because some of the mathematical symbols wouldn't copy into this window. You should be able to understand however. If you would like, I could email you the entire section in pdf form. (It's on polynomials over finite fields.)

    Theorem 6.20 Every finite integral domain is a field.

    Proof Let D be a finite integral domain (which is commutative by definition). We must show that 1 is in D, and that every nonzero a in D has a multiplicative inverse that is also in D. In other words, we must show that for every nonzero a in D there exists b in D such that ab = 1 is in D. Let {x_1, . . . , x_n} denote all the elements of D, and consider the set {ax_1, . . . , ax_n} where a is in D and a ≠ 0. If ax_i = ax_j for i ≠ j, then a(x_i - x_j) = 0 which (since D has no zero divisors) implies that x_i = x_j, contradicting the assumption that i ≠ j. Thus ax_1, . . . , ax_n are all distinct. Since D contains n elements, it follows that in fact we have D = {ax_1, . . . , ax_n}. In other words, every y in D can be written in the form ax_i = x_ia for some i = 1, . . . , n. In particular, we must have a = ax_i_0 for some i_0 = 1, . . . , n. Then for any y = x_ia in D we have

    yx_i_0 = (x_ia)x_i_0 = x_i(ax_i_0) = x_ia = y

    so that x_i_0 may be taken as the identity element 1 in D. Finally, since we have now shown that 1 is in D, it follows that 1 = ax_j for some particular j = 1, . . . , n. Defining b = x_j yields 1 = ab and completes the proof. ˙

    Corollary Z_n is a field if and only if n is prime.

    This last corollary depends on Theorem: The ring Z_n is an integral domain if and only if n is prime.

    I won't take up space with the proof.

    I hope this helps.
     
    Last edited: Apr 15, 2006
  8. Feb 4, 2011 #7
    But an integral domain has unity by definition. So since R has unity, 1 is in R, so since aR contains all n element of R, R=aR, and 1 must be in aR too right? So why do you have to use yx_i_0 = (x_ia)x_i_0 = x_i(ax_i_0) = x_ia = y ?
     
  9. Feb 4, 2011 #8
    Yep! Wedderburn's theorem on Division rings. It's a wee bit more complicated. A more elementary generalization would be that if an integral domain R contains a field K, and is a finite dimensional vector space over it, then it too is a field (this includes MathWonk's examples as well). It's a straightforward generalization.
     
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