Why is every finite Integral Domain a field?

1. Apr 2, 2006

calvino

Any explanation would be great. Also, are there any examples of finite Integral domains that aren't fields?

2. Apr 2, 2006

Hurkyl

Staff Emeritus
Counting!

How many nonzero multiples does a nonzero element r have? How many nonzero elements are there in your domain?

3. Apr 3, 2006

shmoe

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?!?

4. Apr 12, 2006

mathwonk

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.

5. Apr 13, 2006

matt grime

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.

6. Apr 15, 2006

loopgrav

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
7. Feb 4, 2011

tc_11

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 ?

8. Feb 4, 2011

Spartan Math

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.