Difference between revisions of "Ch. 13 - Problem 28" - Rhea

Revision as of 23:04, 11 March 2009

The first thing I did with this problem was find a ring candidate containing exactly 6 elements. There is only on Abelian group with order p, where p is prime. From previous chapters we know Z_6 = Z_2*Z_3 and both Z_2 and Z_3 have prime order, therefore Z_6 is the only Abelian group with 6 elements. Z_6 is an integral domain if it contains no zero-divisors, but Z_6 has 2 zero-divisors, namely 2 and 3, therefore Z_6 is not an integral domain. Similarly we can show that Z_15 = Z_5*Z_3 is a ring with exactly 15 elements that contains 2 zero-divisors, 5 and 3. Lastly we have Z_4 = Z_2*Z_2 however Z_4 does not contain any zero divisors and therefore there is an integral domain with exactly 4 elements. The conclusion I drew from this was that a ring with exactly n elements is not an integral domain if n can be expressed as the product of distinct primes.
--Jniederh 03:33, 11 March 2009 (UTC)

This first part of my proof may seem irrelevant, but we will need it later on, trust me. Anyway, consider a finite integral domain $ \scriptstyle R $. By the definition of an integral domain, $ \scriptstyle R $ is commutative and has no zero-divisors. Now, consider two nonzero elements of $ \scriptstyle R $, $ \scriptstyle x $ and $ \scriptstyle y $, such that $ \scriptstyle\mid x\mid\ =\ n $, $ \scriptstyle\mid y\mid\ =\ m $ (with respect to the addition operation), and $ \scriptstyle n\neq m $.

By the definition of the order of an element, $ \scriptstyle nx\ =\ 0 $ and $ \scriptstyle my\ =\ 0 $.

$ \scriptstyle\Rightarrow\ \ (nx)y\ =\ x(ny)\ =\ 0 $ (because the ring is commutative)

$ \scriptstyle\Rightarrow\ \ \mid x\mid\ =\ ny\ =\ n $

$ \scriptstyle\Rightarrow\ \ y\ =\ 1 $.

Similarly:

   $ \scriptstyle(my)x\ =\ y(mx)\ =\ 0 $

$ \scriptstyle\Rightarrow\ \ \mid y\mid\ =\ mx\ =\ m $

$ \scriptstyle\Rightarrow\ \ x\ =\ 1 $.

(By the way, $ \scriptstyle0 $ and $ \scriptstyle1 $ symbolize the additive and multiplicative identities, respectively.) This implies that $ \scriptstyle x\ =\ y $, $ \scriptstyle\mid x\mid\ =\ \mid y\mid $, and thus, $ \scriptstyle n\ =\ m $. However, this contradicts our initial assumption that $ \scriptstyle n\neq m $. Therefore, it's clear that in all cases, $ \scriptstyle n $ must be equal to $ \scriptstyle m $. Specifically, all the nonzero elements of an integral domain must have the same additive order.

We now know that all the nonzero elements of our finite integral domain $ \scriptstyle R $ must have the same additive order. $ \scriptstyle R $ is finite, so none of its elements have infinite additive order. So by Theorems 13.3 and 13.4, the additive order of all of the elements must be the same prime number; let's call it $ \scriptstyle p $.

Now, return to group theory for a moment. The ring $ \scriptstyle R $ must be isomorphic to an Abelian group $ \scriptstyle G $ under the addition operation. Through inference from the Fundamental Theorem of Arithmetic, there exists at least one prime number that divides $ \scriptstyle\mid G\mid $. Therefore, by Theorem 9.5 (Cauchy's Theorem for Abelian Groups), $ \scriptstyle G $ contains an element of prime order. But since $ \scriptstyle G\approx R $ under addition, all of the nonzero elements of $ \scriptstyle G $ must have the same additive order (which we proved in the first section). This order we showed to be the prime number $ \scriptstyle p $.

By the contrapositive of Theorem 9.5, no other primes besides $ \scriptstyle p $ can divide $ \scriptstyle\mid G\mid $. So (again by the FToA):

$ \scriptstyle\mid G\mid\ =\ \overbrace{\scriptstyle p\cdot p\cdot p\cdots p}^{n}\ =\ p^n $.

Since $ \scriptstyle R $ is isomorphic to $ \scriptstyle G $, $ \scriptstyle\mid R\mid $ also equals $ \scriptstyle p^n $. What we have shown here is that any finite integral domain has order $ \scriptstyle p^n $, where $ \scriptstyle p $ is prime.

Okay, now to the actual problem. There is no integral domain with exactly six elements because $ \scriptstyle6\ \neq\ p^n $ for any prime $ \scriptstyle p $. This argument will not work for an integral domain with exactly four elements because $ \scriptstyle 4\ =\ 2^2 $. It will however work for an integral domain with exactly 15 elements, because $ \scriptstyle 15\ \neq\ p^n $ for any prime $ \scriptstyle p $. The "general result" is what we have just proved, that any finite integral domain has exactly $ \scriptstyle p^n $ elements, where $ \scriptstyle p $ is prime. $ \scriptstyle\Box $

--Nick Rupley 04:04, 12 March 2009 (UTC)