| Line 24: | Line 24: | ||
--[[User:Robertsr|Robertsr]] 19:22, 22 October 2008 (UTC) | --[[User:Robertsr|Robertsr]] 19:22, 22 October 2008 (UTC) | ||
| + | |||
| + | |||
| + | |||
| + | |||
| + | An integral domain is a ring. A ring has the commutative property for addition. | ||
| + | There is only one Abelian group with order(number of elements)=p, where p is prime. Z_6 can be written as Z_2 * Z_3 and Z_2 and Z_3 both have prime order. Z_6 therefore is the only Abelian group with 6 elements. So, Z_6 can be the only ring with 6 elements. Z_6 is an integral domain if there are no zerodivisiors. 2*3=0, 2 and 3 are zero divisors. So, Z_6 is not an integral domain. Similary, Z_15= Z_3 * Z_5, so it is the only Abelian group with 15 elements. 3*5=0, there are zerodivisors, it is not an integral domain. | ||
| + | |||
| + | Try Z_4 = Z_2 * Z_2. There is no zerodivisors, so it is in fact an integral domain. Generally, if the number of elements of a group is the products of different primes, it cannot form an integral domain. However, if the number of elements of a group is the product of the same prime (eg. p^2 or p^3), it can form an integral domain. | ||
| + | |||
| + | -Ozgur | ||
Revision as of 11:22, 24 October 2008
I have no idea where to begin on this problem. I do not know how to prove that there is no integral domain with six elements. A little help would be nice. Thanks
Begin by saying that R is the domain with exactly 6 elements (order of 6). The characteristic of an integral domain is zero or prime, and 6 is the smallest possible integer such that 6*1 = 0 in mod6. Therefore there can not be an integral domain with exactly six elements. Pf: (Char(R))= prime Char(R) = 2, 3, 5 (it is a subgroup of the domain) (R, +) is an Abelian Group
By Lagraunge theorem, the subgroup must be a divisor of the large group so n/6 therefore n must equal either 2 or 3 If n=3 then the subgroup is {0,1,2} contained in R Therefore R = Z_3 * H This is by a theorem that says that every abelian group with G = p1^a1.....pk^ak where p is prime. G is going to be a product G1 *....Gk where the order of Gi = p1^a1.
Therefore, R = Z_3 X Z_2 1 corresponds to (1,0) a corresponds to (1,1) 2 corresponds to (2,0) b corresponds to (2,1) 0 corresponds to (0,0) c corresponds to (0,1) 2 is an element of R, and thus does not equal 0 R domain therefore 2 is not a zero divisor so by induction 2 is not Z_m
Continue with same argument for n=2 vise versa....
--Robertsr 19:22, 22 October 2008 (UTC)
An integral domain is a ring. A ring has the commutative property for addition.
There is only one Abelian group with order(number of elements)=p, where p is prime. Z_6 can be written as Z_2 * Z_3 and Z_2 and Z_3 both have prime order. Z_6 therefore is the only Abelian group with 6 elements. So, Z_6 can be the only ring with 6 elements. Z_6 is an integral domain if there are no zerodivisiors. 2*3=0, 2 and 3 are zero divisors. So, Z_6 is not an integral domain. Similary, Z_15= Z_3 * Z_5, so it is the only Abelian group with 15 elements. 3*5=0, there are zerodivisors, it is not an integral domain.
Try Z_4 = Z_2 * Z_2. There is no zerodivisors, so it is in fact an integral domain. Generally, if the number of elements of a group is the products of different primes, it cannot form an integral domain. However, if the number of elements of a group is the product of the same prime (eg. p^2 or p^3), it can form an integral domain.
-Ozgur
