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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 | 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 | ||
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| + | 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.... | ||
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| + | |||
| + | --[[User:Robertsr|Robertsr]] 19:22, 22 October 2008 (UTC) | ||
Revision as of 15:22, 22 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)
