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| − | I have no idea where to begin on this problem. I do not know how to prove that | + | [[Category:problem solving]] |
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| + | [[Category:math]] | ||
| + | [[Category:algebra]] | ||
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| + | =[[HW7_MA453Fall2008walther|HW7]] (Chapter 13, Problem 28, [[MA453]], Fall 2008, [[user:walther|Prof. Walther]]= | ||
| + | ---- | ||
| + | ==Question== | ||
| + | Prove that there is no integral domain with exactly six elements | ||
| + | ---- | ||
| + | ==Discussion== | ||
| + | |||
| + | *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 Lagrange 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.... | ||
| + | |||
| + | :::--[[User:Robertsr|Robertsr]] 19:22, 22 October 2008 (UTC) | ||
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| + | |||
| + | *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 | ||
| + | |||
| + | ---- | ||
| + | This helped. Thanks. | ||
| + | --[[User:Dakinsey|Dakinsey]] 16:14, 26 October 2008 (UTC) | ||
| + | ---- | ||
| + | [[HW7_MA453Fall2008walther|Back to HW7]] | ||
| + | |||
| + | [[Main_Page_MA453Fall2008walther|Back to MA453 Fall 2008]] | ||
Latest revision as of 08:45, 21 March 2013
HW7 (Chapter 13, Problem 28, MA453, Fall 2008, Prof. Walther
Question
Prove that there is no integral domain with exactly six elements
Discussion
- 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 Lagrange 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
- Therefore, R = Z_3 X Z_2
- 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
This helped. Thanks. --Dakinsey 16:14, 26 October 2008 (UTC)
