(New page: 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 ...) |
|||
| Line 1: | Line 1: | ||
| + | [[Category:MA453Spring2009Walther]] | ||
| + | |||
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.<br> | 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.<br> | ||
--[[User:Jniederh|Jniederh]] 03:33, 11 March 2009 (UTC) | --[[User:Jniederh|Jniederh]] 03:33, 11 March 2009 (UTC) | ||
Revision as of 06:56, 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)
