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Antrax
ESL Student

 Posted: Tue Jul 06, 2004 11:32 pm    Post subject: 1 From my Modern Algebra final: 4) e) Prove/disprove: A multiplication can be defined on Z6 such that it becomes an integral domain. It was a "new" question (meaning nothing similar was ever used on a test), so I'm not sure if I got it right. I showed it to Lili and she didn't solve it in two seconds, so I'm assuming it's of SOME interest. Antrax_________________After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!
Antrax
ESL Student

 Posted: Tue Jul 06, 2004 11:37 pm    Post subject: 2 Here's my answer, invisibled: Incorrect. Proof: If Z6 becomes an integral domain, it has a "1". Consider the expression ("1"+"1")*("1"+"1"+"1"). By the distributive rule we get "1"*"1"+"1"*"1"+"1"*"1"+"1"*"1"+"1"*"1"+"1"*"1"="1"+"1"+"1"+"1"+"1"+"1". Here is the first possibly weak step: This is equivalent to a^6 in the regular (Z6,+), and since 6 is the order of Z6, it means 1+1+1+1+1+1=0. So we found a multiplication of two things from the ring that gives off zero. Now we'll prove neither of them is zero. Let's suppose "1"+"1"=0. Then we get 2+2=2*"1"+2*"1"=0. But 2+2=4, contradiction. Similarily 1+1+1 can't be zero, QED. I strongly doubt this is correct. Where are my errors, and what is the right proof? Antrax_________________After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!
kevinatilusa
Daedalian Member

 Posted: Wed Jul 07, 2004 6:04 am    Post subject: 3 Looks fine, but your weak step can be shortened considerably: If we made Z_6 into an integral domain, it would still have the additive structure of Z_6, so 1+1=2, not 0.
Antrax
ESL Student

 Posted: Wed Jul 07, 2004 10:40 am    Post subject: 4 kevinatilusa, I was walking on eggshells, having never seen something like this. I really didn't want to make a leap that wasn't mathematically valid, so I just wrote out everything. Here's another question from the same test, the other non-trivial question: 4) d) Can there be an integral domain of size 3, 4, 5? Prove. Antrax_________________After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!
Antrax
ESL Student

 Posted: Wed Jul 07, 2004 10:44 am    Post subject: 5 Actually, I think the shortening isn't valid. Nobody said "1" is the 1 in Z6. The multiplication could be defined such that 3="1" (for example, a*b=1/3*a*b. Obviously this wouldn't work, but just as an example). Antrax_________________After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!
kevinatilusa
Daedalian Member

Antrax
ESL Student

 Posted: Wed Jul 07, 2004 12:24 pm    Post subject: 7 Whoops. I used the multiplication table (with elements 0,1,a,b) to prove that it can't be done Antrax_________________After years of disappointment with get rich quick schemes, I know I'm gonna get rich with this scheme. And quick!
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