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Modern Algebra  units and zero divisors 
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#1
Jun809, 10:49 AM

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This is my first assignment in my Modern (Abstract) Algebra class, just to give an idea where I am with tools that I can use.
1. The problem statement, all variables and given/known data Describe all units and zero divisors in Z[i] 2. Relevant equations 3. The attempt at a solution I already know the answers are units = 1, 1, i, i and zero divisors: none I know I can get 1 and 1 by setting them up as multiplicative inverses: 1x = 1(mod i) and 1x = 1(mod i) where x is 1 and i1 respectively. But I am not sure how to go about getting i and i, nor do I agree with them based on my understanding and previously worked problems. One question I have is why the book writes Z[i] with brackets when it never uses brackets when the mod is an actual number or when it is the variable n. Am I missing something special with that notation? 


#2
Jun809, 11:12 AM

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It doesn't mean (mod i). It means the integers extended by i. I.e. all numbers of the form n+m*i where n and m are integers.



#3
Jun809, 01:32 PM

P: 51

I already proved in the previous section that a+bi = i(bai) and that bai is a Gaussian integer, so all Gaussian integers (Z[i]) are multiples of i. I can also say that a+bi = i(b+ai) and b + ai is a Gaussian integer. and a+bi = 1(a+bi) where a+bi is a Gaussian integer and a+bi = 1(abi) where abi is a Gaussian integer so 1, 1, i and i are all units. I believe the above is correct, but I still feel like I am missing a main point here. For instance, to be a zero divisor, you multiply it by a nonzero number and get zero. I am not sure how to think about that in terms of a Gaussian integer. Am I supposed to think of it in terms of "for any a and b, a+bi multiplied by any number can not be zero"? And in this case, should I be thinking in terms of the point (0,0) or the number 0? I feel like it is on the tip of my tongue, so to speak, but I just can not quite grasp it. 


#4
Jun809, 01:42 PM

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Modern Algebra  units and zero divisors
A zero divisor is a case where you have Gaussian integers such (a+bi)*(c+di)=0. A unit is a Gaussian integer such that 1/(a+bi) is a Gaussian integer. To find out whether there are such things, think about multiplying by compex conjugates, or maybe what you might know about multiplication and division of complex numbers.



#5
Jun809, 02:56 PM

P: 51

So 1/(a+bi) * (abi)/(abi) = (abi)/([tex]a^{2}+b^{2}[/tex]) so then (abi)/([tex]a^{2}+b^{2}[/tex]) = a/([tex]a^{2}+b^{2}[/tex])(b/([tex]a^{2}+b^{2}[/tex]))i which is a guassian integer. Am I on the right track with where you are pointing me? Thanks so much for you assistance, I feel like I am close! 


#6
Jun809, 02:59 PM

P: 51

by the way, I am wondering why this does not work for an explanation?
a+bi = i(bai) and bai is another Gaussian integer, so any a+bi is a multiple of i. a+bi = i(b+ai) and –b+ai is another Gaussian integer, so any a+bi is a multiple of i. a+bi = 1(a+bi), so any a+bi is a multiple of 1. a+bi = 1(abi) and –abi is another Gaussian integer, so any a+bi is a multiple of 1. Therefore, i, i, 1, and 1 are all units of Z[i] Is that just completely off the wall? 


#7
Jun809, 03:18 PM

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#8
Jun809, 03:50 PM

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Therefore, a^2+b^2 must equal 1 or 1. For that to be true, then either a^2=1 and b^2=0 or a^2=0 and b^2=1. Then a=1 or 1 and b=0 or b= 1 or 1 and a =0. Plugging those in gives a/(a^2+b^2) = (1+0i)/1 or (1+0i)/1 = 1 or 1 and a/(a^2+b^2) = (0+1i)/1 or (01i)/1 = i or i Is that it? I will have to look at any response and how to show there are no zero divisors later on tonight. Thanks! 


#9
Jun809, 03:56 PM

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Your reasoning is a little sloppy. a<=(a^2+b^2) with equality holding only if a<=1. Right? Similarly, b<=1. Can a and b both be nonzero?



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