Modern Algebra - units and zero divisors


by Proggy99
Tags: algebra, divisors, modern, units
Proggy99
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#1
Jun8-09, 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 i-1 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?
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Dick
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Jun8-09, 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.
Proggy99
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Jun8-09, 01:32 PM
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Quote Quote by Dick View Post
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.
Thank you for that. Your explanation led me back to the previous section where they covered a+bi Gaussian integers and the very last sentence of the section made the comment that Gaussian integers were represented by Z[i].

I already proved in the previous section that a+bi = i(b-ai) and that b-ai 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(-a-bi) where -a-bi 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.

Dick
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Jun8-09, 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.
Proggy99
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Jun8-09, 02:56 PM
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Quote Quote by Dick View Post
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.
okay, the concepts here are being stubborn about not sinking in. By multiplying complex conjugates, I believe you mean (a+bi)(a-bi)= [tex]a^{2}-i^{2}b^{2}=a^{2}+b^{2}[/tex]

So 1/(a+bi) * (a-bi)/(a-bi) = (a-bi)/([tex]a^{2}+b^{2}[/tex])

so then (a-bi)/([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!
Proggy99
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#6
Jun8-09, 02:59 PM
P: 51
by the way, I am wondering why this does not work for an explanation?

a+bi = i(b-ai) and b-ai 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(-a-bi) and –a-bi 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?
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Jun8-09, 03:18 PM
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Quote Quote by Proggy99 View Post
okay, the concepts here are being stubborn about not sinking in. By multiplying complex conjugates, I believe you mean (a+bi)(a-bi)= [tex]a^{2}-i^{2}b^{2}=a^{2}+b^{2}[/tex]

So 1/(a+bi) * (a-bi)/(a-bi) = (a-bi)/([tex]a^{2}+b^{2}[/tex])

so then (a-bi)/([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!
Sure. Now ask yourself how a/(a^2+b^2) and b/(a^2+b^2) can be an integer if a and b are integers. Hint: if |a|>1 then a^2>|a|. Showing for example that any Gaussian integer is a multiple of e.g. i, does show i is a unit, but mostly because then 1 is a multiple of i, and so i is invertible. But it seems simpler just to say i*(-i)=1. That argument shows you that 1,-1,i and -i are units. But it doesn't prove they are the only units.
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Jun8-09, 03:50 PM
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Quote Quote by Dick View Post
Sure. Now ask yourself how a/(a^2+b^2) and b/(a^2+b^2) can be an integer if a and b are integers. Hint: if |a|>1 then a^2>|a|. Showing for example that any Gaussian integer is a multiple of e.g. i, does show i is a unit, but mostly because then 1 is a multiple of i, and so i is invertible. But it seems simpler just to say i*(-i)=1. That argument shows you that 1,-1,i and -i are units. But it doesn't prove they are the only units.
hmmmm, so a/(a^2+b^2) can only be an integer if a^2+b^2 is a, 1 or -1, but b/(a^2+b^2), implies that a^2+b^2 is b, 1 or -1.

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 (0-1i)/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!
Dick
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Jun8-09, 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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