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units...

Thayaden 1

N Oct 31, 2024 by felixgotti

Part 1:
Notice $\mathbb{Z}^{\times}=\{\pm1 \}$ thus for $\mathbb{Z}[i]$ we might enharite this too. This might be easily proven by letting the coefficient of the unreal part be $0$ . Thus we might know that at least $\pm1$ is a unit. Taking that idea once again letting the real part be $0$ we have,
$i\cdot-i=-(i^2)=1$ Thus as we are communitive $\pm i$ is in the set this makes sense from an algebraic perspective on the complex plane when multiplying a unit we have a change of direction likewise this is also true for real numbers thus $\mathbb{Z}[i]^{\times}=\{\pm1,\pm i \}$

1 reply

Thayaden

Sep 25, 2024

felixgotti

Oct 31, 2024

J

H

Exercise 0.6

hkoo 1

N Jun 16, 2024 by felixgotti

What I've tried so far:

<Describe what you have tried so far here. That way, we can do a better job helping you!>

(1) Describe the group of units of the ring of Gaussian integers.

Gaussian integers is set { $a+bi \in C | a,b \in Z$ }.

Case(1) a=0, b is non-zero integer
There exsits invertible (-1/b)i s.t bi*(-1/b)i = 1.
for only b=-1, $(1/b)i\in Z[i]$ .

Case(2) a is non-zero integer, b=0
There exists invertible 1/a s.t a*(1/a)=1
for only a=1, $(1/a) \in Z[i]$ .

Case(3) a,b are non-zero integer
To find inverse of $a+bi$ , think equation $(a+bi)*(c+di)=1$ .
$(a+bi)(c+di)=(ac-bd)+(ad+bc)i=1$ . It means that $ac-bd=1, ad+bc=0$ .
since $a,b$ are non-zero, we can know that
$c=(1+bd)/a, c=(-ad)/b$ . So $(1+bd)/a = (-ad)/b$ .
solve equation and we can know $(c,d)= (-b/(a^2+b^2), a/(a^2+b^2))$

since $|a|,|b| < a^2+b^2$ for non-zero integer a,b, c,d are always rational number.
Therefore we can know that { $1,-i$ } is units of gaussian integers.

(2) For each n, describe the group of units of $Z_n$ .
Existence of inverse of $s\in Z_n=$ { $0,1,...,n-1$ } implies that there exist t s.t st = nQ+1.
we can solve this problem using following lemma.
$gcd(s,n)= 1$ if and only if there exists t s.t st=nQ+1 and t,Q is integer.

so group of units of $Z_n$ = { $i | (i,n)=1, 1<=i<=n$ }. for example n=12, group of units of $Z_n$ ={ $1,5,7,11$ }. done.

proof of lemma

(1) only if
think about $i*s=n*Q_i+r_i$ for i=1,2,...,n s.t $Q_i$ is quotient, $r_i$ is residue and $0<=r_i <=n-1$ . $r_i$ for i=1,..,n are different. if there exist different i,j such that $r_i=r_j$ , $(j-i)*s=n*(Q_j-Q_i)$ .
since (n,s)=1, n divides $(j-i)$ . however n cannot divide (j-i) because $1<=|j-i|<=n-1$ . so it's contradiction.
therefore { $r_i|i=1,...,n$ }={ $0,1,...,n-1$ }. conclusivey we can know that there exits t such that residue of s*t =1.

(2) if
suppos that (s,n)=g such that gs'=s, gn'=n. since st-nQ=g(s't-n'Q)=1, g divides 1. it means that g=1. done.

Where I'm stuck:

<Describe what's confusing you, or what your question is here!>

1 reply

hkoo

May 24, 2024

felixgotti

Jun 16, 2024

J