posted 6 Jan 2022

Z[i]\mathbb{Z}[i] and Z[ω]\mathbb{Z}[\omega] are Euclidean Domains

Notes Number Theory

Z[i]\mathbb{Z}[i] and Z[ω]\mathbb{Z}[\omega] are Euclidean Domains

§ Gaussian Integers

Gaussian Integers

Let i=−1i = \sqrt{-1} and consider Z[i]⊂C\mathbb{Z}[i] \subset \mathbb{C} defined by {a+bi ∣ a,b∈Z}.\{a + bi\,|\,a,b \in \mathbb{Z}\}.

Clearly, Z[i]\mathbb{Z}[i] is closed under addition and attraction.

Also, if

a+bi,c+di∈Z[i],a + bi, c+ di \in \mathbb{Z}[i],

then

(a+bi)(c+di)=ac+adi+bci+bdi2=(ac−bd)+(ad+bc)i∈Z[i].(a + bi)(c + di) = ac + adi + bci + bdi^2 = (ac - bd) + (ad + bc)i \in \mathbb{Z}[i].

So Z[i]\mathbb{Z}[i] is closed under multiplication and is a ring.

Since Z[i]⊂C,\mathbb{Z}[i] \subset \mathbb{C}, it is an integral domain as well.

Z[i]\mathbb{Z}[i] is an Euclidean domain.

For a+bi∈Q[i]a + bi \in \mathbb{Q}[i] define λ(a+bi)=a2+b2.\lambda(a + bi) = a^2 + b^2.

Let α=a+bi\alpha = a + bi and γ=c+di\gamma = c + di and sps. γ≠0.\gamma \neq 0. α/γ=r+si,\alpha/\gamma = r + si, where r,s∈Q.r, s \in \mathbb{Q}.

Choose integers m,n∈Zm,n \in \mathbb{Z} such that ∣r−m∣≤12|r - m| \leq \frac{1}{2} and ∣s−n∣≤12.|s - n| \leq \frac{1}{2}.

Set δ=m+ni.\delta = m + ni.

Then δ∈Z[i]\delta \in \mathbb{Z}[i] and λ((α/γ)−δ)≤12λ(y)<λ(γ).\lambda((\alpha/\gamma) - \delta) \leq \frac{1}{2}\lambda(y) \lt \lambda(\gamma).

It follows that λ\lambda makes Z[i]\mathbb{Z}[i] an Euclidean domain.

§ Eisenstein Integers

The numbers ±1\pm 1 and ±i\pm i are the roots of x4=1x^4 = 1 over the complex numbers. Consider x3=1,x^3 = 1, since x3−1=(x−1)(x2+x+1),x^3 - 1 = (x - 1)(x^2 + x + 1), the roots are 1,(−1±−3)/2.1, \left(-1 \pm \sqrt{-3}\right) / 2. Let ω=(−1+−3)/2,\omega = \left(-1 + \sqrt{-3}\right)/2, then ω2=(−1−−3)/2\omega^2 = \left(-1 - \sqrt{-3}\right)/2 and that 1+ω+ω2=0.1 + \omega + \omega^2 = 0.

Eisenstein Integers

Consider Z[ω]={a+bω ∣ a,b∈Z}.\mathbb{Z}[\omega] = \{a + b\omega\,|\, a, b \in \mathbb{Z}\}.

Clearly, Z[ω]\mathbb{Z}[\omega] is closed under addition and subtraction.

Also, (a+bω)(c+dω)=ac+(ad+bc)ω+bdω2=(ac−bd)+(ad+bc−bd)ω.(a + b\omega)(c + d\omega) = ac + (ad + bc) \omega + bd\omega^2 = (ac - bd) + (ad + bc- bd) \omega. Thusly, Z[ω]\mathbb{Z}[\omega] is a ring.

Again, Z[ω]⊂C\mathbb{Z}[\omega] \subset \mathbb{C} so it is an integral domain.

Additionally, Z[ω]\mathbb{Z}[\omega] is closed under complex conjugation. Since −3‾=3⋅i‾=−3⋅i=−−3\overline{\sqrt{-3}} = \overline{\sqrt{3}\cdot i} = - \sqrt{3}\cdot i = -\sqrt{-3} we get that ω‾=ω2.\overline{\omega} = \omega^2. So if α=a+bω∈Z[ω],\alpha = a + b\omega \in \mathbb{Z}[\omega], then α‾=a+bω‾=a+bω2=(a−b)−bω∈Z[w].\overline{\alpha} = a + b\overline{\omega} = a + b\omega^2 = (a - b) - b\omega \in \mathbb{Z}[w].

Proposition 7

Z[ω]\mathbb{Z}[\omega] is an Euclidean domain.

For α=a+bω∈Z[ω]\alpha = a + b\omega \in \mathbb{Z}[\omega] define λ(α)=a2−ab+b2.\lambda(\alpha) = a^2 - ab + b^2. Then, we can see that λ(α)=aa‾.\lambda(\alpha) = a\overline{a}.

Now, let α,β∈Z[ω]\alpha, \beta \in \mathbb{Z}[\omega] and sps. that β≠0.\beta \neq 0. Then α/β=αβ‾/ββ‾=r+sω,\alpha/\beta = \alpha\overline{\beta}/\beta\overline{\beta} = r + s\omega, where r,s∈Q.r, s \in \mathbb{Q}. We used the fact that ββ‾=λ(β)\beta\overline{\beta} = \lambda(\beta) is a positive integer and that αβ‾∈Z[ω]\alpha\overline{\beta} \in \mathbb{Z}[\omega] since α,β‾∈Z[ω].\alpha, \overline{\beta} \in \mathbb{Z}[\omega].

Next we find mm and nn such that ∣r−m∣≤12|r - m| \leq \frac{1}{2} and ∣s−n∣≤12.|s - n| \leq \frac{1}{2}. Then let γ=m+nω.\gamma = m + n\omega. λ((α/β)−γ)=(r−m)2(r−m)(s−n)+(s−n)2≤14+14+14<1.\lambda((\alpha/\beta) - \gamma) = (r - m)^2 (r - m)(s - n) + (s - n)^2 \leq \frac{1}{4} + \frac{1}{4} + \frac{1}{4} \lt 1.

Let ρ=α−γβ.\rho = \alpha - \gamma\beta. Then either ρ=0\rho = 0 or λ(p)=λ(β((α/β)−γ))=λ(β)λ((α/β)−γ)<λ(β).\lambda(p) = \lambda(\beta((\alpha/\beta) - \gamma)) = \lambda(\beta)\lambda((\alpha / \beta) - \gamma) \lt \lambda(\beta).

Then, Z[i]\mathbb{Z}[i] and Z[ω]\mathbb{Z}[\omega] are PIDs as well, which means the theorem of unique factorization is true.