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Eisenstein Ring's Norm 📂Number Theory

Eisenstein Ring's Norm

Theorem

Let’s consider a function on the Eisenstein ring Z[ω]\mathbb{Z}[ \omega ].

  • [1]: If defined as N(x+ωy):=x2−xy+y2N(x + \omega y) := x^2 - xy + y^2, then NN becomes the multiplicative norm of Z[ω]\mathbb{Z}[ \omega ].
  • [2]: Z[ω]\mathbb{Z}[ \omega ] is a Euclidean domain.
  • [3]: The only unit of Z[ω]\mathbb{Z}[ \omega ] is ±1,±ω,±ω2\pm 1, \pm \omega, \pm \omega^2 .

Description

The study of Eisenstein integers can be greatly facilitated with the help of [abstract algebra](../../categories/abstract algebra). Proving [2] with the norm NN defined in an integral domain makes it almost equivalent to proving the fundamental theorem of arithmetic extended to Eisenstein primes since ED is UFD. This is similar to algebraically explaining the fundamental theorem of arithmetic as Z\mathbb{Z} being UFD.

Proof

[1]

Definition of multiplicative norm:

  • (i): N(α)=0  ⟺  α=0N (\alpha) = 0 \iff \alpha = 0
  • (ii): N(αβ)=N(α)N(β)N ( \alpha \beta ) = N ( \alpha ) N ( \beta )

We need to show that NN becomes a multiplicative norm and that Z[ω]\mathbb{Z}[ \omega ] actually becomes an ID. There is no need to first prove that it’s an ID as the theorem requiring a norm to be defined isn’t needed yet. Let’s say a,b,c,d∈Za,b,c,d \in \mathbb{Z} with respect to α:=a+ωb\alpha := a + \omega b, β:=c+ωd\beta := c + \omega d.

Part (i). N(α)=0  ⟺  α=0N (\alpha) = 0 \iff \alpha = 0

If we set a:=(A+B)a := (A+B), b:=(A−B)b:= (A-B), then N(α)=a2−ab+b2=0  ⟺  (A+B)2−(A+B)(A−B)+(A−B)2=0  ⟺  A2+3B2=0  ⟺  A=B=0  ⟺  a=b=0  ⟺  α=a+ωb=0 \begin{align*} & N(\alpha) = a^2 - ab+ b^2 = 0 \\ \iff & (A+B)^2 - (A+B)(A-B) + (A-B)^2 = 0 \\ \iff & A^2 + 3 B^2 = 0 \\ \iff & A = B = 0 \\ \iff & a = b = 0 \\ \iff & \alpha = a + \omega b = 0 \end{align*}

Part (ii). N(αβ)=N(α)N(β)N ( \alpha \beta ) = N ( \alpha ) N ( \beta )

Since the product of Eisenstein integers is calculated as (a+ωb)(c+ωd)=(ac−bd)+ω(ad−bd+bc)( a + \omega b )( c + \omega d) = (ac - bd) + \omega (ad - bd + bc), N(αβ)=N(ac−bd+ω(ad−bd+bc))=(ac−bd)2−(ac−bd)(ad−bd+bc)+(ad−bd+bc)2=(ac−bd)(ac−ad−bc)+(ad−bd+bc)2=(ac−bd)(ac−K)+(K−bd)2=a2c2−acbd−acK+bdK+k2−2Kbd+bd2=a2c2−acbd−acK−bdK+k2+bd2=a2c2−acbd−(ac+bd)(ad+bc)+(ad+bc)2+bd2=a2c2−acbd−a2cd−abc2−abd2−b2cd+a2d2+2adbc+b2c2+bd2=a2c2−a2cd+ad2−abc2+abcd−abd2+b2c2−b2cd+b2d2=(a2−ab+b2)(c2−cd+d2)=N(α)N(β) \begin{align*} N ( \alpha \beta ) =& N \left( ac - bd + \omega (ad -bd + bc) \right) \\ =& (ac - bd)^2 - (ac -bd)(ad - bd + bc) + (ad - bd + bc)^2 \\ =& (ac -bd)( ac -ad - bc) + (ad - bd + bc)^2 \\ =& (ac -bd)( ac -K ) + (K - bd )^2 \\ =& a^2 c^2 - acbd - acK + bdK + k^2 - 2Kbd + bd^2 \\ =& a^2 c^2 - acbd - acK - bdK + k^2 + bd^2 \\ =& a^2 c^2 - acbd - (ac + bd)(ad + bc) + (ad + bc)^2 + bd^2 \\ =& a^2 c^2 - acbd -a^2 c d - a b c^2 - abd^2 - b^2 c d +a^2 d^2 + 2 adbc + b^2 c^2 + bd^2 \\ =& a^2 c^2 - a^2 c d + a d^2 - abc^2 + abcd - abd^2 + b^2 c^2 - b^2 c d + b^2 d^2 \\ =& (a^2 - ab + b^2)(c^2 - cd + d^2) \\ =& N ( \alpha ) N ( \beta ) \end{align*} where K:=(ad+bc)K := (ad + bc) is.

Part 3. Z[ω]Z[ \omega ] is an ID

Assume α\alpha, β\beta are not 0∈Z[ω]0 \in \mathbb{Z}[ \omega ] yet αβ=0\alpha \beta = 0. Then, according to Part (i)~(ii), N(α)N(β)=N(αβ)=N(0)=0 N(\alpha) N(\beta) = N(\alpha \beta ) = N(0) = 0 As N(α)N(\alpha), N(β)N(\beta) are elements of the integral domain Z\mathbb{Z}, to meet N(α)N(β)=0N(\alpha) N(\beta) = 0, either N(α)N(\alpha) or N(β)N(\beta) must be 00, which contradicts the assumption, thus Z[ω]Z[ \omega ] is also an integral domain.

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[2]

Definition of Euclidean norm:

  • (i): For all a,b∈D(b≠0)a,b \in D (b \ne 0 ), there exist qq and rr such that a=bq+ra = bq + r is satisfied. In this case, r=0r = 0 or ν(r)<ν(b)\nu (r) < \nu (b) must be met.
  • (ii): For all a,b∈D(b≠0)a,b \in D (b \ne 0 ), ν(a)≤ν(ab)\nu ( a ) \le \nu ( ab )

Let’s say ν:=N\nu := N and show that NN becomes the Euclidean norm of Z[ω]\mathbb{Z}[ \omega ]. Assume β≠0\beta \ne 0.

Part (i). ∃σ,ρ:α=βσ+ρ\exists \sigma, \rho : \alpha = \beta \sigma + \rho

For some r,s∈Qr, s \in \mathbb{Q}, let’s say αβ=r+ωs\displaystyle {{ \alpha } \over { \beta }} = r + \omega s. For r,sr, s close to an integer q1,q2∈Zq_{1} , q_{2} \in \mathbb{Z}, let’s consider σ\sigma and ρ\rho as follows. σ:=q1+ωq2ρ:=α−βσ \sigma := q_{1} + \omega q_{2} \\ \rho := \alpha - \beta \sigma

  • Case 1. ρ=0\rho = 0
    Since α=βσ\alpha = \beta \sigma, there’s no more to show. We found σ=αβ=q1+ωq2\displaystyle \sigma = {{ \alpha } \over { \beta }} = q_{1} + \omega q_{2} and ρ=0\rho = 0.
  • Case 2. N(ρ)<N(β)N ( \rho ) < N ( \beta )
    By definition of q1q_{1} and q2q_{2}, ∣r−q1∣≤12∣s−q2∣≤12 | r - q_{1} | \le {{1} \over {2}} \\ | s - q_{2} | \le {{1} \over {2}} thus, N(αβ−σ)=N[(r+ωs)−(q1+ωq2)]=N[(r−q1)+ω(s−q2)]=(r−q1)2−(r−q1)(s−q2)+(s−q2)2≤∣r−q1∣2+∣r−q1∣∣s−q2∣+∣s−q2∣2≤(12)2+(12)(12)+(12)2=34 \begin{align*} N \left( {{ \alpha } \over { \beta }} - \sigma \right) =& N \left[ (r + \omega s) - (q_{1} + \omega q_{2}) \right] \\ =& N \left[ (r - q_{1} ) + \omega ( s - q_{2}) \right] \\ =& (r - q_{1})^2 - (r - q_{1}) (s - q_{2}) + (s - q_{2})^2 \\ \le & |r - q_{1}|^2 + | r - q_{1} | | s - q_{2} | + | s - q_{2} |^2 \\ \le & \left( {{1} \over {2}} \right)^2 + \left( {{1} \over {2}} \right) \left( {{1} \over {2}} \right) + \left( {{1} \over {2}} \right)^2 \\ =& {{3} \over {4}} \end{align*} Accordingly, N(ρ)=N(α−βσ)=N(β(αβ−σ))=N(β)N(αβ−σ)≤N(β)34≤N(β) \begin{align*} N ( \rho ) =& N(\alpha - \beta \sigma ) \\ =& N \left( \beta \left( {{ \alpha } \over { \beta }} - \sigma \right) \right) \\ =& N (\beta) N \left( {{ \alpha } \over { \beta }} - \sigma \right) \\ \le & N(\beta) {{3} \over {4}} \\ \le & N (\beta) \end{align*} ensuring that ρ\rho, σ\sigma exist satisfying either ρ=0\rho = 0 or N(ρ)≤N(β)N(\rho) \le N (\beta).

Part (ii). N(α)≥N(α)N(β)N ( \alpha ) \ge N ( \alpha ) N ( \beta )

Since β≠0  ⟹  N(β)≥1\beta \ne 0 \implies N ( \beta ) \ge 1, N(α)≤N(α)⋅1≤N(α)N(β)=N(αβ) \begin{align*} N ( \alpha ) \le & N ( \alpha) \cdot 1 \\ \le & N(\alpha) N(\beta) \\ =& N ( \alpha \beta ) \end{align*}

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[3]

20190907\_230034.png

According to the properties of multiplicative norm, since N(1)=1N(1) = 1 and if u∈Z[ω]u \in \mathbb{Z}[ \omega ] is a unit, then ∣N(u)∣=1| N ( u ) | = 1, by contraposition, if not ∣N(u)∣=1| N(u) | = 1, then uu is not a unit. The case satisfying u:=x+ωyu := x + \omega y for N(u)=x2−xy+y2=1N(u) = x^2 - xy + y^2 = 1 is e0πi/3=1e1πi/3=−ω2e2πi/3=ωe3πi/3=−1e4πi/3=ω2e5πi/3=−ω \begin{align*} e^{0 \pi i / 3 } =& 1 \\ e^{1 \pi i / 3 } =& - \omega^2 \\ e^{2 \pi i / 3 } =& \omega \\ e^{3 \pi i / 3 } =& -1 \\ e^{4 \pi i / 3 } =& \omega^2 \\ e^{5 \pi i / 3 } =& - \omega \end{align*} i.e., only when u∈{±1,±ω,±ω2}u \in \left\{ \pm 1, \pm \omega, \pm \omega^2 \right\}.

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