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\(\Pi \equiv R - C\) where\(R \equiv P \cdot q\) and\(C \equiv F + V\)
\(MR \equiv \frac{\Delta R}{\Delta q}\)\(AR \equiv \frac{R}{q}\)
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\(\Pi \equiv (AR - AC) \cdot q\)
\(VC \equiv C - F\)\(AVC \equiv \frac{VC}{q} \equiv \frac{C - F}{q}\)
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\(q \equiv q_a + q_b\)
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Title
Division nearrings also called Nearfields constitute a class of skew-fields in which one distributive law is omitted. They were introduced in 1905 by L.E. Dickson, and later named by Hans Zassenhauss. To construct finite proper Dickson nearfields we use the Galois Field (GF(qn), + , ⋅ ) of order qn where (q, n) is a Dickson pair and we change the multiplicative operation ′ ⋅ ′ into a new one ′ ∘ ′. By this kind of construction we get that the finite Dickson nearfield associated with (q, n) is DN(q, n)=(GF(qn), + , ∘ ) such that (GF(qn), + , ∘ ) = GF(qn)ϕ, where ϕ is a coupling map and DN(q, n) is the ϕ−derivation of the finite field GF(qn).
A pairs of numbers (q, n)∈ℕ2 is called a Dickson pair if
q is some power pl of a prime p,
Each prime divisor of n divides q − 1,
If q ≡ 3 mod 4 implies 4 does not divide n.
The set $ \big \lbrace \frac{q^k-1}{q-1}, 1 \leq k < n \big \rbrace $ residues modulo n is the set {i, 0 ≤ i < n} where (q, n) are Dickson pairs. \label{le}
Let $i(k)=\frac{q^k-1}{q-1} $ for k = 1, ⋯, n.
We would like to show that the set {i(1),i(2),⋯,i(n)} residues modulo n is the set {0, 1, ⋯, n − 1}. It suffice to show that the set $ \big \lbrace \frac{q^k-1}{q-1}, 1 \leq k < n \big \rbrace $ are distinct residues modulo n.
Suppose that \begin{align}
& \frac{q^k-1}{q-1} \equiv \frac{q^l-1}{q-1} \quad \text{mod } n \quad 1\leq k
Assume without loss of generality that n = pm.
We know that q ≡ 1 mod p. So we can write q = 1 + pϵ for some ϵ ∈ ℕ.
Assuming that pm divides $ \frac{q^t-1}{q-1},$ we want to show that n = pm divides t leads to contradiction.
In fact \begin{align*} q^t = (1+ p \epsilon)^t = \sum _{k=0}^{t} \Vector {t}{k} (p \epsilon)^k. \end{align*} Hence \begin{align*} \frac{q^t-1}{q-1} = \sum _{k=1}^{t} \Vector {t}{k} (p \epsilon)^{k-1} = \cdots + \Vector{t}{2}p \epsilon + t. \end{align*} For instance
if m = 1, then the assumption is \begin{align*} p \big / \frac{q^t-1}{q-1} \Leftrightarrow p \big / \sum _{k=1}^{t} \Vector {t}{k} (p \epsilon)^{k-1} \Leftrightarrow p /t \end{align*} leads to contradiction since p = n > t.
if m = 2,
\begin{align*} p^2 \big / \frac{q^t-1}{q-1} \Leftrightarrow p^2 \big / \sum _{k=1}^{t} \Vector{t}{k} (p \epsilon)^{k-1} \Leftrightarrow p \big / \Vector{t}{2} p \epsilon +t \Rightarrow p /t \end{align*}
But then $\Vector{t}{2} = \frac{t(t-1)}{2}$, so $ p \big / \Vector{t}{2}$. Hence $p^2 \big / \Vector{t}{2} p \epsilon.$ Thus p2/t leads to contradiction.
By the same approach for some m, $p^m \big / \frac{q^t-1}{q-1} \Rightarrow n=p^m / t $ leads to contradiction.
Therefore the assumption [thd] can not hold. Thus the set $ \big \lbrace \frac{q^k-1}{q-1}, 1 \leq k < n \big \rbrace $ are distinct residues modulo n.
For all Dickson pair (q, n) where n ≠ 1, any Dickson nearfields constructed by the Galois Field GF(qn) are proper finite nearfields.
From finite Dickson construction \begin{align*} DN(q,n):= GF(q^n)^{\phi} = \big (GF(q^n), +, \circ \big ). \end{align*} We would like to show that (GF(qn), + , ∘ ) is not fields i.e., there exists a, b ∈ (GF(qn) such that a ∘ b ≠ b ∘ a. The coupling map is \begin{align*} \phi = \lambda \circ \pi : \quad & F^{ \times} \to \textit{Aut}(F,+,\cdot) \\ & f \mapsto \alpha^k \quad \mbox{for} \quad k=1,\cdots,n. \\ \Leftrightarrow \phi : f \mapsto \begin{cases} \alpha \quad \text{if $f \in Hg^{\frac{q-1}{q-1}} $} \\ \alpha^2 \quad \text{if $ f \in Hg^{\frac{q^2-1}{q-1}}$} \\ \vdots \quad \quad \vdots \\ \alpha^n \quad \text{if $f \in Hg^{\frac{q^n-1}{q-1}}$}. \end{cases} \end{align*} For a, b ∈ (GF(qn) \begin{align*} a \circ _{\phi} b= \begin{cases} \phi_b(a) \cdot b \quad \text{if $ b \neq 0 $} \\ 0 \quad \text{if $b=0$} \end{cases} = \begin{cases} \alpha (a) \cdot b \quad \text{if $ b \in Hg^{\frac{q-1}{q-1}} $} \\ \alpha^2 (a) \cdot b \quad \text{if $ b \in Hg^{\frac{q^2-1}{q-1}} $} \\ \vdots \quad \quad \vdots \\ \alpha^n (a) \cdot b \quad \text{if $ b \in Hg^{\frac{q^n-1}{q-1}} $} \end{cases} = \begin{cases} a^q \cdot b \quad \text{if $ b \in Hg^{\frac{q-1}{q-1}} $} \\ a^{q^2} \cdot b \quad \text{if $ b \in Hg^{\frac{q^2-1}{q-1}} $} \\ \vdots \quad \quad \vdots \\ a^{q^n} \cdot b \quad \text{if $ b \in Hg^{\frac{q^n-1}{q-1}} $} \end{cases} \end{align*} Let $a=g^n \in Hg^{\frac{q^n-1}{q-1}}$ and $ b=g \in Hg^{\frac{q^1-1}{q-1}}$
We have \begin{align*} g^n \circ g & = \alpha^1(g^n)g \\ & = (g^n)^qg \\ & = g^{nq+1}. \end{align*} Also \begin{align*} g \circ g^n & =\alpha ^n (g) g^n \\ & =g^{n+1} \quad \mbox{because} \quad \alpha^n =id. \end{align*} Assume that gnq + 1 = gn + 1, then gn(q − 1) = 1.
But since F× = ⟨g⟩, then ord
(g)=qn − 1.
It follows that if gt = 1 ⇒ qn − 1/t.
Moreover since \begin{align*} g^{n(q-1)}=1 & \Rightarrow q^n-1 / n(q-1) \\ & \Rightarrow 1+q+\cdots+q^{n-1} /n \end{align*} But q = pl > 1 so 1 + q + ⋯ + qn − 1 > n. It follows that 1 + q + ⋯ + qn − 1 does not divides n.
Thus gn(q − 1) ≠ 1. This means that gn ∘ g ≠ g ∘ gn.
There exists $a=g^n \in Hg^{\frac{q^n-1}{q-1}}$ and $ b=g \in Hg^{\frac{q^1-1}{q-1}}$ such that a ∘ b ≠ a ∘ b. Thus the finite Dickson nearfields associated to the pair (q, n) where q ≠ 1 are proper finite nearfields (not fields).
Let (q, n) be a Dickson pair, then GF(q)× is a subgroup of H. \label{thh}
Let GF(q)× = ⟨β⟩ and GF(qn)× = ⟨α⟩. We want to show that $ \beta = \alpha ^ {\frac{q^n-1}{q-1}}$.
We know that ord(α)=qn − 1 and ord(β)=q − 1. Also there exists i ∈ ℕ and β = αi since GF(q)× ⊂ GF(qn)×.
We have $ord(\beta)= ord(\alpha^i)= \frac{q^n-1}{gcd(i, q^n-1)}=q-1.$ It follows that $gcd(i, q^n-1)= \frac{q^n-1}{q-1}.$
We have that $i \geq \frac{q^n-1}{q-1}.$ In fact $i < \frac{q^n-1}{q-1},$ we have 1 = βq − 1 = (αi)q − 1 = αk with k ≤ i and −1 < qn − 1. It follows that αk ≠ 1 since ord(α)=qn − 1 (contradiction.) It follows that $i \geq \frac{q^n-1}{q-1}$.
We take $i=\frac{q^n-1}{q-1}.$ Since $\beta = \alpha^{\frac{q^n-1}{q-1}},$ it follows that $ \beta \in H \alpha^{\frac{q^n-1}{q-1}}=H.$ Whence GF(q)× = ⟨β⟩≤H.
In this section we prove a result that have been stated in the paper and book \cite{wahling1987theorie,boykett2016multiplicative} and is useful in finite nearfield for the representation of the multiplicative group.
Notation \begin{align*} \underbrace{x \circ x \cdots \circ x }_a= x^{\underline{a}} \end{align*} where $"\circ"$ is the new multiplication that come from the finite Dickson nearfield (DN(q, n), + , ∘ ).
\cite{wahling1987theorie} Let (q, n) be a Dickson pair. Then we have the following \begin{align*} DN(q,n)^{\times} \cong \big < x,y \vert x^m=1,x^t=y^n,y^{-1}xy=x^q \big > \end{align*} where $m=\frac{q^n-1}{n}$ and $t= \frac{m}{q-1}$.
We define the mapping, \begin{align*} f:\big < x,y \vert x^m=1,x^t=y^n,y^{-1}xy=x^q \big > \to DN(q,n)^{\times}. \end{align*} We suppose that $ \big < g\big > = GF^{\times}(q^n)$. Let f(x)=gn and f(y)=g. We need to show that f is bijective homomorphism.
By standard Theorem in group presentations f is group homomorphism if and only if f(xm)=1, f(xt)=f(yn) and f(y−1xy)=f(xq).
We have, \begin{align*} f(x^m)=f(x)^{\underline{m}}= (g^n)^{\underline{m}}=g^{nm}= g^{q^n-1}=1 \end{align*} since gqn = g for $ \big < g\big > = GF^{\times}(q^n)$. \begin{align*} f(x^t)=f(x)^{\underline{t}}=(g^n)^{\underline{t}}=g^{nt}=g^{ \frac{q^n-1}{q-1}}, \end{align*} Also, \begin{align*} f(y^n)=f(y)^{\underline{n}}=g^{\underline{n}}=g^{1+q+\cdots+ q^{n-1}}=g^{ \frac{q^n-1}{q-1}}, \end{align*} Hence \begin{align*} f(x^t)=f(y^n). \end{align*} Furthermore f(xq)=gnq \begin{align*} f(y^{-1}xy) & = f(y^{-1}) \circ f(x) \circ f(y) \\ & = g^{-\underline{1}} \circ g^n \circ g \\ & = g^{-\underline{1}} \circ \big ( (g^n)^q \cdot g \big ) \\ & = (g^{-\underline{1}}) \circ g^{nq+1} \\ &=\big (g^{-\underline{1}} \big )^q \cdot g^{nq+1} \\ &= g^{-\underline{1}} g^{nq+1} \quad \mbox{since} \thickspace g^{nq+1} \in Hg \\ &= g^{nq}. \end{align*} Let compute $ g^{-\underline{1}}$.
Since $g^{-\underline{1}} \circ g =1 $ then $\big (g^{-\underline{1}} \big )^q \cdot g =1 .$ So $\big (g^{-\underline{1}} \big )^{q^n} \cdot g^{q^{n-1}}=1$. It follows that $g^{-\underline{1}}= g^{-q^{n-1}} .$
f is surjective:
Given gr ∈ DN(q, n)×. We write $g^r= g^{ \frac{q^s-1}{q-1}} \cdot g^{nt}$ for some s, t. It is possible since $g^r \in H g^{ \frac{q^s-1}{q-1}}$ for some s. Then we consider, \begin{align*} f(y^sx^t)& = f(y)^{\underline{s}} \circ f(x)^ {\underline{t}} \\ & = g^{\underline{s}} \circ g^{nt} \\ &= g^{ \frac{q^s-1}{q-1}} \cdot g^{nt} \\ &= g^r. \end{align*}
We shall show that $ \big \vert \big < x,y \vert x^m=1,x^t=y^n,y^{-1}xy=x^q \big > \big \vert \leq nm = q^n-1= \vert DN(q,n)^{\times} \vert $.
Note that since xy = yxq, the elements y′s can be shifted to the left of x, i.e., elements can be written yaxb for some a, b ∈ ℤ. Note for then that y-exponents can be reduced modn (at the expense of xt) and that x-exponents can be reduced modm. Hence, $ \big \vert \big < x,y \vert x^m=1,x^t=y^n,y^{-1}xy=x^q \big > \big \vert = \vert \lbrace y^ax^b, 0 \leq a< n, 0 \leq b <m \rbrace \vert \leq nm.$
Since f is surjective and $ \big \vert \big < x,y \vert x^m=1,x^t=y^n,y^{-1}xy=x^q \big > \big \vert \leq \vert DN(q,n)^{\times} \vert$ then f is bijective.
We will see the relation between the center and the kernel of a nearfield.
Let N be a nearfield.
The center of (N, ⋅) is defined as follows: \begin{align*} C(N):= \left\lbrace x \in N : xy=yx, \forall y \in N \right\rbrace \end{align*} i.e., it is the set of elements that commute with every element of N.
The kernel of (N, +) is defined as follows: \begin{align*} N_d:= \left\lbrace x \in N : x(y+z)=xy+xz, \forall y,z \in N \right\rbrace \end{align*} i.e., it is the set of distributive elements.
(Subnearfield test)
Let (N, +, ⋅) be a nearfield and M be a subset of N. (M, +, ⋅) is a subnearfield of (N, +, ⋅) if and only if these followings axioms hold:
M× ≠ ∅.
For all x, y ∈ M: x + ( − y)∈M.
For all x, y ∈ M: x ⋅ y ∈ M.
For all x ∈ M×: x−1 ∈ M.
\citep{zemmer1969additive,pilz2011near} Let N be a nearfield. Then Nd is a sub-nearfield of N (with the operations of N). \label{t:1}
1 ⋅ (y + z)=1 ⋅ y + 1 ⋅ z ⇒ 1 ∈ Nd ⇒ Nd ≠ ∅. Also it is clear that Nd ⊆ N. Now let a, b ∈ Nd and α, β ∈ N. We have : \begin{align*} (a-b)( \alpha + \beta) & = a( \alpha + \beta) - b ( \alpha + \beta) = a \alpha + a \beta -b \alpha - b \beta \\ & = a \alpha - b \alpha+ a \beta -b \beta =(a-b) \alpha + (a+b) \beta. \end{align*} Therefore (a − b)∈Nd. Let k ∈ Nd×, so k ≠ 0 and let α, β ∈ N. We have that \begin{align*} k \left[ k^{-1}( \alpha + \beta) \right] = \alpha + \beta = k ( k^{-1} \alpha) + k ( k^{-1} \beta) = k ( k^{-1} \alpha + k^{-1} \beta). \end{align*} Thus k−1(α + β)=k−1α + k−1β, since k ≠ 0, which implies that k−1 ∈ Nd×.
Moreover let a, b ∈ Nd and α, β ∈ N , then \begin{align*} ab ( \alpha + \beta) & = a( b \alpha +b \beta) = a( b \alpha) + a ( b \beta) =(ab) \alpha + (ab) \beta. \end{align*} Thus (ab)∈Nd. Thus Nd is a subnearfield of (N, +, ⋅).
We deduce the following result.
Let N be a nearfield. Then Nd with the operations of N is a skew-field. \label{t:2}
From Theorem [t:1], (Nd, +) is an abelian group and (Nd×, ⋅) is a group. Let u, v, w ∈ Nd. We have that u(v + w)=uv + uw (from the definition of Nd) and (u + v)w = uw + vw (because Nd ⊆ N and property (NF3) holds). Thus (N, +, ⋅) is a skew-field.
Let N be a nearfield. The center of N is a normal subgroup of the multiplicative group of N.
The center of N is always a subgroup of (N×, ⋅). In fact,
C(N) contains the identity element of N, 1, because it commutes with x for all x ∈ N, i.e., 1 ⋅ x = x = x ⋅ 1, by definition.
If x and y are in C(N), then so is xy by associativity: (xy)z = x(yz)=x(zy)=(xz)y = (zx)y = z(xy) for each z ∈ N i.e. C(N) is closed under multiplication.
If x ∈ C(N)×, then so is x−1 for all z ∈ N, x−1 commutes with z : zx = xz ⇒ x−1zxx−1 = x−1xzx−1 ⇒ x−1z = zx−1.
Let y ∈ N× and x ∈ C(N). Let z ∈ N. We have xy = yx ⇒ x = yxy−1 and xz = zx so (yxy−1)z = z(yxy−1). It follows that yxy−1 ∈ C(N). Thus C(N) is a normal subgroup of N.
Let x ∈ C(N), we have xy = yx for all y ∈ N×. It follows that x = yxy−1 for all y ∈ N×, so x ∈ yC(N)y−1. It follows that C(N)⊆yC(N)y−1. We also know that yC(N)y−1 ⊆ C(N). Thus C(N)=yC(N)y−1 for all y ∈ N×.
Notice that (C(N), + , ⋅ ) is a subnearfield of (N, +, ⋅). The center of a nearfield is a field. To see this, for all a, b, c ∈ N, we have c ⋅ (a + b)=(a + b)⋅c = a ⋅ c + c ⋅ b = c ⋅ a + c ⋅ b. Thus the right and left distributive axioms hold.
\citep{pilz2011near} Let M be subnearfield of a nearfield N, then N can be considered as a vector space over M. It’s dimension will be denoted by dimMN. \label{t:6}
Take M = Nd (from Theorem [t:1]). Let u, v ∈ N and α, β ∈ Nd. (N, +) is abelian group. There exists a scalar multiplication N × Nd → N, namely (α, u)↦α ⋅ u which satisfies the following \begin{align*} & (u+v) \alpha = u \alpha + v \alpha, \\ & u (\alpha + \beta) = u \alpha + u \beta, \\ & u( \alpha \beta) = (u \alpha )\beta, \\ & u 1 \cdot =u \end{align*} Thus N is a vector space over Nd. M can be Nd or C(N). So the structures (N, Nd) and (N, C(N)) are vector spaces.
It’s dimension will be denoted dimNd(N) and called the rank of N. If N is a nearfield of rank 2 then any set {1, e} with e ∈ N \ Nd is a basis of N over Nd, hence N = Nd + Nde.
Let N be a nearfield. Then N = Nd if and only if N is a skew-field.
(⇒) Suppose N = Nd, by Lemma $\ref{t:2},$ Nd is a skew-field and since N = Nd, N is also a skew-field.
(⇐) Suppose N is a skew-field. We have that Nd ⊆ N. We need to show that N ⊆ Nd. Let f ∈ N, for all g, h ∈ N we have that f(g + h)=fg + fh, so f ∈ Nd. Thus N ⊆ Nd.
Thus Nd = N.
C(N)⊆Nd. \label{t:3} To see this, let f, g, h ∈ C(N). From Definition [th:t8], (g + h)f = gf + hf = fg + fh, but (g + h)f = f(g + h). Thus f(g + h)=fg + fh ⇒ f ∈ Nd.
There exists the following relation between the center and the kernel of a nearfield.
\citep{wahling1969invariante,andre1963uber,pilz2011near} Let N be a nearfield which is not proper skew-field. Then C(N)=⋂x ∈ N×x−1(Nd)x. \label{sk}
″ ⊆ ″ Let f ∈ C(N), then f ∈ N and fg = gf, ∀g ∈ N. We know that C(N)⊆Nd from Remark [t:3]. So f ∈ Nd, \begin{align*} & f=x^{-1}xf=x^{-1}fx \quad \mbox{for all } x \in N^{\times} \\ & \Rightarrow f \in x^{-1}N_d x \\ & \Rightarrow f \in \bigcap_{x \in N^{\times}} x^{-1} N_d x \\ & \Rightarrow C(N) \subseteq \bigcap_{x \in N^{\times}} x^{-1} N_d x. \end{align*}
″ ⊇ ″ Let a ∈ ⋂x ∈ N×x−1Ndx
By Theorem [th:q] a nearfield has order pn for some p prime and n ∈ ℕ.
Nd is a proper subnearfield of N by Theorem [t:1] because N is not a skew-field.
Let Ca = {x∈N:xa=ax} for a ∈ N. 1 ∈ Ca because 1 ⋅ a = a ⋅ 1 = a. So Ca ≠ ∅. Also Ca ⊆ N.
Let α, β ∈ Ca and a ∈ ⋂x ∈ N×x−1Ndx ⊆ N, then (α − β)a = αa − βa = aα − aβ = a(α − β)⇒(α − β)∈Ca . Also a(αβ)=a(βα)=(aβ)α = β(aα)=β(αa)=(βα)a = (αβ)a ⇒ (αβ)∈Ca. Let k ∈ Ca×, (kk−1)a = k(k−1a)=a = (ak)k−1 = (ka)k−1 = k(ak−1). Since k ≠ 0, then k−1a = ak−1. It follows that k−1 ∈ Ca×. Thus Ca is a subnearfield of N.
If |N×|=pn − 1 then |Ca×|=pk − 1 and |Nd×|=pl − 1 for some k ≤ n, l < n. We know that |N|=pn, So |N×|=pn − 1. Also since Nd is a proper subnearfield of N, there exists l ∈ ℕ such that |Nd|=pl where l < n. Therefore |Nd×|=pl − 1 for l < n. Also Ca is a subnearfield of N. Thus there exists k ∈ ℕ such that |Ca|=pk for some k ≤ n. Therefore |Ca×|=pk − 1.
Since Ca× and Nd× are subgroups of N× by Lagrange’s Theorem |Ca×| divides |N×|. Also |Nd×| divides |N×|.
For all prime p, pr − 1 divides ps − 1 if and only if r divides s.
Let Con(a)={xax−1, x ∈ N×} be the conjugacy class of a. Given xax−1 ∈ Con(a), a ∈ ⋂x ∈ N×x−1Ndx ⇒ ∃λ ∈ Nd such that a = x−1λx. So xax−1 = x(x−1λx)x = λ ∈ Nd ⇒ Con(a)⊆Nd×.
Let us consider the finite group N× that acts on the finite set Ca× by g(a)=gag−1. We have : orb(a)={g(a):g ∈ N×}={gag−1 : g ∈ N×}=Con(a) and stab(a)={g ∈ N× : g(a)=a}={g ∈ N× : gag−1 = a}={g ∈ N× : ga = ag}=Ca×. Therefore |orb(a)||stab(a)| = |N×|⇒|Con(a)||Ca×|=|N×|. It follows that $ \vert Con (a) \vert = \frac{ \vert N^ {\times} \vert}{ \vert C_a^{ \times} \vert }$.
Since pk − 1 divides pn − 1, k divides n, so there exists g ∈ ℕ such that n = gk. In fact $ \vert Con (a) \vert = \frac{ \vert N^ {\times} \vert}{ \vert C_a^{ \times} \vert }=\frac{p^n-1}{p^k-1}=1+p^k+ \cdots + p^{(g-1)k} \leq p^l-1 $ because Con(a)⊆Nd×. So (g − 1)k = n − k < l ≤ n/2. Also k/n < 2n ⇒ k = n. Therefore $ \vert Con (a) \vert =\frac{p^n-1}{p^k-1}=1$. It follows that Con(a)={a}.
Since Con(a)={a}⇒xax−1 = a. It follows that for x ∈ N×, xa = ax. Thus a ∈ C(N).
Thus ⋂x ∈ N×x−1Ndx ⊆ C(N).
Let N be a nearfield which is not proper skew-field. C(N)=Nd if and only if for all x ∈ N× : x−1Ndx = Nd. \label{th}
( ⇒ ) Assume that C(N)=Nd. We would like to show that for all x ∈ N× : x−1Ndx = Nd.
″ ⊆ ″ Let ϵ ∈ x−1Ndx. So there exists λ ∈ Nd such that ϵ = x−1λx. Since x, x−1 ∈ N× and λ ∈ N, x−1λx ∈ N× ⊆ N. Let y, z ∈ N, then \begin{align*} (x^{-1} \lambda x )(y+z)& =x^{-1}x \lambda (y+z) \thickspace \mbox{since} \thickspace C(N)=N_d \\ & = \lambda (y+z) \\ & =\lambda y + \lambda z \thickspace \mbox{since} \thickspace \lambda \in N_d \\ &=x^{-1}x \lambda y +x^{-1}x \lambda z \\ & = (x^{-1} \lambda x)y + (x^{-1} \lambda x)z \thickspace \mbox{since} \thickspace C(N)=N_d. \end{align*} Therefore x−1λx ∈ Nd ⇒ ϵ ∈ Nd. It follows that x−1Ndx ⊆ Nd.
″ ⊇ ″ Let ϵ ∈ Nd. Since C(N)=Nd, ϵ = 1 ⋅ ϵ = x−1xϵ, for all x ∈ N×. So ϵ = x−1ϵx, i.e., ϵ ∈ x−1Ndx. This implies Nd ⊆ x−1Ndx. Thus Nd = x−1Ndx.
( ⇐ ) Assume that ∀x ∈ N×, x−1Ndx = Nd. We want to show that C(N)=Nd.
″ ⊆ ″ From Remark [t:3] we have C(N)⊆Nd.
″ ⊇ ″ By Theorem [sk], C(N)=⋂x ∈ N×x−1Ndx. So ⋂x ∈ N×x−1Ndx ⊆ C(N). It follows that ⋂x ∈ N×Nd ⊆ C(N)⇒Nd ⊆ C(N).
Thus we get C(N)=Nd.
By Theorem [t:8], for all finite Dickson nearfields, we have the following Theorem.
(\citep{pilz2011near}, p.256) Let F be the Galois Field GF(qn)=GF(pln) and N := Fϕ the ϕ−derivation of F where (q, n) are pair of Dickson numbers. Then \begin{align*} C(N)=N_d =GF(q). \end{align*} \label{center}
We will show this fact in the next section.
For Example [th:t10] C(N)=Nd = ℤ3. \label{esss}
In this section we give the proof of the fact that C(N)=GF(q) which mentioned in chapter 9 of Beidleman thesis \cite{beidleman1966near} at page 116.
If K ⊆ F is a field extension (i.e., K is subfield of F), then F is a vector space over K.
Straight-forward.
If K ⊆ F is a field extension and F is finite, then |F|=|K|n for some n ∈ ℕ. \label{c1}
We use that fact that a finite-dimensional vector space over K satisfies F ≅ Kn for some n ∈ ℕ (the dimension of F over K.)
A polynomial equation of degree n has at most n roots over any field.
We use induction and factor out linear terms.
Let (q, n) be Dickson pair. Then n divides $ \frac{q^n-1}{q-1}.$
Similar to the proof of Lemma [le].
The set of elements fixed by a field automorphism is a field.
The subfields of GF(pm) are precisely those GF(pt) with t ∣ m and are unique. \label{l3}
Furthermore, they are the fields fixed by the automorphism ψt : x ↦ xpt, i.e., are the solutions of xpt − x = 0.
Suppose K ⊆ GF(pm) is a subfield. Then ℙ, the prime field of GF(pm) is a subfield of K. Since ℙ ≅ GF(p), we have ∣ℙ ∣ =p, hence ∣K ∣ =pt for some t ∈ ℕ by Corollary [c1].
Since GF(pm) is a vector space over K, we have ∣GF(pm)∣ = pm = ∣K∣r = ptr for some r ∈ ℕ, i.e., t ∣ m.
For uniqueness notice that any x ∈ K satisfy x∣K∣ = x, i.e., xpt − x = 0 ( in K, but also in GF(pm) ). Hence any subfield of size pt coincides with the set of solutions to xpt − x = 0
Let GF(q) be the unique subfield of order q of GF(qn). Then \begin{align*} GF(q) \subseteq C \big ( DN(q,n) \big ). \end{align*}
By Lemma [l3], GF(q) is the solution set to the equation xq − x = 0 in GF(qn). Let g be a generator of GF(qn)× and take x ∈ GF(q) and write x = gl for x ≠ 0. Since x ∈ GF(q), xq = x, i.e., xq − 1 = 1. Then (gl)q − 1 = 1, i.e.,gl(q − 1) = 1. Thus ordGF(qn)×(g)(qn − 1) in GF(qn) divide l(q − 1), i.e., $\frac{q^n-1}{q-1} \mid l.$ Thus $GF(q^n)^{\times} = \big < g^ \frac{q^n-1}{q-1} \big >$. Since $ n \mid \frac{q^n-1}{q-1},$ we have $GF(q)^{\times} \subset \big < g^n \big >= H$. We also see this by Theorem [thh]. Now, ϕ(x)=ϕx = id since $x \in H=Hg^{\frac{q^n-1}{q-1}}$. So by Dickson construction, ϕ(x)=αn = id.
Take any t ∈ DN(q, n). Then x ∘ t = ϕt(x)⋅t = αl(x)⋅t = x ⋅ t since x is fixed by α.
Also t ∘ x = ϕx(t)⋅x = id(t)⋅x = t ⋅ x = x ⋅ t. Whence x ∈ C(DN(q, n)).
(a + b)p = ap + bp for all a, b ∈ F where p is a prime, and F a finite field of characteristic p with order pn where n ∈ ℕ. \label{t:4444}
From the Binomial Theorem, \begin{equation*} (a+b)^p = \sum _{k=0} ^ {p} \Vector{p}{k} a^ {p-k}b ^ k = a^p+ \Vector{p}{1}a^{p-1}b + \Vector{p}{2}a^{p-2}b^2 + \cdots + \Vector{p}{p-1}ab^{p-1} + b^p . \end{equation*} $\Vector{p}{k}$ is a multiple of p for all k ∈ {1, …, p − 1} since $p \Vector{p-1}{k-1}=k \Vector{p}{k} $ and gcd (p, k)=1. Hence $ \Vector{p}{k} =0 $ in F. Thus all the middle terms cancel and we get (a + b)p = ap + bp.
GF(p)(gn)=GF(qn) where GF(p) is the unique subfield of GF(qn) of order p. \label{l4}
In fact we shall prove the stronger GF(p)(gn)=GF(qn). Let f be the smallest positive integer such that (gn)pf = gn. Then gn is a solution to the equation xpf − x = 0. In fact every x in GF(p)(gn) satisfies xpf − x = 0, since if $x= \sum _{i=0}^N a_ig^{nb_i},$ where ai ∈ GF(p) and bi ∈ ℤ. Then \begin{align*} x^{p^f} & = \big ( \sum _{i=0}^N a_ig^{nb_i} \big ) ^{p^f} \\ &= \sum_{i=0}^N (a_ig^{nb_i})^{p^f} \quad \mbox{by Theorem} \thickspace \ref{t:4444} \ \\ &= \sum_{i=0}^N a_i^{p^f} \big ((g^{n})^{p^f} \big )^{b_i} \\ &= \sum_{i=0}^N a_ig^{nb_i} \\ &=x. \end{align*} Thus GF(p)(gn)⊆GF(pf), but note that since f is minimal, GF(p)(gn)=GF(pf).
Since (gn)pf = gn, we have, \begin{align*} (g^n)^{p^f-1}=1 \Leftrightarrow g^{n(p^f-1)}=1, \end{align*} hence \begin{align*} ord(g)= q^n-1 = p^{ln}-1 \mid n(p^f-1). \end{align*} Since GF(p)(gn)=GF(pf)⊆GF(pln), f ∣ ln.
If f ≠ ln, then $f \leq \frac{ln}{2}.$ We show that this leads to a contradiction. \begin{align*} p^{ln}-1 \mid n(p^f-1) \Rightarrow p^{ln}-1 \leq n(p^f-1)\leq n( p^{\frac{ln}{2}}-1). \end{align*} Dividing by $p^{\frac{ln}{2}}-1,$ we get $p^{\frac{ln}{2}}+1 \leq n, $ but \begin{align*} p^{\frac{ln}{2}}+1 \geq 2^{\frac{ln}{2}}+1 \geq 2^{\frac{n}{2}}+1>n. \end{align*} (contradicts)
Thus f = ln, so GF(p)(gn)=GF(qn).
Let GF(q) be the unique subfield of order q of GF(qn). Then \begin{align*} C \big ( DN(q,n) \big ) \subseteq GF(q). \end{align*}
Take x ∈ C(DN(q, n)). Then x ∘ t = t ∘ x for all t ∈ C(DN(q, n)). Let x = gn ∈ H. Then x ∘ t = ϕt(x)⋅t = x ⋅ t since ϕt = id. So t ∘ x = ϕx(t)⋅x. Hence ϕx(t)=t. Also, ϕx fixes GF(q) since GF(q) is fixed by α. Thus ϕx fixes GF(p)(gn), the smallest subfield of GF(qn) that contains GF(q) and gn. By Lemma [l4], ϕx fixes GF(qn), i.e, ϕx = id.
Now take $t=g \in Hg^{\frac{q^1-1}{q-1}},$ so ϕg = α. Then g ∘ x = ϕx(g)⋅x = g ⋅ x. Also x ∘ g = ϕg(x)⋅g = α(x)⋅g. Thus α(x)=x, i.e., xq = x, so x ∈ GF(q).
In this section we aims to show that the additive group of a finite nearfield is always abelian.
Let the characteristic, char N, of a nearfield N be defined as usual in a field, i.e., it is the smallest positive integer n such that $\underbrace{1+1+ \cdots + 1}_{n}=0.$ \label{th:t11}
In Example [th:t10], the finite nearfield (N9, +, ∘) has characteristic 3.
\citep{pilz2011near} Let N be a nearfield. Let ord(1) be the order of 1 in (N, +). Then
If ord (1) is finite then char N = ord (1),
If ord (1) is infinite then char N = 0.
\label{th:r}
Assume that ord(1)<∞. By definition ord(1) is the smallest positive integer p such that p ⋅ 1 = 0 i.e., $\underbrace{1+1+ \cdots + 1}_{p}=0.$ Then it follows that ord(1)=char N from Definition [th:t11].
If ord(1) is infinite then it means that there is not an integer p such that $\underbrace{1+1+ \cdots + 1}_{p}=0.$ Therefore $\underbrace{1+1+ \cdots + 1} _{p} \neq 0.$ It follows that char N = 0.
Let N be a nearfield which is an integral domain and ord(1) be the order of 1 in (N, +). Then the characteristic of N is either 0 or a prime.
If ord(1) is infinite, then all elements 1, 1 + 1, 1 + 1 + 1, …, 1 + ⋯, … are distinct. So there is no p such that $\underbrace{1+1+ \cdots + 1}_{p}=0$. Therefore $\underbrace{1+1+ \cdots + 1}_{p} \neq 0.$ Then char N = 0. Suppose ord(1) is finite. We want to show that char N is prime. Let a, b ∈ ℕ − {1}. Assume that char N = k = ab which is not prime. Then a, b are not the characteristic of N. It implies that $\underbrace{1+1+ \cdots + 1}_{a} \neq 0$ and $\underbrace{1+1+ \cdots + 1}_{b} \neq 0$. Note that $ (\underbrace{1+1+ \cdots + 1}_{a}) \cdot ( \underbrace{1+1+ \cdots + 1}_{b}) =\underbrace{1+1+ \cdots + 1}_{ab}= (1 \cdot a) \cdot (1 \cdot b) = (a \cdot b) \cdot 1 = k \cdot 1 = 0 $. It implies that 1 ⋅ a = 0 or 1 ⋅ a = 0 since N has no zeros divisors. But both a and b are less that n which contradicts the minimality of n. Hence char N = k is prime.
Let N be a nearfield. Then for all n, n′∈N,
0n = 0,
( − n)n′= − nn′.
For n, n′∈N, we have
0n = (0 + 0)n by the right distributive law. \begin{align*} 0n & =(0+0)n \\ & =0n+0n \\ & \Rightarrow 0n=0 \end{align*}
\begin{align*} (-n)n' & = (-n+0) n' \\ &=-nn'+0n \\ &= -nn'. \end{align*} Thus ( − n)n′= − nn′.
A similarity between finite fields and finite nearfields is that the orders of finite nearfields are the same as those of finite fields.
If N is a finite nearfield, then there exists a prime p ∈ ℕ and an integer k ∈ ℕ such that |N|=pk. \label{th:q}
N is a right vector space over ℤp where char N = p. Since N is finite, it is finite-dimensional as a vector space over ℤp. Then N has a basis over ℤp consisting of k elements b1, …, bk. Thus every element of N can be represented in the form a1b1 + ⋯ + akbk where a1, …, ak ∈ ℤp. Since ai can have p values, N has exactly pk elements.
A p-group is a group whose order is some power of a prime.
A non-trivial p-group is the one whose order is pn for some n ∈ ℤ+ and p a prime.
Let p be a prime number. The center of any p−group is non-trivial. \label{thh}
Suppose G is a p-group, so |G|=pn for some n > 0. By the class equation \begin{align*} \vert G \vert & = \vert Z(G) \vert + \frac{\vert G \vert }{\vert Z(a_1) \vert} + \cdots + \frac{\vert G \vert }{\vert Z(a_n) \vert} \\ & = \vert Z(G) \vert + \sum _{i=1}^{i=n} \big [ G : Z(a_i) \big ]. \end{align*} Recall that $ \frac{\vert G \vert}{\vert Z(a_1) \vert}= \big [ G : Z(a_i) \big ]$. For all ai (that are not in the center and represent a conjugacy class), we see that every term $\sum _{i=1}^{i=n} \big [ G : Z(a_i) \big ]$ is divisible by p since ai ∉ Z(G), this implies that [G : Z(ai)] > 1.
Then \begin{align*} \frac{\vert G \vert}{ \vert Z(a_i) \vert } > 1 \Rightarrow \frac{p^n}{\vert Z(a_i) \vert} =p^{r_k} >1. \end{align*} So pn = |Z(G)| + pr1 + ⋯ + prk.
Since p divides pn, pr1, ⋯, prk then p divides |Z(G)|. It follows that |Z(G)| ≠ 1. Thus |Z(G)| > 1.
Notice that Z(ai)=CG(ai) is the centraliser.
Using Theorem [thh] we would like to prove in another way that the additive group of a finite nearfield is abelian.
Let N be a finite nearfield. The additive group (N, +) is abelian.
By Theorem [th:q] |N|=pk for some k ∈ ℕ where p is prime. It follows that (N, +) is p−group. This means that (N, +) has non-trivial center by Theorem [thh]. It follows that there exists y ≠ 0 such that for all x ∈ N, y + x = x + y.
Moreover given a, b ∈ N×, let a′= − b + a + b. We would like to show a′=a.
Now a′= − b + a + b ⇒ b + a′=a + b ⇒ (b + a′) ⋅ (b−1y)=(a + b)⋅(b−1y)⇒bb−1y + a′b−1y = ab−1y + bb−1y ⇒ y + a′b−1y = ab−1y + y.
Since y is in the center, y + a′b−1y = ab−1y + y = y + ab−1y ⇒ a′b−1y = ab−1y ⇒ a′b−1 = ab−1 ⇒ a′=a ⇒ −b + a + b = a ⇒ (b − b)+a + b = b + a ⇒ a + b = b + a. Thus (N, +) is abelian.
Another similarity is that if F is a finite field of cardinality pn (where p is a prime), then the additive group of F is isomorphic to ℤpn, which is also true for finite nearfield.
Let N be a finite nearfield of cardinality pn where p is prime. We have \begin{align*} (N,+) \cong ( \mathbb{Z}_p^n,+). \end{align*}
Let us suppose that char N = p. By the classification of finite abelian groups, there exist a unique tuple (d1, …, dk) of positive integers such that N ≅ ℤd1 × ℤd2 × ⋯ × ℤdk and 1 < d1, d1|d2|…|dk for all di ∈ ℕ where i = 1, …, k. Let αi ∈ N, and ⟨αi⟩ be the generator of the cyclic group ℤdi for i = 1, …, k. Therefore p ⋅ αi = 0, because char N = p. Since ord(αi)=di, di divides p. But |N|=pn = d1 × ⋯ × dk. Since di divides p with p a prime and di > 1, di = p. Therefore, k = n, then \begin{align*} N & \cong \mathbb{Z}_{d_1} \times \mathbb{Z}_{d_2} \times \cdots \times \mathbb{Z}_{d_k} \\ & \cong \mathbb{Z}_{p} \times \mathbb{Z}_{p} \times \cdots \times \mathbb{Z}_{p} \\ & \cong \mathbb{Z}_{p}^n. \end{align*}
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Let’s consider a system with two RC filters followed by one CR filter. The integration step transforms a Dirac’s delta pulse into a Heaviside funtion. This has Laplace transform equal to \begin{equation} F_0(s) = \frac 1 s \end{equation} Each RC filter has a Laplace transform of \begin{equation} F_{RC}(s, \tau) = \frac 1 { 1+ \tau s} \end{equation} While the CR filter has a Laplace transform of \begin{equation} F_{CC}(s, \tau) = \frac {\tau s} { 1+ \tau s} \end{equation} We will name the time constants of the RC filters as a, b and the time constant of the CR filter as c. The response to a pulse in the time domain is therefore \begin{equation} \label{eq:invlaplace1} ILT \left \{ \frac 1 s \times \frac 1 { 1+ a s} \times \frac 1 { 1+ b s} \times \frac {c s} { 1+ c s} \right \} \end{equation} which, in the case of a ≠ b, b ≠ c, a ≠ c corresponds to \begin{eqnarray} f(a, b, c, t) = f_1(a, b, c, t) + f_2(a, b, c, t) + f_2(a, b, c, t) \\ f_1(a, b, c, t) = \frac {a \, e ^{-t/a}} {(a-c)(a-b)} \\ f_2(a, b, c, t) = \frac {b \, e ^{-t/b}} {(b-c)(b-a)} \\ f_3(a, b, c, t) = \frac {c \, e ^{-t/c}} {(c-a)(c-b)} \end{eqnarray} while in the special case of a = b = c = τ corresponds to \begin{equation} f(\tau, t) = \frac 1 2 \left ( \frac t \tau \right )^2 e^{-t/\tau} \end{equation} We remind here that in general a sequence of N RC filters followed by one CR filter, all with the same time constant τ has a response to pulse of \begin{equation} f(\tau, t) = \frac 1 {N!} \left ( \frac t \tau \right )^N e^{-t/\tau} = \frac 1 {\Gamma(N+1)} \left ( \frac t \tau \right )^N e^{-t/\tau} \end{equation} with the latter formulation being apt to be used in a fit with an unknown N.