Intermezzo: Arithmetic Derivative

Basic idea and definition.

Some readers might remember that the analogy between $\mathbb{Z}$$\mathbb Z$ and $\mathbb{C}[T]$$\mathbb C[T]$. One result we already get is analogy between ${\mathbb{Q}}_p$$\mathbb Q_p$ and meromorphic function. Now we will use the analogy define the derivation of $\mathbb{Z}$$\mathbb Z$.

By fundamental theorem of arithmetic, we now that for any $n\in \mathbb{Z}-\{0\}$$n\in\mathbb Z-\{0\}$, $n=\pm p_1^{e_1}{p}_2^{e_2}...p_n^{e_n}$$n=\pm p_1^{e_1}p_2^{e_2}...p_n^{e_n}$.

Similarly, by fundamental theorem of algebra, we now that for any $p(T)\in \mathbb{C}[T],p(T)=\mu (T-{\lambda }_1{)}^{e_1},...,(T-{\lambda }_m{)}^{e_m}$$p(T)\in\mathbb C[T],p(T)=\mu (T-\lambda_1)^{e_1},...,(T-\lambda_m)^{e_m}$.

The derivation of $\mathbb{C}[T]$$\mathbb C[T]$ tell us that

$$\begin{array}{}\text{(1)}& d(p(T)=d_T(\mu (T-{\lambda }_1{)}^{e_1},...,(T-{\lambda }_m{)}^{e_m})=\sum _{i=1}^m{e}_i\frac{p(T)}{(T-{\lambda }_i)}\end{array}$$

Similarly, we could define $d(n)$$d(n)$ by Leibniz Law.

$$\begin{array}{}\text{(2)}& d_{\mathbb{Z}}(n)=d_{\mathbb{Z}}(\pm p_1^{e_1}{p}_2^{e_2}...p_n^{e_n})=n\sum _{i=1}^n\frac{e_i}{p_i}\end{array}$$

However, this is not additive! For example, $\begin{array}{c}d(6)=6(\frac{1}{2}+\frac{1}{3})=5\end{array}$$d(6)=6\left(\frac{1}{2}+\frac{1}{3}\right)=5\\$. But $d(6)=d(3+3)\ne d(3)+d(3)=2$$d(6)=d(3+3)\ne d(3)+d(3)=2$.

Here is some example about counting the derivation.

$$\begin{array}{}\text{(3)}& d(42)=42(\frac{1}{2}+\frac{1}{3}+\frac{1}{7})=21+14+6=41\end{array}$$

Define the constant $\mathrm{Con}(R,d)$$\mathrm{Con}(R,d)$ as usual, $\mathrm{Con}(R,d):=\{n\in \mathbb{Z},d(n)=0\}=\{-1,0,1\}$$\mathrm{Con}(R,d):=\{n\in\mathbb Z,d(n)=0\}=\{-1,0,1\}$.

Some proerties only depend on Leibniz Law inherent from the differential ring.

For example, $d(a^k)=k{a}^{k-1}d(a)$$d(a^k)=ka^{k-1}d(a)$, hence we could extend the derivation on $\mathbb{Q}$$\mathbb Q$ by ${\displaystyle d\left(\frac{a}{b}\right)=\frac{d(a)b-ad(b)}{b^2}}$$\displaystyle d\left(\frac{a}{b}\right)=\frac{d(a)b-ad(b)}{b^2}$.

Definition. $\mathrm{Der}(\mathbb{Z}):\{d:\mathbb{Z}\to \mathbb{Z},d(ab)=d(a)b+ad(b)\}$$\mathrm{Der}(\mathbb Z):\{d:\mathbb Z\to \mathbb Z,d(ab)=d(a)b+ad(b)\}$

Proposition. $\mathrm{Der}(\mathbb{Z})$$\mathrm{Der}(\mathbb Z)$ form a $\mathbb{Z}$$\mathbb Z$-Module.

Proof.

$(d_1+d_2)(ab)=d_1(ab)+d_2(ab)=d_1(a)b+a{d}_1(b)+d_2(a)b+a{d}_2(b)=(d_1+d_2)(a)b+a(d_1+d_2)(b)$$(d_1+d_2)(ab)=d_1(ab)+d_2(ab)=d_1(a)b+ad_1(b)+d_2(a)b+ad_2(b)=(d_1+d_2)(a)b+a(d_1+d_2)(b)$.

$rd(ab)=rd(a)b+rad(b)$$rd(ab)=rd(a)b+rad(b)$. $◻$$\square$

Proposition. $\mathrm{\forall }d\in \mathrm{Der}(\mathbb{Z}),\{-1,0,1\}\subseteq \mathrm{Con}(\mathbb{Z},d)$$\forall d\in\mathrm{Der}(\mathbb Z),\{-1,0,1\}\subseteq \mathrm{Con}(\mathbb Z,d)$.

Proof.

$d(1_R)=d(1_R\cdot 1_R)=d(1_R)+d(1_R)⟹d(1_R)=0$$d(1_R)=d(1_R\cdot 1_R)=d(1_R)+d(1_R)\implies d(1_R)=0$, $d(0_R)=d(0_R\cdot 0_R)=0,d(-1_R\cdot -1_R)=d(1_R)=0=d(-1_R)(-1_R)+(-1_R)d(-1_R)⟹d(-1_R)=0◻$$d(0_R)=d(0_R\cdot 0_R)=0,d(-1_R\cdot-1_R)=d(1_R)=0=d(-1_R)(-1_R)+(-1_R)d(-1_R)\implies d(-1_R)=0\quad \square$

Proposition. Let $c\in \mathrm{Con}(\mathbb{Z},d)$$c\in \mathrm{Con}(\mathbb Z,d)$, then $d(ca)=cd(a),d(ac)=d(a)c$$d(ca)=cd(a),d(ac)=d(a)c$.

Proof.

$d(ca)=d(c)a+cd(a)=cd(a)$$d(ca)=d(c)a+cd(a)=cd(a)$, same for the $d(ac)◻$$d(ac)\quad\square$

Proposition. For $\mathrm{Der}(\mathbb{Z})$$\mathrm{Der}(\mathbb Z)$, the derivation is totally determined by the value of prime element.

Proof. It follows that $d(p^k)=k{p}^{k-1}d(p)$$d(p^k)=kp^{k-1}d(p)$, hence let $n=\pm p_1^{e_1}{p}_2^{e_2}...p_n^{e_n}$$n=\pm p_1^{e_1}p_2^{e_2}...p_n^{e_n}$.

$$\begin{array}{}\text{(4)}& d(n)=d(\pm p_1^{e_1}{p}_2^{e_2}...p_n^{e_n})=n\sum _{i=1}^n\frac{e_i}{p_i}d(p_i)\end{array}$$

and if a function has this form, then it satisfies Leibniz Law (so no matter what value $d(p_i)$$d(p_i)$ has. )

Since $e_i=v_{p_i}(n)$$e_i=v_{p_i}(n)$, let $\begin{array}{c}{a}_i=\frac{d(p_i)}{p_i}\end{array}$$a_i=\frac{d(p_i)}{p_i}\\$.

$$\begin{array}{}\text{(5)}& d(mn)=mn\sum a_i{v}_{p_i}(mn)=mn\sum a_i{v}_{p_i}(m)+mn\sum a_i{v}_{p_i}(n)=d(m)n+md(n)\end{array}$$

Remark. If we let $d(p_i)=p_i$$d(p_i)=p_i$, then $\begin{array}{c}d(n)=n\sum v_p(n)\in \mathrm{Der}(\mathbb{Z})\end{array}$$d(n)=n\sum v_p(n)\in\mathrm{Der}(\mathbb Z)\\$.