📂抽象代数

抽象代数における微分環

Definition

Differential Ring ¹

Let $R$ be an (Abelian) ring. A function $d: R \to R$ that satisfies the following is called an (algebraic) derivation^{algebraic Derivation}. $$ \begin{align*} d \left( x + y \right) =& d (x) + d(y) \\ d \left( x y \right) =& d (x) y + x d(y) \end{align*} $$ The ordered pair $\left( R, d \right)$ is called a Differential Ring^{differential ring}.

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定義

微分環 ¹

$R$を(アーベル) 環とする。次を満たす関数 $d: R \to R$ を （代数的）微分^{algebraic Derivation}という。 $$ \begin{align*} d \left( x + y \right) =& d (x) + d(y) \\ d \left( x y \right) =& d (x) y + x d(y) \end{align*} $$ 順序対 $\left( R, d \right)$ を微分環^{differential ring}と呼ぶ。

Description

Differential Algebra is interested in whether the properties still abstract when we give a ring the function known as differentiation, and what the conditions are. Naturally, the easiest and most familiar example would be the real number $\mathbb{R}$ polynomial ring $\mathbb{R} [x]$, and in this case, the differentiation $d$ $$ d : x^{n} \mapsto n x^{n-1} $$ defined formally, $\left( \mathbb{R}[x] , d \right)$ becomes the differential ring we learned even in high school literature classes. The constant ring of this differential ring is $\mathbb{R}$.

In the definition, especially $d \left( x y \right) = d (x) y + x d(y)$ serves as an important condition that makes ‘differentiation’ differential in this abstraction, and it is also called Leibniz Product^{Leibniz product} after the name of Leibniz, who invented calculus.

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説明

微分代数は、環に我々が知っている微分という関数を与えた時も、その性質が抽象化されるか、その条件は何かに関心を持つ。もちろん、最も簡単で我々に馴染みのある例は、実数 $\mathbb{R}$ 多項式の環 $\mathbb{R} [x]$ であり、この場合の微分 $d$ が $$ d : x^{n} \mapsto n x^{n-1} $$ と形式的に定義されれば、$\left( \mathbb{R}[x] , d \right)$ は高校の文科でも学んだことのある微分環になる。この微分環の定数環は$\mathbb{R}$ である。

定義で特に $d \left( x y \right) = d (x) y + x d(y)$ はこのような抽象化で「微分」を微分たらしめる重要な条件として機能し、微分積分学を創造したライプニッツの名前にちなんで ライプニッツ積^{Leibniz product}とも呼ばれる。

Theorem

Constant Differentiation

• [1]: For the additive identity $0$ of $R$ and the elements $c$ and $r \in R$ of the constant ring, the following holds. $$ \begin{align*} d \left( 0 \right) =& 0 \\ d \left( 1 \right) =& d (c) = 0 \\ d \left( c r \right) =& c d \left( r \right) \end{align*} $$

High-order Term Differentiation

• [2] For $n \in \mathbb{N}$ and $r \in R$, the following holds. $$ d \left( r^{n} \right) = n r^{n-1} d (r) $$

Quotient Differentiation

• [3] For the unit $u$ of $R$ and $r \in R$, the following holds. $$ d \left( r u^{-1} \right) = \left[ d (r) u - r d (u) \right] u^{-2} $$

Proof

[1]

$d \left( 0 \right) = 0$ is obtained as follows. $$ \begin{align*} & d (0) = d \left( 0 + 0 \right) = d \left( 0 \right) + d \left( 0 \right) \\ \implies & d (0) = d (0) + d (0) \\ \implies & 0 = d (0) \end{align*} $$

$d \left( 1 \right) = 0$ is obtained as follows. $$ \begin{align*} d (1) =& d \left( 1 \cdot 1 \right) \\ =& d \left( 1 \right) 1 + 1 d \left( 1 \right) \\ =& d \left( 1 \right) + d \left( 1 \right) \\ \implies d(1) = & 0 \end{align*} $$ Meanwhile, according to the definition, the constant $c$ is thus $d(c) = 0$ is $d(1) = d(c)$.

$d \left( c r \right) = c d (r)$ is obtained as follows. $$ \begin{align*} d (cr) =& d \left( c \cdot r \right) \\ =& d \left( c \right) r + c d \left( r \right) \\ =& 0 + c d \left( r \right) \end{align*} $$

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[2]

Proved by mathematical induction. When $n = 1$,
$$ d (r) = d ( r \cdot 1 ) = d(r) 1 + r d(1) = d(r) = 1 \cdot r^{1-1} d(r) $$ and if the given theorem holds, then $$ \begin{align*} d \left( r^{k} \right) =& d \left( r \cdot r^{k-1} \right) \\ =& d \left( r \right) r^{k-1} + r d \left( r^{k-1} \right) \\ =& d \left( r \right) r^{k-1} + r (k-1) r^{k-2} d \left( r \right) \\ =& d \left( r \right) r^{k-1} + (k-1) r^{k-1} d \left( r \right) \\ =& k r^{k-1} d \left( r \right) \end{align*} $$ thus the given theorem holds for all $n \in \mathbb{N}$.

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[3]

That $u$ is a unit means there exists an inverse for its multiplication $u^{-1} \in R$. If we set $1 = u u^{-1}$ then $$ \begin{align*} d \left( 1 \right) =& d \left( u u^{-1} \right) \\ =& d \left( u \right) u^{-1} + u d \left( u^{-1} \right) \\ \implies 0 =& d \left( u \right) u^{-1} + u d \left( u^{-1} \right) \\ \implies u d \left( u^{-1} \right) =& d \left( u \right) u^{-1} \\ \implies d \left( u^{-1} \right) =& d \left( u \right) u^{-2} \end{align*} $$ thus $d \left( u^{-1} \right) = d \left( u \right) u^{-2}$, and we obtain the following. $$ \begin{align*} d \left( r u^{-1} \right) =& d \left( r \right) u^{-1} + r d \left( u^{-1} \right) \\ =& d \left( r \right) u u^{-1} u^{-1} + r d \left( u \right) u^{-2} \\ =& \left[ d (r) u - r d (u) \right] u^{-2} \end{align*} $$

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