📂Abstract Algebra

Differential Fields in Abstract Algebra

Definition ¹

Let $R$ be an (abelian) ring. A function $d: R \to R$ that satisfies the following is called a derivation, $$ \begin{align*} d \left( x + y \right) =& d (x) + d(y) \\ d \left( x y \right) =& d (x) y + x d(y) \end{align*} $$ and the ordered pair $\left( R, d \right)$ is called a differential ring. Suppose $R$ has a unity $1$. The set of $c \in R$ that satisfies $d (c) = 0$ for the additive identity $0 \in R$ is a subring of $R$ including $1$, and is referred to as the constant ring, in the sense that it is the ring of constants.

If the $F$ of a differential ring $\left( F, \partial \right)$ is a field, it is called a differential field, and its constant ring is called the field of constants.

Explanation

Since every field is a ring, such definitions naturally arise, and indeed, a differential ring $\left( R , d \right)$ being an integral domain is sufficient for its corresponding field of fractions to become a differential field. In fact, there are more general theorems that relax the conditions even further to include fraction rings, ² but let’s just briefly look at the case of integral domains.

Theorem

Suppose $A$ is an integral domain. $$ (a,s) \equiv (b,t) \iff at = bs $$ When defining the equivalence relation $\equiv$ in the Cartesian product $A \times S$ of $A$ and $S$ as above, let’s represent the equivalence class of $(a,s)$ as $a/s$, and represent the set of those equivalence classes as $S^{-1} A := A \times S / \equiv$. When the new two operations $\oplus$, $\odot$ are $$ \begin{align*} {{ a } \over { s }} \oplus {{ b } \over { t }} :=& {{ at + bs } \over { st }} \\ {{ a } \over { s }} \odot {{ b } \over { t }} :=& {{ ab } \over { st }} \end{align*} $$ defined as above, the field $\left( S^{-1} A , \oplus , \odot \right)$ is defined as the field of fractions of $A$.

Suppose the differential ring $\left( R, d \right)$ is an integral domain. Then, there exists an extension function $\partial$ of $d$ for the field of fractions $F$ of $R$ that makes $\left( F, \partial \right)$ a differential field.

Proof

Algebraic differentiation formulas:

• [1]: For the additive identity $0$ of $R$ and elements of the constant ring $c$ and $r \in R$, the following holds. $$ \begin{align*} d \left( 0 \right) =& 0 \\ d \left( 1 \right) =& d (c) = 0 \\ d \left( c r \right) =& c d \left( r \right) \end{align*} $$
• [3] For the unit $u$ of $R$ and $r \in R$, the following holds. $$ d \left( r u^{-1} \right) = \left[ d (r) u - r d (u) \right] u^{-2} $$

Assuming $R$ is an integral domain, the field of fractions $F$ is explicitly determined, and it is sufficient to show that extending $d : R \to R$ to $\partial : F \to F$ satisfies the conditions of a differential field. Let’s start with all $r \in R$ as $$ d \left( r \right) = \partial \left( r \right) $$ When picking any $b \in R$, if $b = 0$, then for any $a \in R$, $$ d \left( a + b \right) = d \left( a + 0 \right) = d (a) + d(0) = d(a) + d(b) $$ while also satisfying $$ d \left( a b \right) = d \left( 0 \right) = 0 = d (a) 0 + a 0 = d (a) b + a d(b) $$ If $b$ is not the additive identity $0$ of $R$, then in the field of fractions $F$, $b$ is a unit hence has a multiplicative inverse $b^{-1} = 1/b$. The original domain of definition $d$ is $R$, so $d \left( b^{-1} \right)$ might not have been defined, requiring a proper extension to $F$. This differentiation can be defined through the derivative $d(1) = 0 = \partial (1)$ of the unity $1 \in R$, guaranteed by the assumption that $R$ is an integral domain, $$ 0 = \partial (1) = \partial \left( b b^{-1} \right) = \partial (b) b^{-1} + b \partial \left( b^{-1} \right) \\ \implies \partial \left( b^{-1} \right) = \partial \left( {{ 1 } \over { b }} \right) = - {{ \partial(b) } \over { b^{2} }} $$ Note that the operations of $F$ use the operations of the field of fractions $\oplus$, $\odot$, but for readability, we continue to write it as $+$, $\cdot$. According to this extension, $$ \begin{align*} \partial \left( a \cdot b^{-1} \right) =& {{ \partial(a) b - a \partial (b) } \over { b^{2} }} \\ =& \partial(a) {{ b } \over { b^{2} }} + a \cdot \left( - {{ \partial(b) } \over { b^{2} }} \right) \\ =& \partial(a) b^{-1} + a \partial \left( b^{-1} \right) \end{align*} $$ and $$ \begin{align*} \partial \left( a + b^{-1} \right) = & \partial \left( {{ ab + 1 } \over { b }} \right) \\ =& \partial \left( ab + 1 \right) b^{-1} + \left[ ab + 1 \right] \partial \left( b^{-1} \right) \\ =& \partial \left( ab \right) b^{-1} - \left[ ab + 1 \right] {{ \partial(b) } \over { b^{2} }} \\ =& \left[ \partial (a) b + a \partial(b) \right] b^{-1} - \left[ ab^{-1} + b^{-2} \right] \partial(b) \\ =& \partial (a) + a \partial(b) b^{-1} - ab^{-1} \partial(b) + b^{-2} \partial(b) \\ =& \partial (a) + \partial \left( b^{-1} \right) \end{align*} $$ In other words, $d$ defined in $R$ is naturally extended to $F$.

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