differential ring | Abstract Algebra

Posts Tagged ‘differential ring’

Integro-differential rings

Throughout, $R$ is a ring with identity.

Definition 1. We say that $(R,\delta)$ is a differential ring if $\delta$ is a derivation of $R.$

If $(R,\delta)$ is a differential ring, then $C:=\ker \delta$ is a subring of $R$ called the ring of constants of $\delta.$ Also, $\delta$ is $C$ -linear.

Definition 2. Let $(R,\delta)$ be a differential ring, and let $C:=\ker \delta.$ An integration on a differential ring $(R,\delta)$ is a $C$ -linear map $\int : R \to R$ such that $\delta \int (a)=a$ for all $a \in R.$ We then say that $(R,\delta, \int)$ is an integro-differential ring.

When exactly does a differential ring have an integration? The following Theorem answers this question.

Theorem. Let $(R,\delta)$ be a differential ring, and let $C$ be the ring of constants of $\delta.$ Then $(R,\delta)$ has an integration if and only if $\delta$ is surjective and there exists a $C$ -module $D$ such that $R=C \oplus D,$ as $C$ -modules.

Proof. Suppose first that $(R,\delta,\int)$ is an integro-differential ring. Then clearly $\delta$ is surjective because, by definition, $a= \delta(\int (a))$ for all $a \in R.$ Now let $D:=\int (R)=\{\int(a): \ a \in R\}.$ Since, by definition, $\int$ is $C$ -linear, $D$ is a $C$ -module. Also, for all $a \in R,$

$\delta(a-\int \delta (a))=\delta(a)-\delta \int \delta (a)=\delta(a)-\delta(a)=0$

and hence $a - \int \delta(a) \in C,$ which then gives $R=C+D,$ because $\int \delta (a) \in D.$ To show that the sum $C+D$ is direct, we just need to show that $C \cap D=(0).$ Well, if $c \in C \cap D,$ then $c=\int (a)$ for some $a \in R$ and hence $0=\delta(c)=\delta \int (a)=a,$ which gives $c=\int (0)=0.$

Suppose now that $\delta$ is surjective and $R=C \oplus D,$ for some $C$ -module $D.$ Let $a \in R.$ Then $\delta(b)=a$ for some $b \in R.$ We can also write $b=c+d,$ for some $c \in C, \ d \in D,$ which then gives $a=\delta(b)=\delta(d).$ So there exists $d \in D$ such that $\delta(d)=a.$ This $d$ is unique because if $\delta(d')=a$ for some other $d' \in D,$ then $\delta(d-d')=\delta(d)-\delta(d')=a-a=0,$ and so $d-d' \in C \cap D=(0).$ So we have shown that for every $a \in R,$ there exists a unique $d_a \in D$ such that $\delta(d_a)=a.$ Define the map $\int : R \to R$ by $\int(a)=d_a.$ Then $\delta \int (a)=\delta(d_a)=a.$ So in order to complete the proof, we just need to show that $\int$ is $C$ -linear. Let $a,b \in R, c \in C.$ Then, since $\delta$ is $C$ -linear, we have