BREAKING NEWS
Differential algebra

## Summary

In mathematics, differential rings, differential fields, and differential algebras are rings, fields, and algebras equipped with finitely many derivations, which are unary functions that are linear and satisfy the Leibniz product rule. A natural example of a differential field is the field of rational functions in one variable over the complex numbers, ${\displaystyle \mathbb {C} (t)}$, where the derivation is differentiation with respect to t.

Differential algebra refers also to the area of mathematics consisting in the study of these algebraic objects and their use for an algebraic study of the differential equations. Differential algebra was introduced by Joseph Ritt in 1950.[1]

## Differential ring

A differential ring is a ring R equipped with one or more derivations, that are homomorphisms of additive groups

${\displaystyle \partial \colon R\to R\,}$

such that each derivation ∂ satisfies the Leibniz product rule

${\displaystyle \partial (r_{1}r_{2})=(\partial r_{1})r_{2}+r_{1}(\partial r_{2}),\,}$

for every ${\displaystyle r_{1},r_{2}\in R}$. Note that the ring could be noncommutative, so the somewhat standard d(xy) = xdy + ydx form of the product rule in commutative settings may be false. If ${\displaystyle M:R\times R\to R}$ is multiplication on the ring, the product rule is the identity

${\displaystyle \partial \circ M=M\circ (\partial \times \operatorname {id} )+M\circ (\operatorname {id} \times \partial ).}$

where ${\displaystyle f\times g}$ means the function which maps a pair ${\displaystyle (x,y)}$ to the pair ${\displaystyle (f(x),g(y))}$.

Note that a differential ring is a (not necessarily graded) ${\displaystyle \mathbb {Z} }$-differential algebra.

## Differential field

A differential field is a commutative field K equipped with derivations.

The well-known formula for differentiating fractions

${\displaystyle \partial \left({\frac {u}{v}}\right)={\frac {\partial (u)\,v-u\,\partial (v)}{v^{2}}}}$

follows from the product rule. Indeed, we must have

${\displaystyle \partial \left({\frac {u}{v}}\times v\right)=\partial (u)}$

By the product rule, we then have

${\displaystyle \partial \left({\frac {u}{v}}\right)\,v+{\frac {u}{v}}\,\partial (v)=\partial (u)}$

Solving with respect to ${\displaystyle \partial (u/v)}$, we obtain the sought identity.

If K is a differential field then the field of constants of K is ${\displaystyle k=\{u\in K:\partial (u)=0\}.}$

A differential algebra over a field K is a K-algebra A wherein the derivation(s) commutes with the scalar multiplication. That is, for all ${\displaystyle k\in K}$ and ${\displaystyle x\in A}$ one has

${\displaystyle \partial (kx)=k\partial x}$

If ${\displaystyle \eta \colon K\to Z(A)}$ is the ring homomorphism to the center of A defining scalar multiplication on the algebra, one has

${\displaystyle \partial \circ M\circ (\eta \times \operatorname {Id} )=M\circ (\eta \times \partial )}$

As above, the derivation must obey the Leibniz rule over the algebra multiplication, and must be linear over addition. Thus, for all ${\displaystyle a,b\in K}$ and ${\displaystyle x,y\in A}$ one has

${\displaystyle \partial (xy)=(\partial x)y+x(\partial y)}$

and

${\displaystyle \partial (ax+by)=a\,\partial x+b\,\partial y.}$

## Derivation on a Lie algebra

A derivation on a Lie algebra ${\displaystyle {\mathfrak {g}}}$ is a linear map ${\displaystyle D\colon {\mathfrak {g}}\to {\mathfrak {g}}}$ satisfying the Leibniz rule:

${\displaystyle D([a,b])=[a,D(b)]+[D(a),b]}$

For any ${\displaystyle a\in {\mathfrak {g}}}$, ad(a) is a derivation on ${\displaystyle {\mathfrak {g}}}$, which follows from the Jacobi identity. Any such derivation is called an inner derivation. This derivation extends to the universal enveloping algebra of the Lie algebra.

## Examples

If A is unital, then ∂(1) = 0 since ∂(1) = ∂(1 × 1) = ∂(1) + ∂(1). For example, in a differential field of characteristic zero ${\displaystyle K}$, the rationals are always a subfield of the field of constants of ${\displaystyle K}$.

Any ring is a differential ring with respect to the trivial derivation which maps any ring element to zero.

The field Q(t) has a unique structure as a differential field, determined by setting ∂(t) = 1: the field axioms along with the axioms for derivations ensure that the derivation is differentiation with respect to t. For example, by commutativity of multiplication and the Leibniz law one has that ∂(u2) = u ∂(u) + ∂(u)u = 2u∂(u).

The differential field Q(t) fails to have a solution to the differential equation

${\displaystyle \partial (u)=u}$

but expands to a larger differential field including the function et which does have a solution to this equation. A differential field with solutions to all systems of differential equations is called a differentially closed field. Such fields exist, although they do not appear as natural algebraic or geometric objects. All differential fields (of bounded cardinality) embed into a large differentially closed field. Differential fields are the objects of study in differential Galois theory.

Naturally occurring examples of derivations are partial derivatives, Lie derivatives, the Pincherle derivative, and the commutator with respect to an element of an algebra.

## Ring of pseudo-differential operators

(Learn how and when to remove this template message)

Differential rings and differential algebras are often studied by means of the ring of pseudo-differential operators on them.

This is the set of formal infinite sums

${\displaystyle \left\{\sum _{n\ll \infty }r_{n}\xi ^{n}\mid r_{n}\in R\right\},}$

where ${\displaystyle n\ll \infty }$ means that the sum runs on all integers that are not greater than a fixed (finite) value.

This set is made a ring with the multiplication defined by linearly extending the following formula for "monomials":

${\displaystyle (r\xi ^{m})(s\xi ^{n})=\sum _{k=0}^{\infty }r(\partial ^{k}s){m \choose k}\xi ^{m+n-k},}$

where ${\displaystyle \textstyle {m \choose k}={\frac {m(m-1)\dots (m-k+1}{k!}}}$ is the binomial coefficient. (If ${\displaystyle m>0,}$ the sum is finite, as the terms with ${\displaystyle k>m}$ are all equal to zero.) In particular, one has

${\displaystyle \xi ^{-1}s=\sum _{k=0}^{\infty }(-1)^{k}(\partial ^{k}s)\xi ^{-1-k}.}$

for r = 1, m = –1, and n = 0, and using the identity ${\displaystyle \textstyle {-1 \choose k}=(-1)^{k}}$