A differential ring is a ring A with a derivation d

1. Differential Rings

Let A be a commutative ring with identity. A derivation on A is a map d : A → A such that

(1) d(a + b) = da + db (2) d(ab) = da · b + a · db.

A differential ring is a ring A with a derivation d. Furthermore, if A is a field (integral domain e.t.c), A is called a differential field (differential integral domain e.t.c.). We write a⁰ for da, a⁰⁰ for d²a and in general a⁽ⁿ⁾ = dⁿa, a ∈ A. In this section, A is always assumed to be a differential ring.

Lemma 1.1. Let a, b ∈ A and n be a natural number. Then (1) (ab)⁽ⁿ⁾=Pn

k=0 n

k
a^(k)b^(n−k) for each n ≥ 0.

(2) (aⁿ)⁰ = naⁿ⁻¹a⁰.

(3) If a is a unit, (2) holds for all n ∈ Z.

(4) (1)⁰ = 0.

Proof. We leave it to the reader. 

Definition 1.1. Let (A, d_A) be a differential ring. If (B, d_B) is a differential ring such that A is a subring of B and dB|_A= dA, we say that (A, dA) is a differential subring of (B, dB) and (B, d_B) is an extension of (A, d_A).

Theorem 1.1. A derivation of an integral domain has a unique extension to its field of fraction.

Proof. Let A be a differential integral domain and Q(A) be its field of fraction. Define

∂ : Q(A) → Q(A) by

∂a b



= a⁰b − ab⁰ b² , a

b ∈ Q(A).

We leave it to the reader to check that ∂ is a well-defined derivation. This proves the existence.

If δ : Q(A) → Q(A) is a derivation such that δ|_Q(A) = d, by Lemma 1.1 (3), given a/b ∈ Q(A) with a, b ∈ A and b 6= 0,

δ(ab⁻¹) = δa · b⁻¹− ab⁻²δb = δa · b − aδb

b² = a⁰b − ab⁰ b² .

Here we use δa = a⁰ and δb = b⁰ since δ|A = d. Thus δ = ∂. This proves the uniqueness of

extension. 

Example 1.1. Let C^∞(a, b) be the ring of all real-valued differentiable functions on the open interval (a, b). Then

D : C^∞(a, b) → C^∞(a, b), f 7→ f⁰ is a derivation. Thus C^∞(a, b) has a structure of a differential ring.

Example 1.2. Let k be a field and k[X] be the polynomial ring over k. Define D : k[X] → k[X],

n

X

i=0

aiXⁱ 7→

n

X

i=0

iaiXⁱ⁻¹.

Then D is a derivation on k[X]. Since k[X] is an integral domain, D can be extended to the field k(X) of rational functions over k by Theorem 1.1.

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Example 1.3. Let U be an open subset of C and O(U ) be the ring of holomorphic functions on U. The map ∂ : O(U ) → O(U ) sending f to f⁰(z) is a derivation. Furthermore, if U is connected, then O(U ) is an integral domain and thus ∂ can be extended to its field of fraction: Q(O(U )) = M(U ) the field of meromorphic functions on U.

Example 1.4. Let A[X] be the polynomial ring over a differential ring A. We define

∂ : A[X] → A[X],

n

X

i=0

a_iXⁱ 7→

n

X

i=0

a⁰_iXⁱ+

n

X

i=0

ia_iXⁱ⁻¹. Then ∂ is an extension of d on A[X].

Example 1.5. Let A be a differential ring and A[{xi : i ≥ 0}] be the polynomial ring over A in the infinite sequence of variables {x_i : i ≥ 0}. Define a D derivation on A[{x_i : i ≥ 0}]

such that

dx_i−1= x_i, i ≥ 1.

We set x = x₀. The resulting differential ring (A[{x_i : i ≥ 0}], d) is denoted by A{x}.

Elements of A{x} are called differential polynomials in x.

Lemma 1.2. Let C be the subset of (A, d) consisting of a ∈ A such that a⁰ = 0. Then C is a subring of A containing the subring of A generated by unit elements of A. The ring C is called the ring of constants.

Proof. One check that if a, b ∈ C, so are a + b and ab. 

Let I be an ideal in a differential ring A. We say that I is a differential ideal of A if a ∈ I implies a⁰ ∈ I.

Lemma 1.3. Let {Iλ} be a family of differential ideals of a differential ring A. ThenT

λIλ

is again a differential ideal.

Proof. Exercise. 

Lemma 1.4. Let A be a differential ring and I be a differential ideal of A. On A/I, we define a derivation d by sending a + I to a⁰+ I. Then (A/I, d) is a differential ring.

Proof. Exercise. 

Definition 1.2. A morphism from differential ring A to differential ring B is a ring homo- morphism ϕ : A → B such that

dB◦ ϕ = ϕ ◦ d_A.

One can easily see that differential rings together with morphisms defined above forms a category.

Lemma 1.5. Let A be a differential ring and I be a differential ideal of A. Then the quotient map π : A → A/I is a morphism of differential rings.

Proof. Exericise. 

Proposition 1.1. Let ϕ : A → B be a morphism of differential rings and I be its kernel.

Then I is a differential ideal of A and the morphism ϕ : A → B induces a monomorphism ϕ : A/I → B of differential rings such that ϕ = ϕ ◦ π, where π : A → A/I is the quotient map.

Proof. The proof is straightforward. 

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Let us recall that an ideal I of A is a radical ideal if aⁿ∈ I implies that a ∈ I.

Lemma 1.6. Let I be a differential radical ideal of a differential ring A. If ab ∈ I, then both ab⁰ and ab⁰ belong to I.

Proof. Since ab ∈ I and I is a differential ideal, then (ab)⁰ = a⁰b + ab⁰ ∈ I. Multiplying the equation by ab⁰, we see that a⁰b⁰(ab) + (ab⁰)² ∈ I. Since ab ∈ I, a⁰b⁰(ab) ∈ I. Thus

(ab⁰)²= (ab)⁰(ab⁰) − a⁰b⁰(ab) ∈ I.

Thus ab⁰ ∈ I. This implies that a⁰b = (ab)⁰− ab⁰ ∈ I.  Lemma 1.7. Let I be a radical differential ideal in a differential ring A and S be any subset of A. Let

T = {a ∈ A : aS ⊂ I}.

Then T is a radical differential ideal in A.

Proof. It is easy to see that T is an ideal of A. Suppose a ∈ T. We want to show that a⁰ ∈ I.

Since a ∈ T, ab ∈ I for all b ∈ S. By Lemma 1.6, a⁰b ∈ I for all b ∈ S. Thus a⁰S ⊂ I, i.e.

a⁰∈ T. This shows that T is a differential ideal.

Now, let us prove that T is radical. Let xⁿ∈ T for some x ∈ A and n ∈ N. Hence xⁿs ∈ I for all s ∈ S. Thus (xs)ⁿ ∈ I for all s ∈ S. Since I is radical, xs ∈ I for all s ∈ S. By

definition of T, x ∈ T. 

In a ring, the intersection of any family of radical ideals is again a radical ideal. By Lemma 1.3, we obtain:

Lemma 1.8. The intersection of any family of differential radical ideals in a differential ring is again a differential radical ideal.

Definition 1.3. Let S be any subset of a differential ring A. The intersection of all differ- ential radical ideals of A containing S is denoted by {S}.

Lemma 1.9. Let S and T be subset of a differential ring A with S ⊂ T. Then {S} ⊂ {T }.

Lemma 1.10. Let a ∈ A and S ⊂ A. Here A is a differential ring. Then a{S} ⊂ {aS}.

Proof. Let I = {aS}. Then I is a differential radical ideal of A. Let T = {x ∈ A : xa ∈ I}.

Then T is a differential radical ideal containing I by Lemma 1.7. Since as ∈ I for all s ∈ S, T contains S. Since T is a differential radical ideal of A containing S, T contains {S}. Hence

a{S} ⊂ I. 

Lemma 1.11. Let S, T be any subset of a differential ring A. Then {S}{T } ⊂ {ST }. Here ST = {st : s ∈ S, t ∈ T }.

Proof. Let I = {ST }. Consider J = {x ∈ A : a{T } ⊂ I}. By Lemma 1.10, s{T } ⊂ {sT } for all s ∈ S. By lemma 1.12, {sT } ⊂ {ST } for all s ∈ S. Thus J contains S. Since J is a differential radical ideal of A containing S, it contains {S}.  Definition 1.4. A Ritt algebra is a differential algebra with identity over Q.

Lemma 1.12. Let I be a differential ideal in a Ritt algebra A. Let a ∈ A such that aⁿ∈ I for some n ∈ N. Then (a⁰)²ⁿ⁻¹ ∈ I.

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Proof. Since I is a differential ideal, (aⁿ)⁰ = naⁿ⁻¹a⁰ ∈ I. Since A is a Q-algebra, aⁿ⁻¹a⁰ ∈ I.

Claim a^n−k(a⁰)^2k−1 ∈ I for all k ≥ 1. We have prove that the statement is true for k = 1.

Assume that the statement is true for k = m, i.e. a^n−m(a⁰)^2m−1 ∈ I. Differentiating a^n−m(a⁰)^2m−1, we obtain

(n − m)a^n−m−1(a⁰)^2m+ (2m − 1)a^n−m(a⁰)^2m−2a⁰⁰∈ I.

Multiplying the above equation by a⁰, we see that

(n − m)a^n−m−1(a⁰)^2m+1+ (2m − 1)a^n−m(a⁰)^2m−1a⁰⁰∈ I.

By induction hypothesis, (2m−1)a^n−m(a⁰)^2m−1a⁰⁰∈ I. Thus (n−m)a^n−m−1(a⁰)^2m+1∈ I. In other words, a^n−m−1(a⁰)^2m+1∈ I. Thus the statement is true for k = m + 1. By induction, we prove that the statement is true for n ∈ N. Taking m = n, we obtain our result.

 Corollary 1.1. In a Ritt algebra, the radical of a differential ideal is a differential ideal.

Proof. Let I be a differential ideal of a Ritt algebra A and √

I be its radical. Let a ∈√ I.

Then aⁿ∈ I for some n ≥ 1. By Lemma 1.12, (a⁰)²ⁿ⁻¹ ∈ I. Hence a⁰ ∈√

I. By definition,

√

I is a differential ideal.