Page 265 - DMTH403_ABSTRACT_ALGEBRA

Abstract Algebra

Notes If R = F [u] and D is the usual u-derivative on F [u], then the polynomial f(X) = (u + u)X + uX 3
3
4
2
2
+ u X + 1 in R[X] has f (X) = (u + 1)X + X . 3
D
2
4
2
Any element of R satisfying D(a) = 0 is called a D-constant, or just a constant if the derivation is
understood. The constants for a derivation form a subring. For instance, from the product rule,
taking a = b = 1, we obtain D(1) = 0.
Example: The set of all constants for X-differentiation on K[X] is K when K has characteristic 0 and
K[Xp] when K has characteristic p.
Example: If D: R  R is a derivation and f  f is the corresponding derivation on R[X], its ring of
D
constants is C[X], where C is the constants for D.
We will generally focus on derivations from R to R, although it will be convenient to allow
R-modules as the target space for derivations in Corollary, which is used in the main text in the
proofs of Theorem 3 and Lemma.
Example: Let’s check that any derivation on K[X] which has the elements of K among its constants
has the form D(f) = hf’ for some h  K[X]. (When h = 1, this is the usual X-derivative.)

When K is among the constants of D, D is K-linear: D(cf) = cD(f) + fD(c) = cD(f). Therefore, D is
determined by what it does to a K-basis of K[X], such as the power functions X . By induction,
n
D(X ) = nX D(X) for all n  1. Therefore, by linearity, D(f) = f’(X)D(X) for every f  K[X]. Set
n-1
n
h = D(X).
Theorem 11: Let R be a domain with fraction field K. Any derivation D: R  K uniquely extends
to D : K  K, given by the quotient rule: D (a/b) = (bD(a) – aD(b))/b .


2
Proof: Suppose there is an extension of D to a derivation on K. Then, if x = a/b is in K (with a, b
 A), a = bx, so
D(a) = bD(x) + xD(b):

Therefore in K,
D(a) – xD(b) bD(a) – aD(b)
D(x) = 
b b 2
To see, conversely, that this formula does give a derivation D on K, first we check it is well-

defined: if a/b = c/d (with b and d nonzero), then ad = bc, so
aD(d) + dD(a) = bD(c) + cD(b).
Therefore,

2
2
bD(a) – aD(b) dD(a) – cD(d) d (bD(a) – aD(b)) – b (dD(c) – cD(d))
b 2  d 2 = b d 2
2
bd(dD(a) – bD(c)) – d aD(b) + b cD(d)
2
2
=
b d 2
2
bd(cD(b) – aD(d)) – d aD(b) + b cD(d)
2
2
=
b d 2
2
(bc – ad)dD(b) – (ad – bc)bD(d)
=
2
b d 2
= 0 since ad = bc.
That D satisfies the sum and product rules is left to the reader to check.

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