Liouville’s Theorem on Integration inTerms of Elementary Functions

R.C. Churchill

Prepared for theKolchin Seminar on Differential Algebra

(KSDA)Department of Mathematics, Hunter College and the Graduate Center, CUNY

September, 2006

(This is a corrected and extended version of a lecture originally given in September 2002.)

This talk should be regarded as an elementary introduction to differen-tial algebra. It culminates in a purely algebraic proof, due to M. Rosenlicht[Ros2], of an 1835 theorem of Liouville on the existence of “elementary”integrals of “elementary” functions. The precise meaning of elementarywill be specified. As an application of that theorem we prove that theindefinite integral

∫ex2

dx cannot be expressed in terms of elementaryfunctions.

Contents

§1. Basic (Ordinary) Differential Algebra§2. Differential Ring Extensions with No New Constants§3. Extending Derivations§4. Logarithmic Differentiation§5. Integration in Finite Terms

References

1

Throughout the notes “ring” means “commutative ring with (multiplicative) iden-tity”. That identity is denoted by 1 when the ring should be clear from context; by1R when this may not be the case and the ring is denoted R. The product rs of ringelements r and s is occasionally denoted r · s.

1. Basic (Ordinary) Differential Algebra

Throughout the section R is a ring.

An additive group homomorphism δ : r ∈ R 7→ r ′ ∈ R is a derivation if theLeibniz rule

(1.1) (rs) ′ = r · s ′ + r ′ · sholds for all r, s ∈ R ; when δ is understood one simply refers to “the derivationr 7→ r ′ (on R)”. The kernel of a derivation refers to the kernel of the underlyinggroup homomorphism, i.e., { r ∈ R : r ′ = 0 }. Note from (1.1) and induction that

(1.2) (rn) ′ = nrn−1 · r ′, 1 ≤ n ∈ Z.

A differential ring consists of a ring R and a derivation1 δR : R → R. When(R, δR) and (S, δS) are such a ring homomorphism ϕ : R → S is a morphism ofdifferential rings when ϕ commutes with the derivations, i.e., when

(1.3) δS ◦ ϕ = ϕ ◦ δR .

Differential rings constitute the objects of a category; morphisms of differentialrings form the morphisms thereof.

An ideal i of a differential ring R is a differential ideal if i is “closed under thederivation”, i.e., if r ∈ i implies r ′ ∈ i. The kernel of any morphism in the categoryof differential rings is such an ideal. The quotient of a differential ring by a differentialideal inherits a derivation in the expected way, and the anticipated factorizations andisomorphisms are valid. In particular, when f : R → S is a morphism with kernel i

and ϕ : R → R/i is the canonical homomorphism there is a factorization

(1.4)

Rf−→ S

ϕ ↓≈↗

R/i

1More generally, one can consider rings with many derivations. In that context what we havecalled a differential ring would be called an “ordinary differential ring”. We have no need for theadded generality.

2

into morphisms of the category.When (R, δR) is a differential ring and r ∈ R one often refers to the evaluation

δRr as “differentiating r”. When r, s ∈ R satisfy δRr = s one refers to s as thederivative of r and to r as a primitive2 of s.

Examples 1.5 :

(a) When x denotes a single indeterminate3 over R the usual derivative d/dx givesthe polynomial algebra R[x] the structure of a differential ring. More generally,when R[x] is the polynomial algebra over R in a single indeterminate x themapping d/dx :

∑j rjx

j 7→ ∑j jrjx

j−1 is a derivation on R[x], and therebyendows R[x] with the structure of a differential ring. This is the usual derivationon R[x].

(b) The mapping r ∈ R 7→ 0 ∈ R is a derivation on R, called the trivial derivation .

(c) Let F be a field, let x be a single indeterminate over F , and give R := F [x]the usual derivation. Then the only proper differential ideal of R is the zeroideal. This follows immediately from the PID property of R and the fact thatthe derivative of any generator of a non-zero ideal, being of lower degree thanthat of the generator, cannot be in the ideal.

(d) The only derivation on a finite field is the trivial derivation. Indeed, when K isa finite field of characteristic p > 0 the Frobenius mapping k ∈ K 7→ kp ∈ Kis an isomorphism, and as a result any k ∈ K has the form k = `p for some` ∈ K. By (1.2) we then have k ′ = (`p) ′ = p`p−1` ′ = 0, and the assertionfollows.

The usual derivation on the polynomial ring R[x] is certainly familiar from ele-mentary calculus, where it is primarily used for the analysis of functions. However,readers should also be aware of the fact that this derivative can be useful for purelyalgebraic reasons, e.g., for investigating the multiplicity of roots.

To recall this application let K be a field, let p ∈ K[x] be a polynomial, and letα ∈ K be a root of p. Then we can write

p(x) = (x− α)mq(x) ,

2One is tempted to refer to primitives as “anti-derivatives” or “indefinite integrals,” but thatterminology is seldom encountered.

3As opposed to an n-tuple x = (x1, . . . , xn) of algebraically independent elements over R.

3

where q ∈ K[x] is relatively prime to x−α and m ≥ 1 is an integer. When m = 1the root is simple; otherwise it is multiple. In either case m is the multiplicity of theroot. Applying the usual derivation to the displayed formula we see that

p ′(x) = (x− α)mq ′(x) + m(x− α)m−1q(x) ,

and therefore

p ′(x) = (x− α)q ′(x) + q(x) when m = 1 ,

p ′(x) = (x− α)m−1(

(x− α)q ′(x) + mq(x))

when m > 1.

Since q(α) 6= 0 we have the following immediate consequence4.

Proposition 1.6 : Let K be a field, let x be a single indeterminate over K,assume the usual derivation on K[x], and suppose p ∈ K[x]. Then an elementα ∈ K is a multiple root of p if and only if p(α) = p ′(α) = 0.

Corollary 1.7 : Suppose L ⊃ K is an extension of fields and p ∈ K[x] is suchthat p and p ′ are relatively prime. Then p has no multiple root in L.

Proof : From the relatively prime assumption we can find elements g, h ∈ K[x] suchthat gp + hp ′ = 1, and if ` ∈ L were a multiple root then evaluating at ` wouldyield the contradiction 0 = g(`) · 0 + h(`) · 0 = g(`)p(`) + h(`)p ′(`) = 1. q.e.d.

Corollary 1.8 : Suppose L ⊃ K is an extension of fields and p ∈ K[x] is irre-ducible. Then p has no multiple root in L if and only if p ′ 6= 0.

Proof :

⇒ If p ′ = 0 and ` ∈ L is a root of p then ` must be a multiple root byProposition 1.6.

⇐ Since p is irreducible p ′ has lower degree than p the two elements p and p ′

must be relatively prime. We conclude from Corollary 1.7 that p has no multipleroot in L.

q.e.d.

We need a characteristic zero consequence of Proposition 1.6 which requires (a)of the following preliminary result. Assertion (b) is never used in our subsequentwork, but contrasting the 0 characteristic and positive characteristic cases can beinstructive.

4The presentation of this result, and the two corollaries that follow, is patterned after [Hun,Chapter III, §6, Theorem 6.10, pp. 161-2].

4

Proposition 1.9 : Suppose K is a field, x is a single indeterminate over K, andp(x) ∈ K[x] has degree n ≥ 1. Assume the usual derivation on K[x]. Then thefollowing statements hold.

(a) When char(K) = 0 the usual derivative p ′(x) of p(x) has degree n − 1. Inparticular, p ′(x) 6= 0.

(b) When char(K) = p > 0 the polynomial p(x) satisfies p ′(x) = 0 if and only ifp(x) has the form p(x) =

∑nj=0 kjx

j, where p|j whenever 0 6= kj ∈ K.

Proof : We have p(x) =∑n

j=0 kjxj and p ′(x) =

∑nj=0 jkjx

j−1, where n > 0 andkn 6= 0.

(a) If char(K) = 0 then nkn 6= 0, hence p ′(x) 6= 0.(b) If char(K) = p > 0 then p ′(x) = 0 if and only if jkj = 0 for j = 1, . . . , n,

i.e., if and only if jkj ≡ 0 (mod p) for j = 1, . . . , n, and this is obviously the case ifand only if p|j whenever 0 6= kj ∈ K.

q.e.d.

Theorem 1.10 : Any algebraic extension of a field of characteristic 0 is separable.

Proof : By Proposition 1.9(a) and Corollary 1.7. q.e.d.

Remark 1.11 : Theorem 1.10 is false when the characteristic 0 assumption isdropped. To see a specific example let t be an indeterminate over Z/2Z, let K =(Z/2Z)(t) and let

√t ∈ Ka be a root of p(x) = x2− t ∈ K[x]. (Ka is used to denote

an algebraic closure of K.) The polynomial is irreducible (otherwise t is easily seento be algebraic over Z/2Z), but (by straightforward verification) factors in Ka[x] as(x − √t )2 (because −√t =

√t and x2 − t = (x +

√t )(x − √t )). The extension

K(√

t ) ⊃ K is therefore not separable. (For generalizations of this example see [Ste,p. 84].)

Suppose R is

• a differential ring with derivation δR,

• a subring of a differential ring S with derivation δS, and

• δS|R = δR;

then R is a differential subring of S, S ⊃ R is a differential ring extension, and thederivation δR on R is said to extend to (the derivation δS on) S. Of course “ring”is replaced by “integral domain” or “field” when R and S have those structures.The subscripts on both δR and δS are generally omitted, e.g., one simply refers tothe extension δ : S → S of the derivation δ : R → R.

5

Examples 1.12 :

(a) The kernel RC of a derivation on R is a differential subring of R, and is adifferential subfield when R is a field. Moreover, RC contains the image of Zin R, i.e., the image of Z under the mapping n ∈ Z 7→ n · 1 := n · 1R ∈ R.In particular, 0,±1,±2, · · · ∈ RC . The verifications of these assertions areelementary. RC is the ring (resp. field ) of constants of R. Example: When theusual derivation on R[x] is assumed one has R[x]C = {constant polynomials} 'R.

(b) Suppose R is an integral domain and QR is the associated quotient field. Thenany derivation δ : r → r ′ on R extends to QR via the quotient rule (r/s) ′ :=(sr ′ − rs ′)/s2, and this is the unique extension of δ to QR. The verifications(that this extension is well-defined and unique) are again elementary.

Proposition 1.13 : Suppose L ⊃ K is a differential field extension and ` ∈ L\K.Then K[`] ⊃ K is a differential ring extension if and only if

(i) ` ′ ∈ K[`].

Proof :

⇒ : Condition (i) is a consequence of the definition of a derivation.

⇐ : If ` is algebraic over K the result follows from K(`) = K[`] (an algebraicresult assumed familiar to readers5). If ` is transcendental over K and p(`) =∑

j kj`j ∈ K[`] then (p(`)) ′ =

∑j k ′j`j + (

∑j jkj`

j−1)` ′ ∈ K[`].

q.e.d.

5For a proof see, e.g., [Hun, Chapter V, §1, Theorem 1.6, p. 234].

6

2. Differential Ring Extensions with No New

Constants

In this section S ⊃ R is an extension of differential rings with derivations denoted,in both cases, by δ and r 7→ r ′.

Note that SC ⊃ RC .

Proposition 2.1 : The following statements are equivalent:

(a) (“no new constants”) SC = RC;

(b) if r ∈ R (already) admits a primitive in R then r does not admit a primitivein S\R; and

(c) if s ∈ S\R satisfies s ′ ∈ R then s ′ has no primitive in R.

Ring (or field) extensions satisfying any (and therefore all) of these conditions arecalled no new constant extensions. They should be regarded as “economical”: theydo not introduce primitives for elements of R which already admit primitives in R.

Proof :(a) ⇒ (b) : When r ∈ R admits a primitive t ∈ R as well as a primitive s ∈ S\R

the element s− t ∈ S\R is a constant, thereby contradicting (a).(b) ⇒ (a) : When (a) fails there is a constant s ∈ S\R, i.e., a primitive for 0 ∈ R.

Since 0 ∈ R is also a primitive for 0 this contradicts (b).The equivalence of (b) and (c) is clear.

q.e.d.

Examples 2.2 :

(a) When K is any differential field the containment K ⊃ KC is a no new constantextension.

(b) Let U ⊂ C be any non-empty open set, pick z0 ∈ U , and let K and L denotethe fields of germs at z0 of meromorphic functions on C and U respectively.The usual derivative d/dz induces derivations on both K and L, and we canthereby regard L ⊃ K as an extension of differential fields. It is a no newconstant extension since LC = KC ' C.

7

(c) For an example of a differential field extension in which the no new constantcondition fails consider L ⊃ R[x], where R[x] is the ring of (real-valued) poly-nomial functions on R and L is any field of complex-valued differentiablefunctions (in the standard sense) of the real variable x containing exp ix. HereR[x]C = R, and from i = (exp ix) ′/ exp ix ∈ L\R[x] we conclude that LC ' Cis a proper extension of R[x]C .

Proposition 2.3 : Suppose L ⊃ K is a no new constant differential extension offields of characteristic 0 and ` ∈ L\K satisfies ` ′ ∈ K. Then:

(a) ` is transcendental over K; and

(b) the derivative (p(`)) ′ of any polynomial p(`) =∑n

j=0 kj`j ∈ K[`] with n > 0

and kn 6= 0 is a polynomial in ` of degree n if and only if kn /∈ KC, and is ofdegree n− 1 otherwise.

Proof :

(a) Let b := ` ′ ∈ K, and note from the no new constant hypothesis that b 6= 0.If the result is false ` is algebraic over K and we can write the corresponding

irreducible polynomial of K[t] as tn+cmtm+· · ·+c0 ∈ K[t], where n > 1, 0 ≤ m < n,and cm 6= 0. (cm = 0 would imply ` = 0, resulting in the contradiction ` ∈ K.)Differentiating

`n + cm`m + · · ·+ c0 = 0

then gives

(i) bn`n−1 + c ′m`m + bcm`m−1 + · · ·+ c ′0 = 0 .

There are three cases to consider.

• n− 1 > m. In this instance division of (i) by bn (which by the characteristic0 hypothesis cannot be 0) results in a monic polynomial relation satisfied by `,involving coefficients in K, in which the associated polynomial has degree lessthan that of the irreducible polynomial. This contradicts the minimality of n.

• n− 1 = m and bn + c ′m 6= 0. Then (i) becomes

(bn + c ′m)`m−1 + · · ·+ c ′0 = 0,

and we can divide by bn + c ′m to achieve the same contradiction as in theprevious item.

8

• n− 1 = m and bn + c ′m = 0. Since bn + c ′m = (`n + cm) ′ and `n + cm ∈ L\K(again by the characteristic 0 assumption), this contradicts the no new constanthypothesis.

These cases are exhaustive, and (a) is thereby established.

(b) The initial assertion regarding kn is seen immediately by writing

(p(`)) ′ = k ′n`n + nkn`n−1` ′ + k ′n−1`

n−1 + · · · + k ′0

in the form

k ′n`n + (k ′n−1 + nbkn)`n−1 + · · ·+ k ′0 = 0 .

If kn ∈ KC and the final assertion fails then 0 = k ′n−1 + nbkn = (kn−1 + nkn`) ′,forcing kn−1 +nkn` ∈ KC ⊂ K. However, in view of the characteristic 0 assumptionthis would imply ` ∈ K, contradicting (a).

q.e.d.

The following immediate consequence is well-known from elementary calculus, butone seldom sees a proof.

Corollary 2.4 : The real and complex natural logarithm functions are transcendentalover the rational function fields R(x) and C(x) respectively, and the real arctangentfunction is transcendental over R(x).

In the subject of differential algebra the characteristic of field under discussionplays a very important role, as is immediately evident from the following result.

Proposition 2.5 : Suppose K is a differential field and k ∈ K\KC. Then:

(a) k is transcendental over KC when K has characteristic 0; and

(b) k is algebraic over KC when K has characteristic p > 0.

Proof :

(a) Apply Proposition 2.3(a) to the no new constant differential field extensionK ⊃ KC .

(b) From (kp) ′ = pkp−1k ′ = 0 one sees that kp ∈ KC for any k ∈ K\KC . Theelement k is therefore a zero of xp − kp ∈ KC [x].

q.e.d.

9

Proposition 2.6 : Suppose L ⊃ K is a no new constant differential extension offields of characteristic 0 and ` ∈ L\K satisfies ` ′/` ∈ K. Then:

(a) ` is algebraic over K if and only if `n ∈ K for some integer n > 1; and

(b) when ` is transcendental over K the derivative (p(`)) ′ of any polynomialp(`) =

∑nj=0 kj`

j ∈ K[`] with n > 0 and kn 6= 0 is a polynomial of degree n,and is a multiple of p(`) if and only if p(`) is a monomial.

Proof : Let b := ` ′/` ∈ K and note from the no new constant hypothesis thatb 6= 0.

(a) Assuming ` is algebraic over K let tn + kmtm + · · ·+ k0 ∈ K[t] be the corre-sponding irreducible polynomial, where n > 1 and 0 ≤ m < n. If all kj vanish then` = 0, resulting in the contradiction ` ∈ K, and we may therefore assume km 6= 0.Differentiating

(i) `n + km`m + · · ·+ k0 = 0

then gives

(ii) bn`n + (k ′m + bmkm)`m + · · ·+ k ′0 = 0 .

Multiplying (i) by bn and subtracting from (ii) results in a lower degree polyno-mial relation for ` unless the two polynomials coincide, in which case k ′m + mbkm =bnkm. This last condition in turn implies k ′m/km = (n−m)b, and it follows that

(km`m−n) ′

km`m−n=

(m− n)km`n−m−1b` + k ′m`m−n

km`m−n

=(m− n)bkm`m−n + k ′m`m−n

km`m−n

= (m− n)b + k ′m/km

= 0 .

This gives km`m−n ∈ LC = KC ⊂ K, hence `n−m ∈ K, and from the minimality ofn we conclude that `n ∈ K.

The converse is obvious.

(b) Write (p(`)) ′ in the form

(k ′n + bnkn)`n + · · ·+ k ′0.

10

If 0 = k ′n + bnkn then (kn`n) ′ = (k ′n + bkn)`n = 0, and we would therefore havekn`n ∈ KC ∈ K. This gives `n ∈ K, contradicting the transcendency of ` over K,and the assertion on the degree of (p(`)) ′ is thereby established.

If p(`) = k`n is a monomial (with k 6= 0, to avoid trivialities) we see from(k`n) ′ = (k ′ + bnk)`n = k ′+bnk

k· k`n that (p(`)) ′ is a multiple of p(`).

Conversely, suppose (p(`)) ′ = q(`)p(`). Then then equality of the degrees of p(`)and (p(`)) ′ implies k := q(`) ∈ K. If p(`) is not a monomial let kn`n and km`m betwo distinct nonzero terms and note from (p(`)) ′ = kp(`) that

k ′j + jkjb = kkj for j = n,m .

This impliesk ′n + nknb

kn

=k ′m + mkmb

km

,

which in turn reduces to

a := (n−m)knkmb + kmk ′n − knk ′m = 0 .

Direct calculation then gives

(kn`

n

km`m

) ′=

a `n+m

(km`m)2= 0 ,

hence kn`n

km`m ∈ KC ⊂ K, and once again we contradict the transcendency of ` overK. We conclude that p(`) must be a monomial when (p(`)) ′ is a multiple of p(`),and the proof is complete.

q.e.d.

Corollary 2.7 : For any non zero rational function g(x) ∈ R(x) the compositionexp g(x) is transcendental over the rational function field R(x).

Proof : This is immediate from Proposition 2.6(a) since no non zero integer powerof exp g(x) is contained in R(x). q.e.d.

11

3. Extending Derivations

Throughout the section L ⊃ K is an extension of fields.

Here we are concerned with extending a derivation on K to a derivation on L,and we first point out that such an extension need not exist.

Example 3.1 : Let t be a single indeterminate over Z/2Z, set K := (Z/2Z)(t).Choose a root

√t of the polynomial x2 − t in some algebraic closure Ka of K.

We claim that the usual derivation p 7→ p ′ cannot be extended to the extension fieldK(√

t ). Indeed, if so then differentiating (√

t )2 = t would give 0 = 2 ·√t ·(√t ) ′ = 1,which is obviously impossible.

As we will see, the obstruction to extending the derivation in this example is thefact that the algebraic field extension K(

√t) ⊃ K is not separable. A proof of this

inseparability is included in Remark 1.11.

To determine when extensions exist it first proves useful to generalize the definitionof a derivation. Specially, let R be a ring, let A be an R-algebra (by which we alwaysmean a left and right R-algebra), and let M be an R-module (by which we alwaysmean a left and right R-module). An additive group homomorphism δ : A → M isa derivation (of A into M) if the Leibniz rule

(3.2) δ(ab) = a · δ(b) + δ(a) · b

holds for all a, b ∈ A. Any derivation δ : R → R provides an example. Additionalexamples can be seen from the discussion of CASE I below. The abbreviations δa anda ′ are also used in this context, and extensions of such mappings have the obviousmeaning.

Proposition 3.3 : Assume, in the notation of the previous paragraph, that A is anintegral domain and M is a vector space over the quotient field K of A. Then anyderivation δ : A → M extends to a derivation δ : K → M via the quotient rule

δ(ab) := b·δ(a)−a·δ(b)

b2, a, b ∈ A, b 6= 0,

and this is the unique extension to a derivation of K into M .

Proof : The proof is by straightforward verification. q.e.d.

12

For the remainder of the section δ : k ∈ K 7→ k ′ ∈ L is a derivation.

Choose any element ` ∈ L\K. We first investigate extending δ to a derivation δ :K(`) → L. We examine the transcendental and separable algebraic cases individually.

CASE I: ` is transcendental over K.

If view of Proposition 3.3 it suffices to extend δ to the polyno-mial algebra K[`], and to this end we choose any m ∈ L and defineδ : K[`] → L by

(i) δ : a =∑

jaj`j 7→ ∑

ja′j`

j + m ·∑jjaj`j−1 .

Note, in particular, that

δ(`) = m.

We claim that δ is a derivation as desired. Indeed, that δ is anadditive group homomorphism extending the given derivation is clear;what requires verification is the Leibniz rule. Let a, b ∈ K[`], a =

∑i ai`

i

and b =∑

j bj`j.

As preliminary observations note that

(∑

iai`i)(

∑jjbj`

j−1) =∑

ijjaibj`i+j−1

=∑

k(∑

i+j=k jaibj)`k−1

=∑

k(∑

i≤k(k − i)aibk−i)`k−1

=∑

k k(∑

i≤k aibk−i)`k−1 −∑

k(∑

i≤k iaibk−i)`k−1

and that

(∑

kiai`i−1)(

∑jbj`

j) =∑

ij iaibj`i+j−1

=∑

k(∑

i≤kiaibk−i)`k−1.

As a consequence we have

(∑

iai`i)(

∑jjbj`

j−1) + (∑

kiai`i−1)(

∑jbj`

j)

=∑

k k(∑

i≤k aibk−i)`k−1 −∑

k(∑

i≤k iaibk−i)`k−1 +

∑k(

∑i≤kiaibk−i)`

k−1

=∑

k k(∑

i≤k aibk−i)`k−1,

13

which is more conveniently expressed as

(ii)

{ ∑k k(

∑i≤k aibk−i)`

k−1

= (∑

iai`i)(

∑jjbj`

j−1) + (∑

kiai`i−1)(

∑jbj`

j) .

Similarly, for any finite collection ci, ej ∈ R we have

(∑

i ci`i)(

∑j ej`

j) =∑

ij ciej`i+j

=∑

k(∑

i+j=k ciej)`k

=∑

k(∑k

i=0 ciek−i)`k.

Taking ci = ai and ej = b ′j this gives

(iii)∑

k(∑

i aib′k−i)`

k = (∑

i ai`i)(

∑j b ′j`

j),

whereas taking ci = a ′i and ej = bj the result is

(iv)∑

k(∑

i a′ibk−i)`

k = (∑

i a′i`

i)(∑

j bj`j).

With the aid of (ii)-(iv) we then see that

(ab) ′ =(∑

k(∑k

i=0(aibk−i))`k) ′

=∑

k(∑k

i=0(aibk−i)′)`k + m

∑k k(

∑ki=0(aibk−i))`

k−1

=∑

k(∑k

i=0(aib′k−i + a ′ibk−i))`

k

+ (∑

iai`i)(

∑jjbj`

j−1m) + (∑

kiai`i−1m)(

∑jbj`

j) (by (ii))

=∑

k(∑k

i=0(aib′k−i))`

k +∑

k(∑k

i=0(a′ibk−i))`

k

+ (∑

iai`i)(

∑jjbj`

j−1m) + (∑

kiai`i−1m)(

∑jbj`

j)

= (∑

i ai`i)(

∑j b ′j`

j) + (∑

i a′i`

i)(∑

j bj`j) (by (iii) and (iv))

+ (∑

iai`i)(

∑jjbj`

j−1m) + (∑

kiai`i−1m)(

∑jbj`

j)

= (∑

iai`i)(

∑jb′j`

j + m∑

jjbj`j−1)

+ (∑

ia′i`

i + m∑

iiai`i−1)(

∑jbj`

j)

= ab ′ + a ′b ,

and the claim is thereby established.In summary, when ` ∈ L\K is transcendental over K any derivation

from K into L extends to a derivation of K(`) into L. Moreover, forany m ∈ L there is such an extension satisfying ` ′ = m.

14

CASE II: ` is separable algebraic6 over K.

Recall once again7 that the algebraic hypothesis on ` implies

(i) K[`] = K(`).

Let p(t) = tn +∑n−1

j=0 kjtj ∈ K[t] denote the associated monic irre-

ducible polynomial. If a given derivation k 7→ k ′ on K can be extendedto a derivation of K[`] into L then from p(`) = 0 we see that

0 = (p(`)) ′

= n`n−1` ′ +∑

jkj`j−1` ′ +

∑k ′j`

j

= ` ′(n`n−1 +∑

jkj`j−1) +

∑k ′j`

j

= ` ′(p ′(`)) +∑

k ′j`j .

From the separability hypothesis and Corollary 1.8 we have p ′(t) 6= 0, andsince p(t) has minimal degree w.r.t. p(`) = 0 it follows that p ′(`) 6= 0.The calculation thus implies

(ii) ` ′ =−∑m

j=0 k ′j`j

p ′(`).

We conclude that there is at most one extension of the given derivationon K to a derivation of K[`] into L, and if such an extension exists (ii)must hold and (as a result) the image must be contained in K[`]. This is instark contrast to the situation studied in CASE I, wherein the extensionswere parameterized by the elements m ∈ L.

To verify that an extension does exist for each derivation k ∈ K 7→k ′ ∈ L it proves convenient to define Dq(t) ∈ K[t], for any polynomialq(t) =

∑j ajt

y ∈ K[t], by Dq(t) =∑

j a ′jtj ∈ K[t]. Notice this enables us

to write (ii) as

(iii) ` ′ =−Dp(`)

p ′(`).

In fact D : q(t) 7→ Dq(t) is a derivation of K[t] into K[t]: it is theextension of k 7→ k ′ obtained from the choice m = 0 in (i) of CASE I(with ` in that formula replaced by t).

6This is, algebraic over K with separable monic irreducible polynomial in K[x].7See Footnote 5.

15

Now note from (i) that we can find a polynomial s(t) ∈ K[t] suchthat

(iv) s(`) =−Dp(`)

p ′(`)∈ K[`] ;

we define a mapping D : K[t] → K[t] (read D as “D check”) by

(v) D : q(t) 7→ Dq(t) + s(t)q ′(t) .

This is a derivation extending the given derivation k 7→ k ′ on K: itcorresponds to the choice m = s(t) in (i) of CASE I (again with ` inthat formula replaced by t).

Let η : K[t] → L denote the “substitution” homomorphism q(t) ∈K[t] 7→ q(`) ∈ L, set i := ker(η), and note that i ⊂ K[t] can also bedescribed as the principal ideal generated by p(t). Any q(t) ∈ i thereforehas the form q(t) = p(t)r(t) for some r(t) ∈ K[t], and from the Leibnizrule we see that

Dq(t) = D(p(t)r(t))

= p(t)Dr(t) + Dp(t)r(t) + s(t)(p(t)r ′(t) + p ′(t)r(t)) .

Evaluating t at ` and using (iv) then gives

Dq(`) = p(`)Dr(`) + Dp(`)r(`) + s(`) (p(`)r ′(`) + p ′(`)r(`))

= 0 · Dr(`) + Dp(`)r(`) + s(`) · 0 · r ′(`) + s(`)p ′(`)r(`)

=(Dp(`) + s(`)p ′(`)

)r(`)

=(Dp(`) + −Dp(`)

p ′(`) p ′(`))

r(`)

= 0 ,

from which we see that D(i) ⊂ i. It follows immediately that D inducesa KC-linear mapping D : K[`] → K[`], i.e., that a KC-linear mappingD : K[`] → K[`] is well-defined by

(vi) Dq(`) := η(Dq(t)) , q(`) ∈ K[`] .

Since D is KC-linear it is, in particular, an additive group homomor-phism. We claim it is a derivation. Indeed, using the derivation properties

16

of D and the ring homomorphism properties of η we see that for anyq(`), r(`) ∈ K[`] we have

D(q(`)r(`)) = η(D(q(t)r(t)))

= η( q(t)Dr(t) + Dq(t)r(t) )

= η(q(t))η(Dr(t)) + η(Dq(t))η(r(t))

= q(`)Dr(`) + Dq(`)r(`) ,

and the claim follows. We conclude that D : K[`] → K[`] is a deriva-tion, which from (v), and (vi), and the fact that η|K : K → K[`] is anembedding, is seen to extend k 7→ k ′.

In summary: when ` ∈ L\K is separable algebraic over K anyderivation k 7→ k ′ on K has a unique extension to the field K(`) =K[`]. Moreover, the derivative of ` within this extension is given by(ii), wherein p(t) = tn +

∑n−1j=0 kjt

j ∈ K[t] denotes the monic irreduciblepolynomial of `.

Theorem 3.4 : When L ⊃ K is an extension of fields of characteristic zero andδ : K → K is a derivation the following statements hold.

(a) δ extends to a derivation δL : L → L.

(b) When ` ∈ L\K is transcendental over K and m ∈ L is arbitrary one canchoose the extension δL so as to satisfy δL(`) = m.

(c) When L ⊃ K is algebraic the extension δL : L → L of (a) is unique.

Proof :(a) When L ⊃ K is a simple extension this is immediate from the preceding

discussion. (By Theorem 1.10 the separability condition required in CASE II is aconsequence of the characteristic zero assumption.) The remainder of the argumentis a routine application of Zorn’s lemma.

(b) Immediate from the discussion of CASE I.(c) Immediate from the discussion of CASE II.

q.e.d.

17

4. Logarithmic Differentiation

In this section L ⊃ K is an extension of differential fields.

Here we isolate two technical results needed for the proof of Liouville’s Theorem.For any element t ∈ L one refers to the element t ′/t ∈ L as the logarithmic

derivative of t. Using induction and the Leibniz rule one sees that for any nonzerot1, . . . , tn ∈ L and any (not necessarily positive) integers m1, . . . , mn one has thelogarithmic derivative identity

(4.1)(Πn

j=1tmj

j ) ′

Πnj=1t

mj

j

=∑n

j=1mjt ′jtj

.

Proposition 4.2 : Suppose L = K(`), where ` is transcendental over K. Supposeα ∈ K and there are constants c1, . . . , cm ∈ KC and elements r(`) and qj(`) ∈L, j = 1, . . . , m, such that

(i) α =∑m

j=1cj(qj(`))

′

qj(`)+ (r(`)) ′.

Then one can assume w.l.o.g. that :

(a) all qj(`) are in K[`]; that

(b) each qj(`) not in K is monic and irreducible; and that

(c) qi(`) and qj(`) are distinct when i 6= j and neither is in K.

What one cannot conclude is that for all 1 ≤ j ≤ m one has (qj(`))′ ∈ K[`]. For

that one needs additional hypotheses.When conditions (a)-(c) hold we say that equality (i) is normalized 8.

Proof : Each qj(`) can be written in the form kjΠnj

i=1(qji(`))nji , where kj ∈ K,

qji(`) ∈ K[`] is monic and irreducible, and the nj and nji are integers with nj > 0(no such restriction occurs for the nij). The logarithmic identity (4.1) then allowsus to assume the qj(`) appearing in (ii) are either non-constant monic irreduciblepolynomials in K[`] or elements of K. This gives (a) and (b); for (c) combine

“like terms,” e.g., write ci(qi(`))

′qi(`)

+ cj(qj(`))

′

qj(`)as (ci + cj)

(qi(`))′

qi(`)when qi(`) = qj(`).

q.e.d.

8This terminology is not standard, but proves convenient.

18

Proposition 4.3 : Assume L = K(`), where ` is transcendental over K and` ′ ∈ K[`]. Suppose α ∈ K, and that p(`) is a monic irreducible polynomial suchthat (p(`)) ′ ∈ K[`] is not divisible by p(`). Then in any normalized equality

(i) α =∑m

j=1cj(qj(`))

′

qj(`)+ (r(`)) ′.

one has qj(`) 6= p(`) for j = 1, . . . , m.

Proof : We first note, from ` ′ ∈ K[`] and Proposition 1.13, that K[`] ⊃ K is adifferential ring extension.

Suppose p(`) ∈ K[`] has the stated properties. In view of the preceding comment(p(`)) ′/p(`) must be in reduced form when regarded as an element of the quotientfield K(`) of K[`]. If for some qj(`) appearing in (i) we have qj(`) = p(`), it follows

that (p(`)) ′/p(`) must be a term of the partial fraction expansion of∑m

j=1cj(qj(`))

′

qj(`).

Indeed, one sees immediately that this partial fraction expansion contains only oneterm with denominator involving p(`), and that denominator is p(`) alone.

Since α has no denominators it is evident from (i) that this last-mentioned termmust be canceled by a term in the partial fraction expansion of (r(`)) ′. This in turnforces the appearance of p(`) as a denominator in the partial fraction expansion ofr(`). Each such occurrence in the latter expansion has the form f(`)/(p(`))e, wherethe degree of f(`) is less than that of p(`). Let d ≥ 1 denote the maximal such e.The corresponding terms of the partial fraction expansion of (r(`)) ′ then consist of

(f(`)

(p(`))d

) ′=

(f(`)(p(`))−d

) ′= −d·f(`)·(p(`)) ′

(p(`))d+1 + (f(`)) ′(p(`))d

together with at most finitely many quotients of the form q(`)/(p(`))h, each with1 ≤ h < d + 1. Arguing as at the beginning of the paragraph we conclude that theterm −d·f(`)·(p(`)) ′

(p(`))d+1 must be canceled by a term in the partial fraction expansion of∑m

j=1cj(qj(`))

′

qj(`). But this is impossible, since we have already noted that in this latter

expansion the relevant denominator only involves p(`) to the first power. The resultfollows. q.e.d.

19

5. Integration in Finite Terms

Throughout the section K denotes a differential field of characteristic 0.

In this section we prove the theorem of Liouville cited in the introduction9. Our firstorder of business is making precise the notion of an “elementary function.”

Let K be a differential field. An element ` ∈ K is a logarithm of an elementk ∈ K\{0}, and k an exponential of `, if ` ′ = k ′/k or, equivalently, if k ′ = k` ′.When this is the case it is customary to write ` as ln k and/or k as e` ; one thenhas the expected formulas

(5.1) (ln k) ′ = k ′/k and (e`) ′ = e` ` ′ .

These definitions are obvious generalizations of concepts from elementary calculus,and examples are therefore omitted.

Let K(`) ⊃ K be a non-trivial simple differential field extension. One says thatK(`) is obtained from K by

(a) the adjunction of an algebraic element over K when ` algebraic over K, by

(b) the adjunction of a logarithm of an element of K when ` = ln k for somek ∈ K, or by

(c) the adjunction of an exponential of an element of K when ` = ek for somek ∈ K.

A differential field extension L ⊃ K is elementary if there is a finite sequence ofintermediate differential field extensions L = Kn ⊃ Kn−1 ⊃ · · · ⊃ K0 = K such thateach Kj+1 ⊃ Kj has one of these three forms, and when this is the case any ` ∈ Lis said to be elementary over K. By an elementary function we mean an elementof an elementary differential field extension L ⊃ R(x) wherein R = R or C, theelements of L are functions10 (in the sense of elementary calculus), and the standardderivative is assumed.

9Our argument is from [Ros2], which is adapted from [Ros1]. A generalization of Liouville’sTheorem 5.2 can be found in [Ros3].

10The algebraic treatment is much cleaner when the relevant differential fields are defined in termsof function germs at a point of R. For example, one can then ignore the fact that a rational functionf ∈ R(x) has a vastly different domain than ln x. Readers familiar with the germ concept shouldhave no trouble formulating such an approach. That framework has been slighted in these notes- it appears only in Example 2.2(b) - since readers are not assumed familiar with function germs,and it is not clear that the benefits would be worth the work involved in presenting the necessarybackground.

20

Theorem 5.2 (Liouville) : Let K be a differential field of characteristic 0 and letα ∈ K. Then α has a primitive within an elementary no new constant differentialfield extension of K if and only if there are constants c1, . . . , cm ∈ KC and elementsβ1, . . . , βm, γ ∈ K such that βj 6= 0 for j = 1, . . . , m and

(i) α =∑

mj=1cj

β ′jβj

+ γ ′ .

Proof (M. Rosenlicht) :⇒ By assumption there is a tower Kn ⊃ Kn−1 ⊃ · · · ⊃ K0 = K of differential field

extensions such that each Kj with j ≥ 1 is obtained from Kj−1 by the adjunctionof an algebraic element over Kj−1, a logarithm of an element of K, or an exponentialof an element in K. Moreover, there is an element ρ ∈ Kn such that ρ ′ = α.

We argue by induction on the “length” n ≥ 0 of an elementary extension, first not-ing that when n = 0 the desired equality holds by taking m = 1, c1 = 0, and γ := ρ.If n > 0 and the result holds for n−1 we can view α as an element of K1 and therebyapply the induction hypothesis to the elementary extension Kn ⊃ K1, obtaining anon-negative integer m, constants c1, . . . , cm ∈ K1 and elements β1, . . . , βm, γ ∈ K1

such that the displayed equation of the theorem statement holds. Since the extensionK1 ⊃ K has no new constants we must actually have cj ∈ KC for j = 1, . . . , m, andwe are thereby reduced to proving the following result: If α ∈ K can be written asdisplayed above, with all cj ∈ KC and γ and all βj in K(`) = K1, then it can alsobe expressed in this form, although possibly with a different m and different cj, γand βj, with all cj again in KC , but with γ and all βj now contained in K.

Case (a) : ` is algebraic over K.

Choose an algebraic closure Ka for K containing K(`) and let σi : K(`) → Ka

be the distinct embeddings of K(`) into Ka over K, i = 1, . . . , s, where w.l.o.g. σ1

is inclusion. Then the images `i := σi(`) are the roots of the monic irreduciblepolynomial p ∈ K[x] of `, and the σi permute these roots.

Extend the given derivation to Ka. This extension is unique by Theorem 3.4(c),and will again be denoted δ. Since each σi ◦ δ ◦ σ−1

i : Ka → Ka is also a derivationextending δ|K this uniqueness forces σi ◦ δ ◦ σ−1

i = δ, i.e.,

σi ◦ δ = δ ◦ σi, i = 1, . . . , s.

Recall that in this case11 K(`) = K[`]. For any q(`) ∈ K[`] we obviously have

11See Footnote 5.

21

σi(q(`)) = q(`i), and from the displayed formula of the previous paragraph we seethat σi((q(`))

′) = (q(`i))′ ∈ K[`] holds as well, i = 1, . . . , s.

Choose polynomials q1, . . . , qn, r ∈ K[x] such that

βj = qj(`), j = 1, . . . , n, and γ = r(`).

Write (i) accordingly, i.e., as

α =∑

jcj(qj(`))

′

qj(`)+ (r(`)) ′;

then apply σi to obtain

α =∑

j cjσi(β

′j )

σi(βj)+ σi(γ

′)

=∑

j cj(σi(βj))

′

σi(βj)+ (σi(γ)) ′

=∑

j cj(qj(`i))

′

qj(`i)+ (r(`i))

′.

Summing over i, dividing by s (which requires the characteristic 0 hypothesis), andappealing to the logarithmic derivative identity (4.1) yields

α = 1s

∑i

(∑j cj

(qj(`i))′

qj(`i)

)+ 1

s·∑i(r(`i))

′

=∑n

j=1cj

s

(∑i

(qj(`i))′

qj(`i)

)+ 1

s(∑s

i=1 r(`i))′

=∑n

j=1cj

s

(Πsi=1qj(`i))

′

Πsi=1qj(`i)

+(Ps

i=1 r(`i)

s

) ′.

By construction each of the products and sums on the right-hand-side of this equalityis fixed by each σi, and as a result must belong K (the usual symmetric polynomialargument). This last expression for α therefore has the required form.

Having established Case (a) we may assume, for the remainder of the proof, that` is transcendental over K. We can then find qj(`), r(`) ∈ K(`) such that βj = qj(`)and γ = r(`), and thereby write

(ii) α =∑n

j=1 cj(qj(`))

′

qj(`)+ (r(`)) ′.

Indeed, we can assume this equality is normalized in the sense of Proposition 4.2.

22

Case (b) : ` is a logarithm of an element of K, i.e., ` ′ = k ′/k for some k ∈ K.

First note from ` ′ = k ′/k ∈ K ⊂ K[`], Proposition 1.13 and Example 1.12(b)that K(`) ⊃ K is a differential field extension.

Let p(`) ∈ K[`] be non-constant, monic and irreducible. From monotonicity wesee that (p(`)) ′ has lower degree than p(`), and as a result that p(`) cannot divide(p(`)) ′ ∈ K[`]. It follows from Proposition 4.3 that all qj(`) appearing in the sum∑n

j=1 cj(qj(`))

′

qj(`)within (ii) are in K, and we therefore only need be concerned with

r(`).To that end observe from (r(`)) ′ ∈ K and Proposition 2.3(b) (which requires

the no new constant hypothesis) that r(`) must have the form r(`) = c` + c, wherec ∈ KC and c ∈ K. Equality (ii) is thereby reduced to

α =∑n

j=1cjq ′jqj

+ c k ′k

+ c ′ ,

precisely as desired.

Case (c) : ` is an exponential of an element of K, i.e., ` ′/` = k ′ for some k ∈ K.

In this case we see from ` ′ = `k ′ ∈ K[`], Proposition 1.13 and Example 1.12(b)that K(`) ⊃ K is a differential field extension.

As in Case (b) let p(`) ∈ K[`] be non-constant, monic, and irreducible. FromProposition 2.6 (and the irreducibility of p) we see that for p(`) 6= ` we have (p(`)) ′ ∈K[`], and that p(`) does not divide (p(`)) ′; we can then argue as in the previouscase to conclude that all qj := qj(`) in (ii) are in K, with qj(`) = ` as a possibleexception.

On the other hand, in all cases the quotients (qj(`))′/qj(`) are in K, and the

same therefore holds for (r(`)) ′. A second appeal to Proposition 2.6(b) gives r :=r(`) ∈ K.

If qj(`) 6= ` for all j we are done, so assume w.l.o.g. that q1(`) = `. We can thenwrite

α = c1` ′`

+∑n

j=2 cjq ′jqj

+ r ′ =∑n

j=2 cjq ′jqj

+ (c1k + r) ′ ,

which achieves the required form.

⇐ When (i) holds we have α =(∑m

j=1 cj ln βj + γ) ′

.

q.e.d.

23

Corollary 5.3 (Liouville) : Suppose E ⊂ K = E(eg) is a no new constant dif-ferential extension of fields of characteristic 0 obtained by adjoining the exponentialof an element g ∈ E. Suppose in addition that eg is transcendental over E and letf ∈ E be arbitrary. Then feg ∈ K has a primitive within some elementary no newconstant differential field extension of K if and only if there is an element a ∈ Esuch that

(i) f = a ′ + ag ′ .

Proof : To ease notation write eg as `.⇒ By Theorem 5.2 the element f` ∈ K has a primitive as stated if and only if

there are elements cj ∈ KC = EC and elements γ and βj ∈ K, j = 1, . . . , m, suchthat

(ii) f` =∑m

j=1 cjβ ′jβ

+ γ ′.

Write γ as r(`) and βj as gj(`) and assume, as in the discussion surroundingequation (ii) of the proof of Theorem 5.2, that the qj(`) are either non-constant monicirreducible polynomials in K[`] or elements of E. Arguing as in Case (c) of that proof(again with K replaced by E) we can then conclude that ` is both the only possiblemonic irreducible factor in a denominator of the partial fraction expansion of r(`)as well as the only possibility for a monic irreducible gj(`) ∈ K[`]\E. But this gives(gj(`))

′/gj(`) ∈ E for all j as well as r(`) =∑t

j=−t kj`j, where t > 0 is an integer

and the coefficients kj are in E. In particular, (ii) can now be written

f` = c +∑t

j=−tk′j`

j + g ′∑t

j=−t jkj`j = c +

∑tj=−t(k

′j + jg ′kj)`

j.

and upon comparing the K-coefficients of equal powers of ` we conclude that f =k ′1 + k1g

′. Equation (i) then follows by taking a = k1.⇐ When (i) holds aeg ∈ K is a primitive of feg.

q.e.d.

Corollary 5.4 : For R = R or C the function x ∈ R 7→ ex2 ∈ R has no elementaryprimitive.

By an elementary primitive we mean a primitive within some elementary no newconstant differential field extension of R(x)(ex2

).

Proof : By Corollaries 2.4 and 5.3 the given function has an elementary primitive ifand only if there is a function a ∈ R(x) such that 1 = a ′ + 2ax.

24

There is no such function. To see this assume a = p/q ∈ R(x) satisfies thisequation, where w.l.o.g. p, q ∈ R[x] are relatively prime. Then from 1 = qp ′−q ′p

q2 +

2 · pxq⇒ q − 2px− p ′ = −−q ′p

qwe see that q|q ′p. Now choose s, t ∈ R[x] such that

sq + tp = 1, whereupon multiplication by q ′ gives sqq ′ + tpq ′ = q ′. From q|q ′pwe conclude that q|q ′, hence q ′ = 0, and q is therefore a constant polynomial, i.e.,w.l.o.g. a = p. Comparing the degrees in x on the two sides of 1 = a ′ + 2ax nowresults in a contradiction. q.e.d.

Remarks 5.5 :

(a) Additional examples of meromorphic functions without elementary primitives,including sin z/z, are treated on pp. 971-2 of [Ros2]. The arguments are easy(but not always obvious) modifications of the proof of Corollary 5.4.

(b) One can (easily) produce an elementary differential field extension of the ra-tional function field R(x) containing arctan x (we are assuming the standardderivative), but not one with no new constants. Indeed, it is not hard to showthat 1/(x2 + 1) cannot be written in the form (i) of the statement Theorem5.2 (see p. 598 of [Ros2]). This indicates the importance of the no new constanthypothesis in the statement of Liouville’s Theorem.

25

References

[Hun] T.W. Hungerford, Algebra, GTM 73, Springer-Verlag, New York, 1974.

[Ros1] M. Rosenlicht, Liouville’s Theorem on Functions with Elementary Inte-grals, Pac. J. Math., 24, (1968), 153-161.

[Ros2] M. Rosenlicht, Integration in Finite Terms, Am. Math. Monthly, 79,(1972), 963-972.

[Ros3] M. Rosenlicht, On Liouville’s Theory of Elementary Functions, Pac. J.Math., 65, (1976), 485-492.

[Ste] I. Stewart, Galois Theory, 2nd-edition, Chapman and Hall, London, 1989.

R.C. ChurchillDepartment of MathematicsHunter College and Graduate Center, CUNY695 Park AvenueNew York, New York 10021, USA

Summer address:Department of Mathematics and StatisticsUniversity of CalgaryCalgary, Alberta T2N1N4, Canada

e-mail address (at all times):

rchurchi@math.hunter.cuny.edu

26