For the last version of this text, typedidrygaillard on Google. Date of this version: Tue Dec 23

09:13:38 CET 2008. Jean-Marie Didry and Pierre-Yves Gaillard

Around the Chinese Remainder Theorem

Contents

1 Introduction 2

2 Laurent Series 3

3 Generalized Rational Fractions 4

4 Chinese Remainder Theorem 6

5 Exponential 9

6 Matrices 10

7 Inductive Sequences 11

8 Differential Equations 12

9 Euclid 13

10 The Case of an Arbitrary Ring 18

11 Universal Remainder 19

12 An Adjunction 21

13 Wronski 22

1

1 Introduction

The purpose of this text is to provide a convenient formulation of some folkloreresults. The five main statements are Theorem 28 page 15, Theorem 30 page 16,Theorem 31 page 17, Theorem 32 page 17, and Theorem 33 page 19.

By “polynomial” we mean “complex coefficient polynomial in the indetermi-nateX”.

We show that

• the computation of the quotient and the remainder of the euclidean divisionof a polynomial by a nonzero polynomial,

• the partial fraction decomposition of a rational fraction,

• the computation of an inductive sequence,

• the exponentiation of a matrix,

• the integration of an ordern constant coefficient linear differential equation,

follow from a unique, simple and obvious formula: Formula (3) page 7, which wecall Taylor-Gauss Formula.

The field C(X) of rational fractions and the ringE of entire functions aretwo important examples of differential rings containingC[X]. The subringC[X]“controls” E in the sense that an entire function can be euclideanly divided bya nonzero polynomial, the remainder being a polynomial of degree strictly lessthan that of the divisor. Among the rings having this property, one contains all theothers: it is the differential ring

∏

a∈C

C[[X − a]]. (1)

Any torsionC[X]-module is a module over the ring (1), and this ring is universalfor this property. In particular any elementf of (1) can be evaluated on a squarematrixA, the matrixf(A) being by definitionR(A), whereR is the remainder ofthe euclidean division off by a nonzero polynomial annihilatingA. Since thereis an obvious formula for this remainder (the Taylor-Gauss Formula), we are done.An important example consists in taking forf the exponential function, viewedas the element of (1) whosea-th component is the Taylor series ofeX at a. Werecover of course the usual notion of exponential of a matrix, but cleared form itsartificial complications.

2

It is handy to embedC(X) andE into the differential ring

∏

a∈C

C((X − a)),

which we use as a huge container. A side advantage of this ringis that it short-cuts the usual (particularly unilluminating) construction of the field of rationalfractions (as a differential field) from the ring of polynomials.

2 Laurent Series

Let a be a complex number. ALaurent series in X − a is an expression of theform

f = f(X) =∑

n∈Z

fa,n (X − a)n,

where(fa,n)n∈Z is a family of complex numbers for which there is an integerna

such thatn < na impliesfa,n = 0.

We define the operations of addition, multiplication and differentiation on theLaurent series inX − a by

(f + g)a,n = fa,n + ga,n,

(f g)a,n =∑

p+q=n

fa,p ga,q,

(f ′)a,n = (n + 1) fa,n+1,

and we check that these operations have the same properties as on polynomials.

Theorem 1. Let f be a Laurent series inX − a. If f 6= 0, then there is a uniqueLaurent seriesg in X − a such thatf g = 1.

Proof. Exercise.

Putg = 1/f = 1f

andh g = h/f = hf

if h is a Laurent series inX − a.

3

3 Generalized Rational Fractions

A generalized rational fraction is a family f = (fa)a∈C each of whose mem-bersfa is a Laurent series inX − a. The complex numbersfa,n are called thecoefficientsof f . The generalized rational fractions are added, multipliedanddifferentiated componentwise.

To each polynomialP is attached the generalized rational fraction(

∞∑

n=0

P (n)(a)

n!(X − a)n

)

a∈C

. (2)

As the formal sum in the parenthesis contains only a finite number of nonzeroterms, it can be viewed as a polynomial. As such, it is of course equal to thepolynomialP . Hence the sums, products and derivatives of polynomials aspoly-nomials coincide with their sums, products and derivativesas generalized rationalfractions. These facts prompt us to designate again byP the generalized rationalfraction (2).

We can now define arational fraction as being a generalized rational fractionobtained by dividing a polynomial by a nonzero polynomial. The sums, productsand derivatives of rational fractions are rational fractions. If a nonzero generalizedrational fractionf = (fa)a∈C is a rational fraction, thenfa is nonzero for alla.

For all generalized rational fractionf and all complex numbera put

µ(a, f) := inf {n ∈ Z | fa,n 6= 0}

with the conventioninf ∅ = +∞, and say thatµ(a, f) is themultiplicity of a aszero, or root, off . We have

µ(a, fg) = µ(a, f) + µ(a, g).

If µ(a, f) ≥ 0 say thatf is defined ata, and denotefa,0 by f(a). If a generalizedrational fractionf is defined at a complex numbera, then we have

fa =∞∑

n=0

f (n)(a)

n!(X − a)n.

Any sum, product or derivative of generalized rational fractions defined ata is ageneralized rational fraction defined ata.

4

For all generalized rational fractionf , all complex numbera, and all integerµset

Jµa(f) :=

∑

n≤µ

fa,n (X − a)n,

and say that this Laurent series inX − a is theorder µ jet of f at a. If f andgare generalized rational fractions defined ata, we have

Jµa(f + g) = Jµ

a(f) + Jµa(g), Jµ

a(f g) = Jµa

(Jµ

a(f) Jµa(g)

).

Let a be a complex number, letµ be an integer, and letf andg be two general-ized rational fractions.

Exercise 2.Show that the following conditions are equivalent

1. µ(a, f − g) ≥ µ,

2. (X − a)−µ (f − g) is defined ata,

3. Jµ−1a (f) = Jµ−1

a (g).

If these conditions are satisfied and iff andg are defined ata, then we say thatf andg arecongruent modulo(X − a)µ and we write

f ≡ g mod (X − a)µ.

We thus haveJµ−1

a (f) ≡ f mod (X − a)µ,

as well as

f1 ≡ g1 mod (X − a)µ

f2 ≡ g2 mod (X − a)µ

⇒

f1 + f2 ≡ g1 + g2 mod (X − a)µ

f1 f2 ≡ g1 g2 mod (X − a)µ.

Exercise 3.Assumef is defined ata andµ ≥ 0. Let R be a degree< µ polyno-mial. Show that the following conditions are equivalent.

1. R = Jµ−1a (f),

2. R ≡ f mod (X − a)µ,

3. (X − a)−µ (f − R) is defined ata.

In other words (f − Jµ−1

a (f)

(X − a)µ, Jµ−1

a (f)

)

is the unique pair(q, R) such that

5

1. q is a generalized rational fraction defined ata,

2. R is a polynomial of degree< µ,

3. f(X) = (X − a)µ q(X) + R(X).

If we wish to extend the result of Exercise 3 to the division byan arbitrarypolynomialD, it is natural to restrict to the generalized rational fractions definedat all point ofC. We thus define ageneralized polynomialas being a generalizedrational fraction defined at all point ofC. Any sum, product or derivative ofgeneralized polynomials is a generalized polynomial.

Notation 4. Fix a nonconstant polynomialD and, for all complex numbera, letµa be the multiplicity ofa as a root ofD.

Exercise 5.Let f be a generalized polynomial. Show that the generalized rationalfractionf/D is a generalized polynomial if and only iff ≡ 0 mod (X − a)µa forall a.

We admit

Theorem 6 (Fundamental Theorem of Algebra). There a nonzero complex num-ber c satisfying

D(X) = c∏

D(a)=0

(X − a)µa .

Exercise 7.Let P be a polynomial. Show that the following conditions are equiv-alent

1. P ≡ 0 mod (X − a)µa for all a,

2. P/D is a generalized polynomial,

3. P/D is a polynomial,

4. P/D is a degreedeg P − deg D polynomial.

4 Chinese Remainder Theorem

Theorem 8(Chinese Remainder Theorem). For all generalized polynomialf andall nonconstant polynomialD there is a unique couple(q, R) such that

1. q is a generalized polynomial,

2. R is a polynomial of degree< deg D,

6

3. f = D q + R.

In Notation 4 page 6 we have theTaylor-Gauss Formula (see [1])

R(X) =∑

D(a)=0

Jµa−1a

(f(X)

(X − a)µa

D(X)

)D(X)

(X − a)µa(3)

Say thatR is theremainder of the euclidean division of f by D, and thatqis itsquotient.

Proof. Uniqueness. Assume

f = D q1 + R1 = D q2 + R2

(obvious notation). We get

R2 − R1

D= q1 − q2

and Exercise 7 (2.⇒ 4.) impliesR1 = R2 and thusq1 = q2.

Existence. Put

s(a)(X) := Jµa−1a

(f(X)

(X − a)µa

D(X)

)1

(X − a)µa, s :=

∑s(a),

and note thatfD− s(a) is defined ata by Exercise 3, thatf

D− s is a generalized

polynomialq, and thatsD is a degree< deg D polynomialR. QED

Example. Let a andb be two complex numbers. Forn ≥ 0 write sn for the sumof the degreen monomials ina andb. The remainder of the euclidean division ofthe polynomial

∑n≥0 an Xn by (X − a)(X − b) is∑

n≥1

an sn−1 X + a0 − a b∑

n≥2

an sn−2.

Corollary 9. (First Serret Formula — see [1].)The principal part ofg := f/D ata is

Jµa−1a

(g(X)(X − a)µa

)(X − a)−µa

There are partial analogues to Theorem 8 and Corollary 9 overan arbitrarycommutative ring (see Section 10), but the uniqueness of thepartial fraction de-composition disappears. For instance we have, in a product of two nonzero rings,

1

X − 1−

1

X=

(1,−1)

X − (1, 0)+

(−1, 1)

X − (0, 1).

7

Theorem 10. LetP be a polynomial, letf be the rational fractionP/D, let Q bethe quotient, which we assume to be nonzero, of the euclideandivision ofP byD, and letq be the degree ofQ. ThenQ is given by theSecond Serret Formula(see [1]) :

Q(X−1) = Jq0(f(X−1) Xq) X−q

Proof. If d is the degree ofD and ifR is the remainder of the euclidean divisionof P by D, then the rational fraction

g(X) := X−1(f(X−1) − Q(X−1)

)=

Xd−1 R(X−1)

Xd D(X−1)

is defined at 0, and we have

Xq f(X−1) − Xq Q(X−1) = Xq+1 g(X) ≡ 0 mod Xq+1,

which, asXq Q(X−1) is a polynomial of degree at mostq, implies

Jq0

(Xq f(X−1)

)− Xq Q(X−1) = 0.

QED

In Notation 4 page 6 fix a roota of D, denote byB the set of all other roots,and, for any mapu : B → N, b 7→ ub, denote by|u| the sum of theub.

Theorem 11. The coefficientca,k of (X − a)k in Jµa−1a ( (X−a)µa

D(X)) is

ca,k = (−1)k∑

u∈NB

|u|=k

∏

b∈B

(µb − 1 + ub

µb − 1

)1

(a − b)µb+ub. (4)

Proof. It suffices to multiply the jets

Jµa−1a

(1

(X − b)µb

)=

µa−1∑

n=0

(µb − 1 + n

µb − 1

)(−1)n (X − a)n

(a − b)µb+n.

QED

Corollary 12. The polynomialR(X) in the Chinese Remainder Theorem (Theo-rem 8 page 6) is

R(X) =∑

D(a)=0k+n<µa

ca,k

f (n)(a)

n!(X − a)k+n,

whereca,k is given by (4).

8

Corollary 13. The coefficients of the partial fraction decomposition of a rationalfraction are integral coefficient polynomials in

• the coefficients of the numerator,• the roots of the denominator,• the inverses of the differences of the roots of the denominator.

These polynomials are homogeneous of degree one in the coefficients of the nu-merator and depend only on the multiplicities of the roots ofthe denominator. (Weassume the denominator is monic.)

5 Exponential

The most important example of generalized polynomial is perhaps the exponential

eX :=

(

ea

∞∑

n=0

(X − a)n

n!

)

a∈C

,

which satisfiesd

dXeX = eX

as a generalized polynomial.

More generally we can, for any real numbert, define the generalized polyno-mial

etX :=

(eat

∞∑

n=0

tn (X − a)n

n!

)

a∈C

, (5)

and observe the identity between generalized polynomials

etX euX = e(t+u)X .

We would like to differentiateetX not only with respect toX but also withrespect tot. To do this properly we must allow the coefficients of a generalizedpolynomial to be not only complex constants, butC∞ functions fromR to C. Letus call variable coefficient generalized polynomiala generalized polynomialwhose coefficients areC∞ functions fromR to C. We then have the followingidentity between variable coefficient generalized polynomials

∂etX

∂t= X etX .

9

One is often lead to compute jets of the form

Jµa(e

tXf(X)),

wheref(X) is a rational fraction defined ata. This jet is the unique degree≤ µpolynomial satisfying

Jµa(etXf(X)) ≡ eat Jµ

a(f(X))

µ∑

n=0

tn (X − a)n

n!mod (X − a)µ+1. (6)

Let us sum this up by

Theorem 14. In Notation 4 page 6 the remainder of the euclidean division of etX

byD(X) is given by theWedderburn Formula (see [1])

∑

D(a)=0

Jµa−1a

(etX (X − a)µa

D(X)

)D(X)

(X − a)µa(7)

the jet being given by (6).

6 Matrices

Let A a complex square matrix.

Exercise 15.Show that there is a unique monic polynomialD which annihilatesA and which divides any polynomial annihilatingA.

Say thatD is theminimal polynomial of A.

Exercise 16.Let f be a generalized polynomial,D a polynomial annihilatingA, andR the remainder of the euclidean division off by D. Show that thematrix R(A) does not depend on the choice of the annihilating polynomialD.[Suggestion: use Exercises 7 and 15.]

It is then natural to setf(A) := R(A).

Exercise 17.Let f, g be two generalized polynomials andD the minimal polyno-mial of A. Showf(A) = g(A) ⇐⇒ D dividesf − g.

Exercise 18.Show(f g)(A) = f(A) g(A).

Theorem 19. The matrixetA := f(A), wheref(X) = etX (see (5)), dependsdifferentiably ont and satisfiesd

dtetA = A etA, e0A = 1.

10

Exercise 20.Prove the above statement.

Exercise 21.Let

D(X) = Xq + aq−1 Xq−1 + · · ·+ a0

be a monic polynomial,(ej) the canonical basis ofCq, andA the q by q matrixcharacterised by

j < q ⇒ A ej = ej+1, A eq = −a0 e1 − · · · − aq−1 eq.

Show thatD is the minimal polynomial ofA.

Let D be a degreeq ≥ 1 polynomial, letf be a generalized polynomial, andbq−1 Xq−1 + · · ·+ b0 the remainder of the euclidean division off by D.

Exercise 22.Showf(A) e1 = b0 e1 + · · ·+ bq−1 eq.

7 Inductive Sequences

Let D be a degreeq ≥ 1 monic polynomial; letCN be the set of complex num-ber sequences; let∆ be the shift operator mapping the sequenceu in CN to thesequence∆u in CN defined by(∆u)t = ut+1; let f be inCN; let c0, . . . , cq−1 bein C; let y be the unique element ofCN satisfying

D(∆) y = f, yn = cn for all n < q ;

for (n, t) in N2 denote bygn(t) the coefficient ofXn in the remainder of theeuclidean division ofX t by D.

Theorem 23. If t ≥ q thenyt =∑

n<q

cn gn(t) +∑

k<t

gq−1(t − 1 − k) fk.

Proof. Introduce the sequence of vectorsxt := (yt, . . . , yt+q−1). We have

xt+1 = B xt + ft eq, x0 = c,

whereeq is the last vector of the canonical basis ofCq, andB is the transpose ofthe matrixA in Exercise 21. Hence

xt = Btc + f0 Bt−1 eq + f1 Bt−2 eq + · · · + ft−1 eq.

It suffices then to take the first component of the left and right hand sides, and toinvoke Exercise 22. QED

11

8 Differential Equations

Let D be a degreeq ≥ 1 monic polynomial andy the unique solution to thedifferential equation

D

(d

dt

)y = f(t), y(n)(0) = yn ∀ n < q := deg D, (8)

wheref : R → C is a continuous function.

Theorem 24. We have theCollet Formula (see [1])

y(t) =∑

n<q

yn gn(t) +

∫ t

0

gq−1(t − x) f(x) dx

wheregn(t) is the coefficient ofXn in the remainder of the euclidean division ofetX byD.

[For D(X) = Xq we recover the Taylor Formula with integral remainder.]

Proof. Puttingvn := y(n−1), v0n := yn−1 for 1 ≤ n ≤ q, and denoting byeq thelast vector of the canonical basis ofCq, Equation (8) takes the form

v′(t) − B v(t) = f(t) eq, v(0) = v0,

whereB is the transpose of the matrixA in Exercise 21. Applyinge−tB we get

d

dte−tB v(t) = e−tB f(t) eq, v(0) = v0,

whence

v(t) = etB v0 +

∫ t

0

f(x) e(t−x)B eq dx.

In view of the Wedderburn Formula (7), it suffices then to takethe first componentof the left and right hand sides, and to invoke Exercise 22. QED

Let h be a variable coefficient generalized polynomial, letf and y be twocontinuous functions fromR to Cq, let y0 be a vector inCq, andA a q by qcomplex matrix. Ify is differentiable and satisfies

y′(t) + h(t, A) y(t) = f(t), y(0) = y0, (9)

then

H(t, A) :=

∫ t

0

h(u, A) du ⇒d

dteH(t,A) y(t) = eH(t,A) f(t),

whence

12

Theorem 25. The unique solution to (9) is given by theEuler Formula (see [1])

y(t) = exp

(∫ 0

t

h(u, A) du

)y0 +

∫ t

0

exp

(∫ v

t

h(u, A) du

)f(v) dv

9 Euclid

By “ring” we mean in this text “commutative ring with 1”. For the reader’s con-venience we prove the Chinese Remainder Theorem.

Theorem 26(Chinese Remainder Theorem). LetA be a ring andI1, . . . , In idealssuch thatIp +Iq = A for p 6= q. Then the natural morphism fromA to the productof theA/Ip is surjective. Moreover the intersection of theIp coincides with theirproduct.

Proof. Multiplying the equalitiesA = I1 + Ip for p = 2, . . . , n we get

A = I1 + I2 · · · In. (10)

In particular there is ana1 in A such that

a1 ≡ 1 mod I1, a1 ≡ 0 mod Ip ∀ p > 1.

Similarly we can find elementsap in A such thatap ≡ δpq mod Iq (Kroneckerdelta). This proves the first claim. LetI be the intersection of theIp. Multiplying(10) byI we get

I = I1I + II2 · · · In ⊂ I1 (I2 ∩ · · · ∩ In) ⊂ I.

This gives the second claim, directly forn = 2, by induction forn > 2. QED

Let A be a PID (principal ideal domain). Construct the ringA∧ as follows. Afamily (ad)d6=0 of elements ofA indexed by the nonzero elementsd of A representsan element ofA∧ if it satisfies

d | e ⇒ ad ≡ ae mod d

(whered | e means “d dividese”) for all pair (d, e) nonzero elements ofA. Twosuch families(ad)d6=0 and(bd)d6=0 represent the same element ofA∧ if and only if

ad ≡ bd mod d ∀ d 6= 0.

The ring structure is defined in the obvious way. EmbedA into A∧ by mappinga in A to the constant family equal toa. By abuse of notation we often designateby the same symbol an element ofA∧ and one of its representatives. LetP be arepresentative system of the association classes of prime elements ofA.

13

Lemma 27. Let a = (ab)b6=0 be in A∧ and d a nonzero element ofA such thatad ≡ 0 mod d. Then there is aq in A∧ such thata = d q.

Proof. Let p be inP andi the largest integerj such thatpj dividesd. In otherwords there is an elementd′ of A which is prime top and satisfiesd = pid′. Forall nonnegative integerj we have

api+j ≡ 0 mod pi.

As a result there is an elementa′j of A such thatapi+j = pi a′

j. We have

pi a′j+1 ≡ api+j+1 ≡ api+j ≡ pi a′

j mod pi+j

and thusa′

j+1 ≡ a′j mod pj .

For all j chooseqpj such that

d′ qpj ≡ a′j mod pj

and thusd qpj ≡ apj mod pj.

We getd′ qpj+1 ≡ a′

j+1 ≡ a′j ≡ d′ qpj mod pj

and thusqpj+1 ≡ qpj mod pj.

Let b be inA, b 6= 0, andPb the (finite) set of those elements ofP which divideb.Forp in Pb denote byi(p) the largest integerj such thatpj dividesb and choose asolutionqb in A to the congruence system

qb ≡ qpi(p) mod pi(p), p ∈ Pb,

solution which exists by the Chinese Remainder Theorem (Theorem 26 page 13).One then checks that the familyq := (qb)b6=0 is in A∧ and that we do haved q = a.QED

See Theorem 32 page 17 for a generalization.

Let a be inA∧, let d be a nonzero element ofA, let µ(p) be the multiplicity ofp in P as a factor ofd, let

Jµ(p)−1p

(a

pµ(p)

d

)

14

be an element ofA satisfying

Jµ(p)−1p

(a

pµ(p)

d

)d

pµ(p)≡ apµ(p) mod pµ(p),

and put

ρ :=∑

p | d

Jµ(p)−1p

(a

pµ(p)

d

)d

pµ(p)∈ A.

Let B be a ring andA an integral domain contained inB. Say thatA is prin-cipal in B if for all b in B and all nonzerod in A there is aq in B and anr in Asuch thatb = dq + r.

Theorem 28(First Main Statement). Let A be a PID, leta be an element ofA∧,and letd be a nonzero element ofA. In the above notation, the elementsa, ad andρ of A∧ are congruent modulod. Moreover, ifA is principal inB, then there is auniqueA-algebra morphism fromB to A∧.

Proof. Modulo d we havea ≡ ad by the Lemma, andad ≡ ρ by the ChineseRemainder Theorem (Theorem 26 page 13). This proves the firstclaim. Let uscheck the second one, starting with the uniqueness. Letf be anA-linear mapfrom B to A∧. If b andq are inB, and ifd 6= 0 andr are inA, then the equalityb = d q + r impliesf(b) = d f(q) + r, and thusf(b)d ≡ r mod d. Consequentlythere is at most one such map. The existence is proved by putting f(b)d := r inthe above notation, and by checking that this formula does define anA-linear mapfrom B to A∧. QED

An A-module istorsion if each of its vectors is annihilated by some nonzeroscalar.

Theorem 29. If A is principal in B, then any torsionA-module admits a uniqueB-module structure which extends itsA-module structure. Moreover anyA-linearmap between torsionA-modules isB-linear.

Proof. Let us check the uniqueness. Letb be inB ; let v be in our torsion moduleV ; let d 6= 0 in A satisfydv = 0 ; let q in B andr in A verify b = dq + r.We then getbv = rv, which proves the uniqueness. The existence is obtained bysettingbv := rv in the above notation, and by checking that this formula doesdefine aB-module structure onV which extends theA-module structure. The lastassertion is clear. QED

[Let Homct(G, H) be the group of continuous morphisms from the topologicalgroupG to the abelian topological groupH, andZ the profinite completion of

15

Z. EquipQ/Z with the discrete topology or with the topology induced by that ofR/Z. Equip alsoQ/Z with theZ-module structure provided by Theorem 29. Wethen have

Homct(Z, R/Z) = Homct(Z, Q/Z) = HombZ(Z, Q/Z) = Q/Z.

We recover the well known fact thatQ/Z is the dual ofZ in the category of locallycompact abelian groups. See Weil’sL’int egration dans les groupes topologiques,pp. 108-109.]

Let L be a sublattice of the lattice of ideals of an arbitrary ringA. Here aresome examples:

1. L is the set of nonzero ideals of an integral domain,

2. L is the set of powers of a fixed ideal of an arbitrary ring,

3. letY be a set of prime ideals ofA, andL the set of those idealsI of A suchthat any prime ideal containingI is in Y ,

4. L is the set of open ideals of a topological ring.

Construct the ringA∧ as follows. An element ofA∧ is represented by a family(aI)I∈L of elements ofA satisfying

I ⊂ J ⇒ aI ≡ aJ mod J

for all I, J in L. Two such families(aI)I∈L and(bI)I∈L represent the same ele-ment ofA∧ if and only if

aI ≡ bI mod I ∀ I ∈ L.

The ring structure is defined in the obvious way. By abuse of notation we oftendesignate by the same symbol an element ofA∧ and one of its representatives.MapA to A∧ by sendinga in A to the constant family equal toa.

Call torsion module any A-module each of whose vector is annihilated bysome element ofL. [In the case of Example 3 above, anA-module is torsion ifand only if its support is contained inY — Bourbaki,Alg. Com.II.4.4.] Assumeto simplify that the intersection of the elements ofL reduces to zero, and considerA as a subring ofA∧.

Theorem 30(Second Main Statement). Any torsionA-moduleV admits a uniqueA∧-module structure such that

a ∈ A∧, v ∈ V, I ∈ L, Iv = 0 ⇒ av = aI v.

ThisA∧-module structure onV extends theA-module structure.

16

Proof. The uniqueness being obvious, lets us check the existence. For v in V andI, J in L such thatIv = 0 = Jv, we haveaI v = aI+J v; as this vector dependsonly ona andv, it can be denotedav. Let us showa (v + w) = av + aw. If a isin A∧, and ifI, J are inL and verifyIv = 0 = Jw, we have

a (v + w) = aI∩J (v + w) = aI∩J v + aI∩J w = a v + a w.

QED

Theorem 31 (Third Main Statement). Let B a ring containingA. Assume thateach torsionA-module is equipped with aB-module structure which extends itsA-module structure, and that eachA-linear map between torsionA-modules isB-linear. Then there is a uniqueA-algebra morphismf from B to A∧ whichsatisfiesb v = f(b) v for all vectorv of any torsionA-module, and allb in B.

Proof. For all I in L denote by1I the element 1 ofA/I. Let us verify theuniqueness. Letb be inB andI in L. There is ana in A satisfying

a · 1I = b · 1I = f(b) · 1I = f(b)I · 1I ,

and thusf(b)I ≡ a mod I. To check the existence, we putf(b)I := a in the abovenotation, and verify that this formula does define anA-linear mapf from B to A∧

which satisfiesb v = f(b) v for all vectorv of any torsionA-module, and allb inB. [To check thatf(b)I andf(b)J are congruent moduloJ for I ⊂ J in L, we usethe assumption that the canonical projection fromA/I to A/J is B-linear.] QED

Let A be a Dedekind domain; letL be the set of nonzero ideals ofA; and letMbe the set of maximal ideals ofA. For anyA-moduleV , let V ∧ be the projectivelimit of the V/IV with I in L; and, for anyP in M , let V ∧

P be the projectivelimit of the V/P nV . ThenA∧ andA∧

P areA-algebras; by the Chinese RemainderTheorem (Theorem 26 page 13)A∧ is the direct product of theA∧

P ; theA-moduleV ∧ is anA∧-module; theA-moduleV ∧

P is anA∧P -module;V ∧ is the direct product

of theV ∧P . Here is a generalization of Lemma 27 page 14.

Theorem 32(Fourth Main Statement). The ringA∧ is A-flat. If V is a finitelygeneratedA-module, then the natural morphism fromA∧ ⊗A V to V ∧ is an iso-morphism. In particular, ifI is in L, thenIA∧ is the kernel of the canonicalprojection ofA∧ ontoA/I.

Proof The first claim follows from Proposition VII.4.2 ofHomological Alge-bra by Cartan and Eilenberg (Princeton University Press, 1956). The secondclaim follows from Exercise II.2 of the same book in conjunction with Propo-sition 10.13 ofIntroduction to Commutative Algebraby Atiyah and Macdonald

17

(Addison-Wesley, 1969). The third claim is obtained by tensoring the exact se-quence

0 → I → A → A/I → 0

with A∧ overA. QED

10 The Case of an Arbitrary Ring

(Reminder: by “ring” we mean “commutative ring with 1”.) LetA be a ring,Xan indeterminate, anda an element ofA. For any formal power series

f =∑

n≥0

an (X − a)n ∈ A[[X − a]]

and any nonnegative integerk put

f (k)

k!:=∑

n≥k

(n

k

)an (X − a)n−k ∈ A[[X − a]].

We then get theTaylor Formula

f =∑

n≥0

f (n)(a)

n!(X − a)n.

We regard theA-algebraA[[X−a]] as anA[X]-algebra via the morphism fromA[X] to A[[X − a]] which mapsP to

∑

n≥0

P (n)(a)

n!(X − a)n.

For anyf in A[[X − a]] and any nonnegative integerk we callk-jet of f at athe polynomial

Jka(f) :=

∑

n≤k

f (n)(a)

n!(X − a)n.

ThenJka induces a ring morphism

Jka : A[[X − a]] −→

A[X]

(X − a)k.

Let a1, . . . , ar be inA; let m1, . . . , mr be positive integers; letD be the productof the(X−ai)

mi; and letu be the natural morphism fromA[X]/(D) to the productB of theA[X]/(X − ai)

mi :

u :A[X]

(D)−→ B :=

r∏

i=1

A[X]

(X − ai)mi.

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Assume thatai − aj is invertible for alli 6= j. In particular

Di(X) :=D(X)

(X − ai)mi

is invertible inA[[X − ai]]. Let v be the morphism

v :(Pi mod (X − ai)

mi

)r

i=17→

r∑

i=1

Jmi−1ai

(Pi

Di

)Di mod D

from B to A[X]/(D).

Theorem 33(Fifth Main Statement). The mapsu andv are inverse ring isomor-phisms. Moreover Corollary 12 page 8 remains true (with the obvious changes ofnotation).

Proof The compositionu ◦ v is obviously the identity ofB. Since both rings arerankdeg D freeA-modules, this proves the statement. QED

11 Universal Remainder

In this Section we work over some unnamed ring (reminder: by “ring” we mean“commutative ring with 1”). The remainder of the euclidean division of Xr by

(X − a1) · · · (X − ak)

is a universal polynomial ina1, . . . , ak, X with integral coefficients. To computethis polynomial first recall Newton’s interpolation.

As a general notation set

f(u, v) :=f(v) − f(u)

v − u,

define theNewton interpolation polynomial

N(f(X); a1, . . . , ak; X)

of an arbitrary polynomialf(X) ata1, . . . , ak by

N(f(X); a1, . . . , ak; X) := f(a1) + f(a1, a2)(X − a1)

+f(a1, a2, a3)(X − a1)(X − a2) + · · ·

+f(a1, . . . , ak)(X − a1) · · · (X − ak−1),

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and putg(X) := N(f(X); a1, . . . , ak; X),

h(X) := N(f(a1, X); a2, . . . , ak; X).

Theorem 34 (Newton Interpolation Theorem). We haveg(ai) = f(ai) for i =1, . . . , k. In particular g(X) is the remainder of the euclidean division off(X)by (X − a1) · · · (X − ak). Moreoverf(a1, . . . , ai) is symmetric ina1, . . . , ai.

Proof. Argue by induction onk, the casek = 1 being easy. Note

g(X) = f(a1) + (X − a1) h(X).

The equalityg(a1) = f(a1) is clear. Assume2 ≤ i ≤ k. By induction hypothesiswe haveh(ai) = f(a1, ai) and thus

g(ai) = f(a1) + (ai − a1) f(a1, ai) = f(ai).

Sincef(a1, . . . , ak) is the leading coefficient ofg(X), it is symmetric ina1, . . . , ak.QED

Assume thatbi :=

∏

j 6=i

(ai − aj)

is invertible for all i. By Lagrange interpolation we have, forf(X) = Xr =fr(X),

fr(a1, . . . , ak) =∑

i

ari

bi

.

The generating function of the sumsn = sn(a1, . . . , ak) of the degreen monomi-als ina1, . . . ,ak being

1

(1 − a1 X) · · · (1 − ak X)=∑

i

ak−1i b−1

i

1 − ai X=∑

n,i

ak−1+ni

bi

Xn,

we get

sn =∑

i

ak−1+ni

bi

, fr(a1, . . . , ak) = sr−k+1

with the conventionsn = 0 for n < 0, and the remainder of the euclidean divisionof Xr by (X − a1) · · · (X − ak) is

k∑

i=1

sr−i+1(a1, . . . , ai) (X − a1) · · · (X − ai−1)

(even when some of thebi are not invertible).

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12 An Adjunction

Recall that aQ-algebra is a ring containingQ (“ring” meaning “commutativering with 1”), and that aderivation on aQ-algebraA is aQ-vector space endo-morphisma 7→ a′ satisfying(ab)′ = a′b + ab′ for all a, b in A. A differentialQ-algebra is aQ-algebra equipped with a derivation. We leave it to the readerto define the notions ofQ-algebra and of differentialQ-algebramorphism. ToeachQ-algebraA is attached theQ-algebraA[[X]] (whereX is an indetermi-nate) equipped with the derivationd

dX. Let A be aQ-algebra andB a differential

Q-algebra. Denote respectively by

A(B, A) and D(B, A[[X]])

theQ-vector space ofQ-algebra morphisms fromB to A and of differentialQ-algebra morphisms fromB to A[[X]]. The formulas

f(b) = F (b)0, F (b) =∞∑

n=0

f(b(n))

n!Xn,

whereF (b)0 designates the constant term ofF (b), set up a bijectiveQ-linearcorrespondence between vectorsf of A(B, A) and vectorsF of D(B, A[[X]]).Let us check for instance the following point. LetF be inD(B, A[[X]]), let f inA(B, A) be the constant term ofF , and letG in D(B, A[[X]]) be the extension off . Let us showG = F . We have

G(b) =∞∑

n=0

G(b)(n)(0)

n!Xn by Taylor

=

∞∑

n=0

G(b(n))(0)

n!Xn becauseG(x)′ = G(x′)

=∞∑

n=0

f(b(n))

n!Xn by definition ofG

=

∞∑

n=0

F (b(n))(0)

n!Xn by definition off

=∞∑

n=0

F (b)(n)(0)

n!Xn becauseF (x′) = F (x)′

= F (b) by Taylor.

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13 Wronski

Here is a brief summary of Section II.2 in H. Cartan’s bookCalcul differentiel. IfE andF are Banach spaces, letL(E, F ) be the Banach space of continuous linearmaps fromE to F . Let E be a Banach space,I a nonempty open interval, andAa continuous map fromI to L(E, E). Consider the ODE’s

x′ = A(t) x, (11)

where the unknown is a mapx from I to E, and

X ′ = A(t) X, (12)

where the unknown is a mapX from I to F := L(E, E). Let t0 be in I. Thesolution to (12) satisfyingX(t0) = IdE is called theresolvent of (11) and isdenotedt 7→ R(t, t0).

Let B be a continuous map fromI to E andx0 a vector ofE. The solution to

x′ = A(t) x + B(t), x(t0) = x0

is

x(t) := R(t, t0) x0 +

∫ t

t0

R(t, τ) B(τ) dτ.

Here is an additional comment. Fort in I let A(t) be the left multiplicationby A(t). In particularA is a continuous map fromI to L(F, F ). Let R be itsresolvent. ThenR(t, t0) is the left multiplication byR(t, t0).

Corollary. IfW is a differentiable map fromI to F satisfyingW ′ = A W , then

W (t) = R(t, t0) W (t0).

Corollary to the Corollary. If in additionW (t0) is invertible, then

R(t, t0) = W (t) W (t0)−1.

*

[1] For terminological justifications, typedidrygaillard on Google.

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