Appendix A: On Power Series Rings

APPENDIX A

ON POWER SERIES RINGS

A.l

Power Series Rings

(A.l.1) Multi-indices Let K be an index set, finite or infinite. A multiindex indexed by K is a function n: K -+ N u (0) such that n ( i ) # 0 for only finitely many i E K (i.e., n has finite support). Given multi-indices n, k, w&define (A. 1.2) n Ik

(A.1.3)

(A.t.4) (A. 1.5)

n

+k =1

n(i) I k(i)

0

o

n
for all i E

I(i) = n(i) + k(i) o

K

for all i

E K

n s k and I n ( < ( k (

We write 0 for the multi-index O ( i ) = 0 for all i E K . Then n > k iff n = k + I for some multi-index I # 0. We write I, for the set of all multi-indices indexed by K ; and if K = {I, . . . , m}, we simply write I, or even I for the set of multi-indices indexed by { 1, . . ., m), i.e., I, = {(nl, n2, . . ., n,) I tii E N u (0)). (A.1.6) Power series rings For each i For each n E I,, we define X " as

E K,

let X ibe an indeterminate.

Let A be a ring (commutative, 1 E A ) . Then the power series ring A [ X i \ i defined as the set of all formal sums

, f ( X )=

c a,X",

a,

E

A

n E Ir

with addition and multiplication defined by (A. 1.7) (A. 1.8)

1 u,X" + 1b,X" = 1 (a, + b,)X" (1anXn)(c bnXn) == 1 cn = 1 cnXn9

k + I=n

517

akb,

E

rc] is

518

APPENDIX A

There are two obvious ring homomorphisms (A.1.9)

A

c

I:

-+

t:: A [ [ X , l iE t i p

A[X,li E

A

&(I

defined by ~ ( a=) a n X nwith a, = 0 if ( n1 2 1, a, = a, and anXn)= a(,. Of course E I = id,. Note that (A.1.7), (A.1.8) make A [ X , I i E t i 1 a commutative ring with unit element I ( 1). We shall usually identify a E A with its image ~ ( aand ) thus view A as a subring of A [ X , I i E t i ] . 1

(A.l.10) Lemma An element & ( f ( X ) E) A is invertible in A .

f ( X ) E A [ X ,) iE ti]

is

invertible

iff

Proof Since r. is a ring homomorphism, the "only i f " part is trivial. So suppose that &(f(X)) = a, E U ( A ) , the group of invertible elements of A, where f ( X ) = anXn. Let g ( X ) = h, X" with h, E A yet to be determined. f ( X ) g ( X )= 1 gives us the equations

C

1

(A.l.ll) (A.1.12)

C

O=

kf I=n

soh,

1

anbn=aob, +

C

arb,

k t I=n

I
Since a, is invertible, we can solve ( A . l . l l ) with h , E A ; and given h , E A with 1 < n, we can solve (A.1.12) with h, E A again because a,, E U ( A ) . Q.E.D. Notation and convention If ti = {l,..., nn], we shall write A C X , , ..., X,] for A [ X , l i E (1, ..., m)]and even A [ X ] if m is clear from the context. From now on in this appendix A [ X 1will alwajls be a power series ring in a jinite number of uariables.

(A.1.13)

(A.1.14)

Exercise A W , , ..., x,w,+lli= A V , , ..., Xn+ln

A.2

Filtration and Topology

Let R be a ring. A ringjltration on R is a function. such that (A.2.1) (A.2.2) (A.2.3)

o: R -+ N

u {OO] u (0 )

u(0) = co,

Im(i1) # {co)

[>(a- h ) 2 min{o(a), r(b)]

v(ab) 2 u(a)

+ c(b)

For each m E N u (a}, we define I , = ( a E R 1 u(a) 2 m}. Then I , is an ideal of R and we have I , I , c I n l + k . The ideals I , define a topology on A as follows:

519

FORMAL POWER SERIES RINGS

the sequence {a,} converges to a E A iff lim,,+= u(a, - u ) = 0. This topology turns R into a topological ring (i.e., addition, subtraction, and multiplication are continuous). The topology is Hausdorff iff I x , = (0). (A.2.4) Example Let A be any ring, R

=

(A.2.5)

for all n with ( n1 < s

(1 a,X")

2s

o

a,

=0

A[X]. Define u : R

-+

N u {0}by

One easily checks that this is a ring filtration. Moreover, the induced topology on R is Hausdorff and complete (exercise). (A.2.6) Exercise Suppose that A has no zero divisors. Show that then o(f(X)g(X)) = u(f(X))u(g(X)) and conclude that A[X] has no zero divisors. (Hint: use induction and (A.1.14).) (A.2.7) Exercise Let R be as in (A.2.4). Let M , be the subgroup of R of all f ( X ) E R such that a, = 0 for In I 2 s . Then R = M , 01, as an abelian group. (A.2.8) Example Let A be a local ring with maximal ideal 111. Define u : A -+ N u {a) by u ( x ) 2 m o x E i l l r n . This is a ring filtration on A. (A.2.9) Theorem Let R and A be as in (A.2.4). If A is noetherian, so is R. For a proof, see e.g., [511, Volume 11, Chapter VII, Theorem 4, p. 1381. A.3

Formal Weierstrass Preparation Theorem

Let now A be a complete Hausdorff local ring with maximal ideal 111 (cf. example (A.2.8)), and let R = A [ X ] = A [ X , l J ;i.e., the index set has one element in this section. (A.3.1) Definitions An element ,f(X) E A[X] is called a distinguished polynomial if it is of the form f ( X ) = a. a , X . . . a,- X"- ' + X" with a,, . . ., a,- E 111.An elementf(X) E A[X] is said to have Wrierstrass degree n if f ( X ) = a i X i with a, E 111 for i < n and a, E U ( A ) = A\iii. We write W-degree(f(X)) = n. If f ( X ) E iii[X$ we have W-degree(f(X)) = 00 (and inversely).

+

+ +

,

(A.3.2) Weierstrass preparation theorem Letf(X) E A[X] and suppose that W-degreef(X) = n < a. Then there exist a unique invertible power series u(X) E A[XlJ and a unique distinguished polynomial g(X) of degree n such that f ( X ) = u(X)g(X). Proof We shall construct power series ucm)(X) and distinguished polynomials g(")(X)such that

(A.3.3)

f(x)= dm)(X)gcm)(X)mod(llrm[XIJ)

and we shall show that these ucm)(X) and g'")(X) are unique mod iii"[X].

520

APPENDIX A

I??,,

First take m = 1. Take g")(X) = X", u ( , ) ( X )= a i X i - " . Then clearly (A.3.3) holds for m = 1, and one easily checks that u ' , ) ( X ) and g ( l ) ( X )are unique mod i i i [ X ] . Indeed, g")(X) being distinguished must be of the form X" mod n t [ X ] . Now suppose we have found u("')(X),g(")(X),m 2 1. Write

g ( m + l ) ( X=) g'""(X) + bo + b , X

dm+y x ) = U'"'(X) + h ( X ) ,

+ ... + b , - , X " - '

,

bi E id"

h ( x )E

ltIm[xq

Suppose that -g(")(X)u(")(X)+ f ( X ) E I ( X ) E ni"[X] mod iii"+'[X]. Then we see that bo, b,, ..., b,- h ( X ) must satisfy

,,

+

(A.3.4) I ( X ) = X " h ( X )+ bou'"')(X) b,Xu'"'(X)

+ ... + b , - l X " - l u ' m ) ( X )

[ X ] g ( " ) ( X ) h ( X )= X " h ( X ) mod iii"'+'[X] since g'"'(X) is mod I T I ~ + ~because X ] all distinguished of degree n and h ( X ) E iii"[X] and b i h ( X ) E i ~ i ' " + ~ [for i = 1, . . . , n - 1. Because u('")(X)is a unit there exist bo, . . ., b,- E iitm and h ( X ) E iii"[X] such that (A.3.4) holds (if I ( X ) E iiim[XI]),moreover such bi and h ( X ) are unique modulo iiim+ Now let u ( X ) ,g ( X ) be the unique elements of A [ X ] such that u ( X ) = u'"'(X), g ( X ) = g'")(X) mod id"[XI] for all m E N. Q.E.D.

,

A.4

Homomorphisms and Isomorphisms. Formal Inverse Function and Implicit Function Theorems

(A.4.1) Continuous homomorphisms Let R , and R 2 be two rings with ring filtrations u,, 21,. Then a ring homomorphism 9: R , + R , is continuous (with respect to the topologies defined by u , and u2 iff u,(a,,)-+ 00 as n + co implies u2 ($(a,)) -+ 00 as n -+ co. (A.4.2) Proposition Let R = A I X 1 , , . ., X , ] as in Example (A.2.4). Let R' be an A-algebra with a ring filtration u such that R' is complete and Hausdorff in the topology defined by u. Let a,, ..., a, E R' be n elements such that u(ai) 2 1, i = 1, . .., m. Then there exists a unique continuous A-algebra homomorphism 9: R + R' such that 9 ( X i )= ai. Proof

(A.4.3)

Define 9: R + R' by the formula

S(Ia, X " ) =

a, a;'

. . . a$

n

where we note that the sum on the right converges to a unique element of R' because R' is complete and Hausdorff. Now because 9 must be an A-algebra homomorphism, 9 is necessarily unique on the sub-A-algebra A [ X , , ..., X,] c A [ X , , ..., X , ] and given by (A.4.3) in this case (the sum is then finite). Also 9 is an A-algebra homo-

52 1

FORMAL POWER SERIES RINGS

morphism on A [ X ] . By continuity of 9 it follows that (A.4.3) is the only possibility for 9 because every element of A [ X ] is a limit of elements of A [ X ] . This approximation argument also shows that 9 as defined by (A.4.3) is an A-algebra homomorphism since 9 is an A-algebra homomorphism on A [ X ] .

rn (A.4.4) Jacobian matrix Let X

= ( X l , . .., X m ) , Y = (Yl, . . ., Y,,) be two sets of indeterminates, and let 9: A [ X ] - A [ Y ] be a continuous homomorphism of A-algebras. By Proposition (A.4.2) 9 is uniquely determined by giving the m power series 9(Xl ), . . ., 9 ( X m ) E A[Y]; i.e., 9 corresponds t o an m-tuple a s ( Y )of power series in . . . , Let J(9)be the unique m x n matrix with coefficients in A such that

x, x .

in A [ Y r . The matrix J(9) is called the Jacobian matrix of 9 and also the Jacobian matrix of the m-tuple of power series as( Y).

rn (A.4.5) Proposition (formal inverse function theorem) A continuous Aalgebra homomorphism $: A [ X ] -+ A [ Y ] is an isomorphism iff J(9) is an invertible matrix.

r1

Proof If 9 is an isomorphism, there is an inverse isomorphism 4, A [ Y ] + A [ X ] ; let M be the unique matrix such that

i"") 4(K)

=M

mod(degree 2)

Xm

then because 9 o 4 = id, 4 o 9 = id, we must have J ( 9 ) M = I,, M J ( 9 ) = I , S O that J(9) is invertible. Conversely, suppose that J(9) is invertible. Then n = m and identifying A [ X ] with A [ Y ] by means of Xi++ we find the following reformulation of (A.4.5).

(A.4.6) Let a ( X ) be an n-tuple of elements of A I X 1 , . . ., X , ] such that a ( X ) = J X mod(degree 2) where J is an invertible n x n matrix in A""" and Then there exists a unique where X is short for the column vector ( X 1, . . ., X,,). n-tuple of elements B ( X ) such that B ( a ( X ) )= X = u ( B ( X ) ) . Proof Take a ( ' ) ( X )= J - ' X . Then P")(u(X))= X mod(degree 2) and P'"(X) is uniquely determined mod(degree 2) by this condition. Suppose we ) that P'")(u(X))= X mod(degree m 1) and suppose have found p c m ) ( Xsuch that p c m ) ( Xis) uniquely determined mod(degree m 1) by this condition. Let B(,)(&(X))- X = N ( X ) mod(degree m 2) where N ( X ) is an n-tuple of homo) R(X) geneous polynomials of degree m 1. Now take /?"")(X) = p c m ) ( X-

+

+

+

+

522

APPtNDlX A

where R ( X ) is such that R ( J X ) = N ( X ) , i.e., R ( X ) = N ( J - ' X ) . Then [~("+')(cc(X)) E X mod(degree tn + 2) and p'"'')(X) is unique modulo(degree tn + 2) with this property. Let o ( X ) be the unique power series such that p ( X ) = p("')(X)mod(degree tn + 1) for all m. ( p ( X )exists and is unique because A [ X ] is complete and Hausdorff.) Then /l(cc(X))= X . Similarly, one shows the existence of a & X ) such that cc(&X)) = X , and then p ( X ) = /j(cc(D(X))= (p r @ ) ( f l ( X )=) & X ) . Q.E.D. (A.4.7)

Formal implicit function theorem

Let

F ( X ; Y ) E A [ X , , ..., X",; Yl, . . ., Y,] be an n-tuple of power series in X I , . . . , X,, and Y,, . . . , Y, such that F ( 0 ; 0) = 0 and such that the unique (partial Jacobian) matrix J such that F ( X , Y ) = J Y mod(Xl, ..., X,; degree 2 ) is invertible (in particular there are as many variables as there are power series). Then there exists a unique n-tuple of power series u ( X ) E AFX,, . . ., X , , r such that F ( X ; ct(X))= 0. Proof We can write

F ( X ; Y )= M X

(A.4.8)

+ JY + G(X;Y )

where every monomial occurring in G ( X ; Y ) is at least of degree 2 and where M is some n x t n matrix with coefficients in A . Now take cx")(X) = - J - ' M X . Then F ( X , cc(')(X))= 0 mod(degree 2) and a ( ' ' ( X ) is uniquely determined mod(degree 2) by this condition. Suppose we have found an a'*)(X),unique mod(degree r f l), such that F ( X , a'*'(X)) 0 mod(degree r 1). Suppose F ( X , u " ) ( X ) )= N ( X ) mod(degree r 2) with N ( X ) an n-tuple of homogeneous polynomials of degree r 1. Now take d * + ' ) ( X )= a " ) ( X ) - J - ' N ( X ) , then F ( X , cx("')(X)) = 0 mod(degree r 1) (cf. (A.4.8)) and dri ' ) ( X )is uniquely determined mod(degree r 2) by this condition because d " ( X ) is unique mod(degree r + 1). Let a ( X )be the unique element of A [ X $ such that cx(X) = cx'")(X) mod(degree r + 1) for all r E N. Q.E.D.

+

+

+

+

+