norming c(Ω) and related algebras

Norming C(Ω) and Related AlgebrasAuthor(s): B. E. JohnsonSource: Transactions of the American Mathematical Society, Vol. 220 (Jun., 1976), pp. 37-58Published by: American Mathematical SocietyStable URL: http://www.jstor.org/stable/1997632 .

Accessed: 30/04/2014 06:24

Your use of the JSTOR archive indicates your acceptance of the Terms & Conditions of Use, available at .http://www.jstor.org/page/info/about/policies/terms.jsp

.JSTOR is a not-for-profit service that helps scholars, researchers, and students discover, use, and build upon a wide range ofcontent in a trusted digital archive. We use information technology and tools to increase productivity and facilitate new formsof scholarship. For more information about JSTOR, please contact [email protected].

.

American Mathematical Society is collaborating with JSTOR to digitize, preserve and extend access toTransactions of the American Mathematical Society.

http://www.jstor.org

This content downloaded from 92.232.40.208 on Wed, 30 Apr 2014 06:24:55 AMAll use subject to JSTOR Terms and Conditions

http://www.jstor.org/action/showPublisher?publisherCode=ams

http://www.jstor.org/stable/1997632?origin=JSTOR-pdf

http://www.jstor.org/page/info/about/policies/terms.jsp


TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 220, 1976

NORMING C(Q) AND RELATED ALGEBRAS

BY

B. E. JOHNSON

ABSTRACT. The first result of the paper is that the question of defining a submultiplicative seminorm on the commutative unital C * algebra C(QI) is equivalent to that of putting a nontrivial submultiplicative seminorm on the algebra of infinitesimals in some nonstandard model of C. The extent to which the existence of such a norm on one C(Q1) implies the existence for others is investi- gated. Using the continuum hypothesis it is shown that the algebras of infinitesimals are isomorphic and that if such an algebra has a submultiplicative norm (or, equivalently, seminorm) then, for any totally ordered field I containing R, the

R-algebra of infinitesimals in I has a norm. A result of Allan is extended to show that in the particular case when t is a certain field of Laurent series in several

(possibly infinitely many) unknowns then the infinitesimals have a submultiplicative seminorm.

1. Introduction. In this paper we prove a number of results related to the question of whether C(Q), the algebra of continuous complex valued functions on the compact space Q, has a submultiplicative norm not equivalent to the usual sup norm. Bade and Curtis [21 showed that this is equivalent to the problem of putting a nontrivial seminorm on the algebra C(f2)IZs(F) where F is a finite subset of Q2 and A(F) is the set of all functions in C(Q) which are zero in some neighbourhood of F. We extend this by replacing A (F) by A(E) where E is a finite subset of the complement of Q\F in its Stone-Cech compactification and A (E) is the set of f E C(Q) which, when restricted to Q\F and then extended to g(Q2\F),are zero in a neighbourhood of E. Since C(f2)/Zs(E) is a direct sum of the C(&2)/Z,(co), co E E, the question is one of norming C(Q)/s(co). When Q2 is the Stone-Cech compactification of the integers the algebras C(Q)/ Z (c) are the complexification of the algebra of finite elements of a nonstandard model of the real numbers.

In ?3 we see the extent to which the existence of a nontrivial seminorm on one C(Q)/ Z (co) implies the same for all others. We see that if C(Q) has an incomplete submultiplicative seminorm for some compact Q2 then it has such a seminorm for all compact metric E. The results in this section were inspired by some unpublished work of A. M. Sinclair. The central result in ?4 extends a result of Erdos, Gillman and Henriksen [4, Theorem 13.13] which states that,

Received by the editors August 12, 1974 and, in revised form, January 6, 1975. AMS (MOS) subject classifications (1970). Primary 46J99, 12J 15, 12 L0.

Copyright C) 1976, Americani Mathematical Society

37



38 B. E. JOHNSON

under the continuum hypothesis, all real-closed 7 -fields of cardinality 28? are isomorphic. The extension is that if the fields both contain R then the isomorphism can be chosen to be the identity on R. As a corollary to this the results in ?2 can be improved; e.g. if any C(Q) has an incomplete norm so does every other infinite dimensional C(Q). Another corollary is that the algebra of infinitesimals in a nonstandard model of R has a norm if and only if the algebra of infinitesimals in any totally ordered field containing R has a norm. In ?5 we extend the result of G. R. Allan [1, Theorem 2] to show that certain algebras of power series with several variables (possibly infinitely many) can be normed. The general method is that of [1] . These algebras are algebras of infinitesimals in a totally ordered field and so, by the results of ?4, norming these algebras is necessary if C(Q) is to have an incomplete norm. In ?6 we embed the power series algebra in the algebra of finite elements in an ultrapower of R without using the continuum hypothesis.

2. Extension of the results of Bade and Grtis. Let Q2 be a compact topological space and v an algebra homomorphism C(Q2) -*B where e, is a Banach algebra. We shall extend Theorem 4.3 of [2] by making a more detailed analysis of the open sets G such that the restriction of v to

{f; f E C(QE), f(co) = Vco E Q\G} = R(M\G)

is continuous. We shall denote the set of all these sets by G (this is not the same object as (2 in [2] ).

LEMMA 2.1. If G1, G2, . . . are disjoint open sets in Q2 then Gi E G for all but a finite number of values of i.

This is merely a corollary of [2, Corollary 5.3].

LEMMA 2.2. If G1, G2 are cozero sets in G then G, U G2E

A cozero set is a set of the form f -1 (C\{0}) where f E C(Q2).

PROOF. If g, h E C(Q), G1 =g-1(C\{0}),G2 =h-(C\{0})andfE

R(R\(Gl U G2)) thenf = Igl(lgl + lhl)-lf + Ihl(lgl + Ihl)'ffon G1 U G2. Put g1 = IgI( Igi + I h-L)f, g2 = I h i(Igl + Ih I)- lf on G1 U G2 and extend to Q by putting g, = 92 = 0 on M\(G1 U G2). Then g E C(iQ) and so g, E

R(E2 \\Gi); the map f-+ g1 is continuous so the map f - (g1, g2) - (v(g1), v(g2)) + V(f) = V(g1) + v(g2) is continuous UD(2\(G1 U G2)) B-

LEMMA 2.3. SUPGE GlI v IIl(Q\G)II < X .

PROOF. Let p(G) = IIv1(E2\G)II for G E G. The proof of Lemma 2.2 shows that for cozero sets G1, G2, p (G1) + p(G2) > p(G1 U G2). Since Z (E2\G = { f;f E C(2), supp f C G} is dense in )(M \G) we have, for G E G, p(G) =



NORMING C(Q) AND RELATED ALGEBRAS 39

sup p(G *) where G* ranges over the cozero sets with closure in G. If the lemma is false we choose a sequence {G,} of cozero sets from G inductively by taking Go = 0, G?'+1 a cozero set with p(G+ 1 )>i + p(Gi), Gi+ 1 = Gi U G+ 1, and hence { G,} is increasing, p(Gi+ 1) > i + p(G,). Now choose Go = 0 and for each i > 0, G* a cozero set in G with G*cG, C G C G.* p(G*) >p(Gi)-1 . We have

p(G*\Ui* 1) > P (Gi )-p(Gi_ 1 -I-p(Gi-

1) > i-2,

which is impossible by [2], Corollary 5.3 as the sets G *\G-1 are disjoint.

LEMMA 2.4. Let QO be an open subset of Q. Then I3Q0 contains a finite subset E such that if G is open in Q then either G n E is nonvoid or G E G.

Here G denotes the closure of G in , nO. PROOF. Let E = { co; co E Ono) N n nO I G for all open neighbourhoods

N of co}. E is finite because if it were not we could find a sequence {co,} in

OBQ0 and an open neighbourhood Ni of coi for each i with N1 n Nj = 0 for i $ j. We would then have {N1 n QO} as a disjoint sequence of open sets in 2, none of which is in G, contradicting Lemma 2.1.

If G n E = 0 choose an open neighbourhood Ne, for each co E G with Ne, n Q E G. Let f<, E C(fQO3 ) have f,,(co) = 1, f, = 0 outside N., and put N,

= f,) 1(C\f0}), so that co E Nw C N. and, in particular, N,, n QO GG. By compactness there are cot,..., C E G with G C N,, U * * * U N' . Put G

= n nlN,, . Let G' C G be a cozero set in Q. Because G' is a cozero set in

Q with G' C nO and each G, is a cozero set in nO) Gi n G' is a cozero set in Q (it is (fg)-'(C\{O}) wheref= 0 on C\Qo2 f =fci on no) gE C(2Q) with G'

=g91(C\{O});fg is in fact continuous on ). Thus G' = (G1 n G') U * U

(Gn n G') E G by Lemma 2.2. Hence, using Lemma 2.3, we see G E G. One might seek to determine the collections G' of open sets in 2 which

satisfy the conclusion of Lemma 2.1 and other, more elementary, properties of G, for example any open subset of a set in G is also in G and any finite open set in 2 is in G. If nO = N, and Q is the one point compactification of N then the alternatives in Lemma 2.4 are mutually exclusive and only possible collections G' are the set of all open subsets of Q and

[S; SC N, S Ui i = 1,. ... , n}

where U1, . . , Un are free ultrafilters on N. For more general spaces the situ- ation is not so simple and the conclusion of Lemma 2.4 with nO = U G= Q\F, where F is the singularity set of Bade and Curtis, gives a sufficient condition for a set to be in G which is by no means necessary.



40 B. E. JOHNSON

We now introduce some more notation. As F does not contain any isolated points of 12, 12 is a compactification of Q\F. Thus there is a continuous map a: ,B(Q\F) - 12 which is the identity on Q\F. Applying Lemma 2.4 with 20 =

Q\F we have a E = F. Let U E E and put

. (F)= {f;fEC(Q2),f(co)=Ofor all coEF},

I(U) = {f;; f E fR(F), f o a = 0 on some neighbourhood of U in B(Q2\F)},

1(U) = J(F)/I(U) = (T(aU)/5(U).

I (U) depends on U but is unaltered if F is replaced by any other fmite set containing a U.

Lemmas 2.3 and 2.4 and the same methods as are used in [2, Theorem 4.31 enable us to prove

THEOREM 2.5. Let v be an algebra homomorphism of C(Q2) into a dense sub- subalgebra of a Banach algebra B8. Then there is a closed subset 12' of 12, a finite set F C 12', a finite set E c P(12\F)\(Q\F) and an algebraic and topological isomorphism o of X onto NB such that v = qpo, where

(i) For U E E, RU is the completion of 1 (U) in a nontrivial submultiplicative seminorm, i.e., the completion of the quotient of 1 (U) by the kernel of such a seminorm.

(H) (E=CQ9) (D Rul (1 (1 ?Un

where U1,..., U,, are the points of E, the Banach space structure is the direct sum and multiplication is the commutative operation determined by

(f, 0, **. , 0) (0, r, ... , rn) = ( fgO,f,(U 1))rl,.. U (O,~~~~~~~~~~~~~~~ rl, *, Si * , un O 1s* n (sr 1 **sr

(iii) i(f) =(f IQ'r1* rn)

ri being the element of Ru. corresponding to any element of g (F) which is

equal to f - f (a(Ui)) in some neighborhood of a(U,).

The important feature of this result is that it demonstrates the equivalence between the problem of putting an incomplete algebra norm on C(Q2) and the problem of putting a nontrivial algebra seminorm on the algebras I (U).

3. Norming C(Q2) and norming algebras of infilnitesimals. Let V be a free ultrafilter on N, s the algebra of all sequences of complex numbers with point- wise operations, A (V ) = {a; a E s, a- l(0) E V } a maximal ideal in s, *C(V ) =



NORMING C(a) AND RELATED ALGEBRAS 41

s/ZS(V ), i (V ) the image in *C(V ) of those elements a of s with lim v a = 0 and

i 0(V ) the image of co in *C(V ). We shall frequently write merely i, i 0X *C for i(V),io(V), *C(V).

LEMMA 3.1. Let &2 be an infinite metrisable compact space, co a noniso- lated point of &2 and U E f3(&\ {Co0 })\(Q\ {C0 }). Then there is an algebra

homomorphism p: co - 1(U) with l(U). Im p = I(U).

PROOF. Let d be a metric in &2 giving the topology and let

G. = {CD; C E &2,(2n)-1 > d(C, wo0)> (2n + 3)-1},

Go = {co; co E &2, d(co, co) > 3-1}

and put G?=U= 2+n Ge = U-n0=o G2n. U can be considered as an ultrafilter of relatively closed sets in &\{coo}. We have Go = G n (2\{wO}) for some neighbourhood G of U in 3(&2\{ coo}) if and only if Go contains a set from

U. Thus either Go = G n (&2\{coo}) or Ge = G n (2\{co0}) for some open set G in 13(&A{co0}) containing U. For definiteness we assume Ge = G n

(Q2\{ WoD}- Let Pn E C(Q&2) with Pn = 1 on Gn, Pn = 0 on Gm for m = n - 1, n,

n + 1. Let q be the quotient map RI(co0) I-+ (U) and v the map co -CQ ) defined by ij(a) = z an P2n. Then l is linear and, on Ge, 4i(ab) = (a) (b), a, b E co. Thus p = q P is the required algebra homomorphism. If f E U(o0)

put bn = supweG2n If(co)I (bn = 0 if G2n = 0). Then b E co and there exists

c, d Ec coo c ? > ,dn>O with bn = cndn. Let go = dn lf on G2n, go(co) = O so that go EC((U G20)-) and has a continuous extension g E C(Q&2). We have

g E 2R(co0) and on G2n, 4I(d)g = dndn lf = f so p(d)q(g) = q(f). The following result is related to [8, Corollary 4.4].

THEOREM 3.2. (i) If &2 is a compact topological space such that C(Q&2) has an incomplete submultiplicative seminorm then i o(V ) has a nontrivial submultiplicative seminorm for some ultrafilter V on N.

(ii) If o(V ) has a nontrivial submultiplicative seminorm for some ultra-

filter V on N then C(Q&2) has an incomplete submultiplicative norm for every in-

finite compact metric space &2, and more generally, for every compact topological space with a nontrivial convergent sequence.

PROOF. (i) If C(Q&2) has an incomplete seminorm then there is a discontinuous homomorphism : C(Q&2) -+ F [2, Theorem 4.4]. Hence there is a separable closed * subalgebra I of C(Q&2) containing 1 with PI U discontinuous. We have

U t C(&2*) where Q* is a compact metric space. Now apply Theorem 2.5 giving co E Q* and U E 0(&2*\{co }) such that l(U) has a nontrivial seminorm p. Let



42 B. E. JOHNSON

ep be the map in Lemma 3.1. Then p o <p is an algebra seminorm on co giving rise to a homomorphism v': co )- B' for some Banach algebra V'. The kernel of p is a proper ideal in l(U) whereas Im <p lies in no proper ideal. Thus v' 0 0. However if a E co has finite support then 41(a) is 0 in a neighbourhood of coo so

v'(a) = 0. This shows that v' is not continuous and so, applying Theorem 2.5 we see that there is an ultrafilter V on N such that io(V ) carries a nontrivial multi-

plicative seminorm. (ii) Let {wi) I be a sequence of distinct points of Q2 with limit co where,

for all n, X o wn. The map 0: f - { f (COn)} is a continuous algebra homomor-

phism of C(Q2) onto c. If i o(V ) has a nontrivial submultiplicative seminorm then the construction in Theorem 2.4 gives a discontinuous homomorphism z: c -+B

for some Banach algebra B. vO is then a discontinuous homomorphism of C(Q&2). into B.

THEOREM 3.3. If there is a nontrivial submultiplicative seminorm p on i (V) for some V and Q2 is an infinite compact topological space then there is a discontinuous homomorphism of C(Q2) into a Banach algebra.

PROOF. First of all we show that there is an ultrafilter V' on N and a semi-

norm on i (V ') which does not vanish on io(V '). Let a E 1X(N) with llall < 1, lim v an = 0 and pqa 0 0 where q is the quotient map s *C(V ). Put Vj =

{n; n E N, la'lI < -1} and let o: N N be defined by o(n) = i if n E

Vi\ Vj . Define (az*b)n = bo(n) so that c* is a unital homomorphism l"(N) - l(N). a V = {a V; V E V } is the base for an ultrafilter V ' on N which is

free because V1 E V and a (Vj) C { , i + 1, i + 2, . . . }. pqa * is an algebra semi-

norm on the subalgebra of l-(N) of elements b with lim v b = 0 which is zero for any b with b- 1(0) E V '. Thus it induces a seminorm p' on i (V '). As in the proof of Lemma 3.1 we show a E qoc*co i (V ) so p is not trivial on q c*c .

Thus p' is not trivial on i O(V'). Now choose a sequence {G,} of disjoint open nonvoid sets from QI and for

each i, coi E Gi. Let A(f)n = f(con). Then p is a continuous algebra homomor-

phism C(Q2) I- l'(N) with closed range containing c. Writing 131 = { f; f E C(Q&), lim V <p(f) = 0}, pqoe* p is a seminorm on 13 which is not continuous with re-

spect to the sup norm as pqa* is not continuous on co and <p is an open map-

ping with range containing co. As C(&2) is the algebra obtained by adjoining a

unit to 91, pqa*<p extends to a submultiplicative seminorm on C(Q2) which is dis-

continuous.

4. Results using the continuum hypothesis. We now turn to a study of

the algebras Z(U), i (V ). Since each is the complexiflcation of a real subalgebra

J(U) or i(V), the subalgebra of elements corresponding to real valued functions

or sequences, and a real algebra carries a nontrivial seminorm if and only if its



NORMING C(a) AND RELATED ALGEBRAS 43

complexification does, we study J and i instead. If t is a totally ordered field containing R as a subfield then

to= {k;kE t,-q<k<qforallqEQ+}

is the algebra of infinitesimals in f. We use Q+ rather than R+ in the definition because Q can be embedded in t in only one way whereas the embedding of R is not unique. t 0 is a real algebra if we take agk, ae E R, k E t? to be a prod- uct in t. Similar remarks apply to

t# = {k; k E , -q < k < q for some q E Q+}

the algebra of finite elements of t. t# is the algebra obtained by adjoining a unit to t 0 so every element a of t # can be expressed uniquely as f (a) + b where f (a) E R, b E f O. f is an order preserving homomorphism of t # onto R. When V is an ultrafilter on N, i(V) = *R(V )O. If U is a free ultrafilter of zero sets in Q\ {coo}, and

~(U)= {f;fE CR(Q\{CO}),f '(0)E U}

then *R(U) = CR(Q\{coo})I~(U) is a nonstandard model of R and there is a homomorphism J(U) -* *R(U)o because every real valued function in Ai(U) is in ~(U).

We now show that the various algebras i (V ) are isomorphic. It follows from [4, Theorem 13.13] that they are isomorphic as rings but to show that they are isomorphic as algebras we need to show that there is an isomorphism between the fields *R(V) which leaves R invariant.

LEMMA 4.1. Let t, L be totally ordered fields containing R and let u be an order preserving isomorphism t -+ t. Then gu = f where f is the map I

R defined above and g is the corresponding map L# -+ R.

PROOF. For a E t#, La = {q;q E Q. q <a} andRa = {q;q E Q, q >a} define a Dedekind cut which is the cut associated with f (a). As u is the identity on Q, uLa = La, uR = R but because u is order preserving uLa = L , uR = a' a a a ua' a Rua so Lap Ra is the same cut as Lua' Rua' which defines guta. Thus fa = gua.

If t is a field containing R and t' is a subfield then t' is full if t' n t # is closed under f. If A, B, . . . are subsets of a field then K[A, B, . . . ] is the

subfield generated by A, B, .... If ' is a subfield of t then alg d' is the set of all elements of t which are algebraic over t '.

LEMMA 4.2. Let t be a totally ordered field containing R Let E' be a full subfield of , w E R Then t " = K [ t 'w] is full.

PROOF. We suppose w C t' as otherwise there is nothing to prove. Let z E t " n t #. z can be expressed as a rational function of w with coefficients in



44 B. E. JOHNSON

t ' and, if w is algebraic over t', we can assume that the degrees of the denominator and numerator are less than the degree of w. Dividing numerator and denominator by the coefficient of the denominator with largest absolute value we have

P(w) ao +aw + +akw

Q(w) b +b w+ +b w1

where -1 bi<+1 and one b1is 1. We havefQ(w) =f(bo) +f(b1)w + * * + f(bl)w O because the coefficients are in t ', are not all zero and if w is algebraic over t ' it is of degree > 1. Thus Q(w)- 1 E E 0. Similarly we can write P(w) =c(aO +a'w + - * - +a'wk) =cR(w)where ce E ',a'E e'n E and one a; is 1, unless P(w) = 0 in which case clearly f(z) E t". We have R(w) and R(w)- 1 E t #. Thus c = zQ(w)- 1 E t and hence P(w) = cR(w) E E 0. This shows

f(a0) + + f(ak)wk f(z) =

f(bo) + * * * + f (bl)w

where the coefficients are in t ' so f(z) E U".

LEMMA 4.3. Let t be a totally ordered field containing R, and t 'a subfield of t. Then there is a smallest full extension t " of e'

t" =K(t'," n R)

and if t ' is denumerable so is t ".

PROOF. For any subfield t of t put E( )=K( L, f(f n )). Take

U " = j Ui P ifC).

LEMMA 4.4. Let t be a totally ordered field containing R and t' a full subfield of t. Then ailg t' is full.

PROOF. If z E: t# n alg E ' then we have a nontrivial relation aO +

a z + * * * + a?zn = 0, a E t '. Dividing by the coefficient of largest absolute

value we can assume ai E E # n t' and one of the a, is 1. Thus

f(ao) + f(a1)f(z) + * * * + f(an) (f(z))n = 0

where f (ai) E t' and some f (ai) is 1 so f (z) E alg '.

LEMMA 4.5. Let f, L be ordered fields containing R, t' a full subfield of t and u an order preserving isomorphism of t'into t with u = identity on n R Then u(f'n R) = u(t') n R.

PROOF. Suppose a E t' with u(a) E R. If p, q E Q with p < u(a) < q then p <a <q so a E t # and p <f(a) <q. Thus f(a) = u(a). Since f(a) E E'




fn R, u(f(a)) = f(a) we have u(a) = u(f(a)) so a = f(a) e t ' n R.

LEMMA 4.6. Let t be a real-closed n1 -field containing R Then e has a

dense transcendence base over R.

This is an extension of [4, Lemma 13.111. Our terminology is that of [41, in particular ? ? 13.5-13.8.

PROOF. First of all we show that if N is the cardinal of t then the trans- scendence degree of t over R is S. If N > 2? 0 then any transcendental extension of R of degree less than N has cardinality less than N and any algebraic extension of an infinite field has the same cardinality as the field [4, p. 1721.

Suppose N < 2 ?. Since R C t we have N = 2?0. Because t is an field, t 0 has strictly positive elements. Let y be one and put x = y - . The map

p -+ xP where, if p = rs- 1, r, s E Z, s > 0, then xP is the unique positive solution z

of zS = xr, is an order preserving map of Q into t with xp E t \ t # if p > 0. Let A C R+ be a basis for R considered as a vector space over Q and for a E A choose x(a) E t with xP < x(a) < xq whenever p, q E Q with p < a < q. We shall show that the elements {x(a); a E A} are algebraically independent over R. Suppose a1, ... , ak EA, ml ... I, mk are nonnegative integers and p E Q with p < mlal + * + mkak. Then there exists p1,. . - , Pk in Q with p1 < a, and

p <m lp+ + + mkpk SoXP < (XP1)ml ... (XPk9.

k <X(ai)ml

x(ak)mk. Similarly if m1a, + * * * + mkak < q E Q then x(a 1)I * * * x(ak) k

<Xq. Thus putting v(M) = m1al + * * * + mkak if M= X(al)m1 *. * *x(ak)mk

we have xP < M < Xq whenever p, q G Q with p < v(M) < q. In particular, as the a, are rationally independent, v is well defined. Suppose there is a nonzero polynomial P with real coefficients and P(x(aj), . . . , x(ak)) = 0 for distinct al, . . ., ak E A. Omitting all terms with zero coefficient and dividing by the coeffilcient of the monomial M for which v takes its highest value (as the a, are rationally independent M is uniquely determined) we can assume that M has coefficient 1. Let p, q E Q with v(M) > p > q > value of v at all other monomials in P, and let - , be the sum of the negative coefficients in P. We have xP < M = M - P(x(a1), .. . , x(ak)) < pxq < Xp-qXq = x P, a contradiction.

We complete the proof by taking the smallest ordinal o* of cardinality X, indexing the intervals (a, b), a, b E f, a < b by the ordinals < .* and choosing

for each co < co* an element x., of I,, which is transcendental over K [R, x ; a < co] by transfinite induction. This choice is possible because K [R, x<>; ca < co] and hence its algebraic closure in t, is of transcendence degree <* over R whereas t is of transcendence degree X, so t 0 alg K [R, xa,; a < co] = C'. Since t' # t we cannot have I,_ C t'. {x<,; co < co*} is a dense set of ele-

ments of t which are algebraically independent over R and so can be completed to form a dense transcendence base.



46 B. E. JOHNSON

The following result was first obtained by a different method by N. J. Block (thesis, University of Rochester, 1965, but otherwise unpublished; see also [5, p. 145] ). The author wishes to thank the referee for drawing his attention to this reference.

THEOREM 4.7. Let f, t be real-closed T71-fields containing R and of cardinality N1. Then there is an isomorphism u of t onto L. with u IR = identity.

PROOF (cf. [4, Theorem 13.13]). Let co, be the first uncountable ordinal, S, T dense transcendence bases of t, f over R. Index S, T by the countable ordinals so S = {sX>; a < co}, T = {tXx; a < co}. We shall define for each a <

cI, by transfinite induction, countable full algebraically closed subfields t ,x , t a of e, t and an isomorphism u. of t a onto t a with u. = identity on E a n R and,for<cx< col,so Etc,lto E t, to ct to La)uJit=up. We start the induction by taking t 0 = 120 = algebraic numbers, uo = identity. If ai < co1 is a limit ordinal and E t, , u, have been defined for <3 <awe put tE

= Up<x a , t =U<a= rV and takeu. to be the unique map to a clwith ultI =u, for 3< c.

Suppose ax = , + 1 and t ., t Y , uy have been defined for 'y?1,. First of all we shall extend u, to t', the smallest full algebraically closed subfield of t

containing to and so. If s, E E, then t ' = E, so we put u' = u,. Suppose

so 0 E , and let r1, r2, .. . , be a transcendence base of t ' n R over E l n R (the base is countable because ' is countable, the base might be finite or even void). We shall put E n = alg K[ to, r , * * rn I, f'n = alg K[Lo, r1 ,. .* . rn] and define bijective isomorphisms un; E n t ' with u 0 = up, u n+1 1t n = un un = identity on E " n R, un(rn) = rn by ordinary induction. Suppose u n has been defined. t n and t n are full (Lemmas 4.2 and 4.4) so, by Lemma 4.5, t n n R =

un(En n R) = n n R. As rn +1 is transcendental over t n it is transcendental over t n n R = n n R and the proof of Lemma 4.4 with z = rn + 1 = f(Z) shows that rn+ 1 is transcendental over L n. Thus u n has a unique extension to an isomorphismvofK[t n,rn+1] ontoK[Ln,rn+1] withv(rn+1)=rn+1. IfxE tn#, x> rn+ 1 then f(x) > rn+1. As t n is full,f(x) E t n whereas rn+ I t n

sof(x)>rn+1. Thusthereisq E Qwithftx)>q>rn+j andsox>q>rn+1. This conclusion holds trivially if x E f n \ t n andx >rn+l so ifx Efn,X>

rn + 1 then un (x) > un (q) = q > rn + 1. Similarly if x < rn + 1 then un(x) < rn+ 1 showing that v is order preserving on t n U {rn+ 1 } and hence on K[ t n, rn + 1] [4, Lemma 13.12]. By the Artin-Schreier theorem [7, p. 285] v has an extension un + 1 which maps t n + 1, the real-closure of K [ t n, rn + 1 ] onto

Ln + 1, the real-closure of K [ t n, rn + 1 ] . The proof of Lemma 4.2 shows that if

a E K[t n, rn+ 1] n R then a E K[ t n n R, rn+ 1] = 5so v(a) = a. If a E

t n + 1 n R then the proof of Lemma 4.3 shows that a E alg ' . Both u n + 1 and




the identity are order preserving isomorphisms alg c -- t which agree on r so by [7, Corollary, p. 172] they agree on alg r . Thus u'+1 = identity on t I+ 1 n R.

The u ' give an extension u of u, to t= U t =alg K[t, r1,...]. If s E t then t' = t and u' = u. If so X t then it is transcendental over t Now consider the sets

A = {a; a E , a < s,X} B = {a; a E,a > s}.

Because t is algebraically closed in t it is real-closed, hence L = u (t is real- closed and so algebraically closed in L. Put

A'= u (A) U {t; t E T n T , t < u(a) for some a E A},

B'=u(B)U {t;tETn L,t>u(a)forallaEA}.

Then A', B' are countable sets. Let t E T with A' < t < B'. Since t E L, t is transcendental over L and u extends to an isomorphism u' of K[ [, so] onto K[ t, t] with u'(sd) = t. u' is clearly order preserving on t U {s,} so by [4, Lemma 13.12] on K [ t , s] . Again using the Artin-Schreier theorem we extend u' to the algebraic closure of K[ , sp] in E, which is t '. If a E t ' n R then a is algebraic over K[ n R, r1, r2, . . . ] so a E t and u'(a) = u(a) = a. Put '

= u'( t'). By the argument used above to show t I n R = tn n R we see t ' nl

R = L'n RR. By Lemma 4.1 gu'=f so L'n R = 'n R =f(t'#) g(t#) and L ' is full. L' is algebraically closed for the same reason that t was.

We now repeat the whole procedure with t ,, L,, s,, u, replaced by L',

t, t,, (u')f 1 and take u. to be the inverse of the resulting map. This completes the inductive step.

We define u: U ax< by u It 0 = UO ' Ua< Y is

full and contains S. Since S is order dense, for each a E R there is s E S n t #

with f (s) = a. Thus U < w 1 t a contains R. Since it is also algebraically closed

in * it is t . Similarly t = U x< tLa

COROLLARY 4.8. Let E be an ordered field containing R and of cardinality t. Let T be a real-closed 7t1-field containing R. Then there is an order preserving isomorphism u of t into T with u IR = identity.

PROOF. Remarks similar to those at the end of ? ? 13.9 and 13.5 of [4] apply. By the Artin-Schreier theorem we can assume that t is real-closed. In the proof we take S to be any transcendence base for t over R and when a = ,3 + 1 we take t a = t' and the induction proceeds as before except that if S is countable, U<x<< < tx is replaced by Un=1 n. For the final extension to t we

use the same procedure as was used to extend from , to t ', applied transfinitely if necessary.



48 B. E. JOHNSON

COROLLARY 4.9. If i (V 0) has an algebra norm for one free ultrafilter V 0 on N then, under the continuum hypothesis, i (V *) has an algebra norm for every free ultrafilter V on N.

PROOF. i(V) - *R(V )O and *R(V ) is a real-closed 7, -field of cardinality 2 0 containing R.

COROLLARY 4.10. If V is a free ultrafilter on N and i (V ) has a nontrivial algebra seminorm then, under the continuum hypothesis, it has an algebra norm

(cf. [9, ?4]).

PROOF. Let t = K[*R(V), X], the algebra of rational expressions in the in- determinate X with coefficients from *R(Vl). We order the polynomial ring lexicographically, P> 0 if the term of lowest degree with nonvanishing coefficient has a positive coefficient, and take the induced order on the quotient field E. t is a totally ordered field containing *R(V) as the polynomials of zero degree, and hence containing R. The cardinality of E is 2N . Put t = *R(V) and let p be a nontrivial seminorm on L . Suppose to E Lo with p(to) 0 0 and to > 0. In Corollary 4.8 we can take so = X. K[t 0, so] is the field of rational functions with coefficients from the field of algebraic numbers. Any nonzero element z can be written in the form

Xm(aO + al X + * * * + akXk)

(bo + b1X + * * . + b1X')

withai, bi toe, ao 0, bo #0,mEZ,ze *#ifandonlyifm>0andif m = O,f(z) = ao/bo E % ,if m > O,f(z) =0 E to, Thus K[t %,so] is full and hence, by Lemma 4.3, t1 = alg K[t0, so]. If z e fln R then the proof of 4.3 shows that z = f(z) is algebraic over f(K [ t o, so] #) which we have just shown to be t %. Thus E I n R = t 0 n R and t 0 = t . The sets A, B are the set of non- positive algebraic numbers and the set of positive algebraic numbers respectively, also A = A', B = B'. Thus we may begin our induction by taking u1(sO) = to, that is,there is an order preserving isomorphism u; e *R(V) with u = identity on R, u(X) = to.

We show that p o uj is really a norm. Suppose a Ej, a > 0. AsX < qa for all q E Q+, we have a- 'X E i . Thus 0 < p(to) < p(u(a)) p(u(a- 'X)) showing p(u(a)) # 0. A similar argument applies if a < 0.

COROLLARY 4.11. Let V be a free ultrafilter on N. Under the continuum hypothesis, if j 0 (V) has a nontrivial algebra seminorm p then so does i (V ).

PROOF. Let E = K [ *R(V ), X, Y] t = *R(V ) where X, Y are indeter- minates. First of all we put an order on the ring of polynomials in Y with coefficients from *R. We say P > O if P = c(ao + a, Y + * * * + ak yk) where c E




R, c > 0, a1 E *R# for all i, a, q *RO for some i and the first coefficient not in *RO is positive. This gives an order on the quotient field K [ *R, Y] and we order t = K [K [ *R, Y ] , X] lexicographically as in 4.10.

Let tE iO with t >0,p(t) >0. There is w io with w > t'n for all n E N because for each sequence {an} of elements from co there is b E co with

bj > an, for all but a fmite number of values of j . Define uo I A = identity, uo(X) = t, uo(Y) = w we have an isomorphism of K[A, X, Y] into t = *R where A is the set of algebraic numbers. As uto is order preserving it extends to t 0 = alg K [A, X, Y] . t 0 is full. Applying the induction in 4.7 and 4.8 starting with this definition of t 0' uo we extend uo to an order preserving homomorphism u; t 1. Ifa Ejwehave - Y<a< Yin t so -w<u(a)<wwhichimpliesuiC I 0. Thus pu j is a seminorm on j and an argument similar to that at the end of 4.10 shows that it is a norm.

5. Norming algebras of power series. Although they involve the continuum hypothesis, the results of the previous section suggest the problem of norming t for any ordered field t containing R. One such algebra, the algebra of formal power series is one variable, has been normed [1]. In this section we extend these results. The field Q (X, t ) of formal Laurent series in X with coefficients from a field t is the set of all expressions 2n= an Xn where j E Z, an E e. If i < j and an = 0, i < n < j then n= a Xn and n= an Xn are identified. We define

E an Xn + E bn Xn= E (an+ bn)Xn n =j n =j n=j

E amXm E bnXn = E( E ambn) XP. m =j n=k p=j+k m+n=p

With these operations Q (X) is a field containing a copy of t, the series with an = ? for n * 0, which we identify with f. If t is an ordered field tl(X) is also an ordered field under the lexicographic ordering. If t is ordered and contains R we put z(X) = {i aX; ai Et* aO E tEE }, (X) is an algebra over

R and g (X) = V (X)#. We now define ordered fields 2af for each ordinal a., corresponding algebras

a and projections p,c,; ;a E-i1 for a. > 3. Put U o-R = '0, d + 1 =

W(Xa,I tj), &1+i = ?k(Xa, 2) where for each ordinal a, Xc, is an indetermin- ate. We define P,a+ = identity on Sa+1 Pa+j,a (i-o a, X.') = aO,

PI+1,P = PUP"+l Ia: for a. > 3. For a limit ordinal at, &, is the projective limit of the system {95p; ph,, a. > , > y}, an integral domain because each , is, by transfinite induction, and 2af is the quotient field of g. If a. > ,, p<,, is the projection of ga, onto the ??,l coordinate. pco, is the identity on Sa.



50 B. E. JOHNSON

For a > ( we define injections ia of p into & by transfinite induction; i<> +1 is the injection Qae C Q (XQc) restricted to ig, 1j3cx+1 = 1i,+1ij3,c

and for a a limit ordinal, ,B < a, A E 0p, ig,B (A) is the element B of , with

poly(B) = ip^ (A) if a > y > , poa (B) = p(A) if , > y. We have pcgipci = identity on '0p for (3 < a. Since pa is the quotient field of &, for all a , i,, extends to an isomorphsim Q --> 2,,. We shall identify 4p with its image under this map so that the 2 , are an expanding system of fields. We put Q =

Ua < w a, = U< w1 g where .o, is the first uncountable ordinal.

Denote the complexification of 2C, ), a , MS by PC, 0a, R, (R. These can

be defined in the same way as the 9 's and 42's but starting from C in place of R. They are not totally ordered. We are going to map these algebras into a complex Banach algebra K and it seems more natural to work with the complex algebras

f throughout. We shall use p to denote the map 9, - obtained by complexifying the original projections.

The term monomial is defined by transfinite induction. 1 is a monomial. An element of P. + 1 is a monomial if it is of the form aX, where a is a monomial in R at and n E Z. An element of P, where a is a limit ordinal is a monomial if it is a monomial in P for some A < a. Every monomial can be written X, Xn2 n.Xk weeten Xa n * * * 2 k where the ni are integers.

LEMMA 5.1. Let a be a countable ordinal and let A E g;0. Then A is regular in (c if andl only if pyo0(A) # 0.

PROOF. As pg0 is a unital homomorphism, if p,,O(A) = 0 then A is not regular. We prove the converse implication by induction on a. It is obvious for a. = 0. If it is true for some value of a and A E (c+ 1 with p. + 1,0(A) # 0 then A = S- aiXcwhere a0 E ( and pOt0(a0) = pa,o&p+ 1,? (A) = p+1 ,0(A) #

0. Thus aO is regular in ( and so the inverse of A in Ra+ 1 is an element of ( + 1 . If oz is a limit ordinal and the result is true for all < a, let A E ( , . For (3 < a, pgop,,A = P<,loA / 0 so (pcgA)f 1 exists in (Lp. The (p,,A)- form the inverse of A in a

LEMMA 5.2. Let a be a countable ordinal and let A E CSf, A # 0. Then there is a monomial M in Z and a regular element B in ;0e with A = MB.

PROOF. The result is obvious for a = 0. Suppose it holds for all (3 < oa

and let 'y be the first value of ( with p,g(A) # 0. Sy is not a limit ordinal so pO ,,(A) = SO ai X_. Let j be the first value of i with a1 0 0. j is not O. Since P'- is the quotient field of (E,_ we have a1 = M'B' where B' is regular in LY ,- 1 and M' is a monomial in Rq,- Thus pO,, (A) = MB" where M =




M'X' is a monomial in and B" = B' + Ej 1 M' 11aiX>,'-7 is regular in

(E 7 because p7 OB" = py -1 o B'

We now show that for y < ,B < a there is a regular element B, in (i, with pop(A) = MBO. We have shown this for ,B = ,y. If it holds for one value of ,B

less than a then p + (A) = pp(A) + YEn bn,YP = MB + I ' 1 bn J =

MO+ I bwhere Bp1 = Ba + E is invertible in If 6 is a limit ordinal with -y < 5 S o and the result holds for all ,B with -y < , < 5 then

Mppp (Bp) = ppp (MBp) = MBO' where y < ,B' S ,B < 8 so, since fi , is an integral

domain, pgg (Bg) = Bp' and the Bg define a regular element Bs of (S 8 with MBS

= Pa (A). Since R.> is the quotient field of (yaf m if A is a nonzero element of R c then

A = MB where M is a monomial in PC, and B is regular in ( a .

21 is the algebra L1 (0, 1) with convolution multiplication and Ul the algebra obtained by adjoining a unit to U. Every element of 2t is a topological nilpotent so U = Radical U I. An element a of Et is a zero divisor if and only if

f I a(t) I dt = 0 for some e > 0. It is easy to see from this that a is a zero divisor if and only if it is properly nilpotent. If a is not a zero divisor then a U = Kt and conversely. The nontrivial parts of these results follow from the Titchmarsh convolution theorem by methods similar to those of Donoghue [3]. We now select elements xaf of Ut which will be the images of the elements X<> of (E .

LEMMA 5.3. There is a system {xa; a. < co } of elements of U indexed by

the countable ordinals with xaQi = 2t and xa 2 C 'x Efor>,B,nE Z+.

PROOF. We proceed by transfinite induction. Let xo be any nonzero- divisor in VI. Suppose xa has been defined.

As [Xo = [xaf [xon we see (xn = for all n E Z+. Put

Ia+ 1 = fl,= x t. Application of [8, Lemma 3.1] with Rn as left multiplication by xa shows Ia+ 1 is dense in 21. On Ia+ 1 we define I Ix I In = II yll where y is the element of a with x n y = x. y is unique because x n is not a zero divisor. We put llxllo = lix II. This sequence of norms makes I 1 a complete metric space. The set En = {x; x E I, x = 0 on [0, n1] } is a closed linear subspace in Ut which is not dense in 21. Hence Ia+ 1 n En is a closed proper subspace of Ia+1 and, in particular, is nowhere dense in Ia+ 1, Thus, by the category theorem,

U0 1 (Ica+ 1 n En) # icf+ 1 so Ic+ 1 contains an element x,+ 1 which is not a zero divisor in VI.

If a. is a limit ordinal we make a similar construction. Writing the ordinals < a. in a sequence { On} and putting Ia, = n,T<af xg EDno,

which is dense in Ut by [8, Lemma 3.1] because x,1 ... xp n Uis, we define norms

ll ll, on Ia, by lix 11I = 11 yll where x = x, y. Again Iaf is a complete metric space



52 B. E. JOHNSON

and so contains an element xa which is not a zero divisor.

COROLLARY 5.4. If 3 < oa then y F-+ xp y maps IO one to one onto itself.

PROOF. Since the I,, are ideals in Wt, y-+ xy maps Ic, into I,,. Because

x, is not a zero divisor the map is one to one. Suppose a is not a limit ordinal and x E I,a . Then for each n E Z+ there isYn E U with x = xi1y Yn. As there is z E Ul with xpz = x,..1 we have x = x nxaT1' y,z = x,ylz. Thus, because

x, is not a zero divisor, y1z = x yn - 1z, n = 2, 3,.. ., so y1z E Ic . Suppose

a is a limit ordinal and x E I,,,. Then x = xpw for some w E 2t. Also for each

y with a. > y > 3 there are y), z) E .C with x = x2y )X = xpz . Thus x w x= xx zy yyso that w = XyyYz for all y with a > y > ,. This shows

wEIC,

LEMMA 5.5. Let a be a countable ordinal and 0. an algebra isomorphism , O-.+ f 1 with O,(Xp) = xp for 3 < ot. Let it denote the quotient map -

9 /I + 1. Then irr Of has an extension t to an algebra homomorphism fi o+ I II+ 1 with 4 (XO) = ir(x,).

PROOF. Let a C Rc,. By Lemma 5.2 we have a = X-'1 * X - 'B whe7re

3i < a and B E C. If , is the largest of the j3i then x, f1 c x. n x, * X

921 so there is an element 0 (aXe) of Kt with x * x03 O(aXa) = xaO(B).

0 (xX() is independent of the choice of 1, . .. , On. We have

O(aXa) + 0(bX,)= 0((a + b)X0),

0 (aXc)0 (bXa) = 0 (abXcl)xa

0 (aX, ), (c) = 03 (acXO,), a, b E ga c E CE-c from which if we define

~~~~~~~~0 '(,. + (amX,)xm-1 0(ao +a, Xm+ + am Xm a (ao) 0ax+ + (a7 )am:

where ao E a a,. . , am E $2 then we have a homomorphism O from the polynomials in Ia+ into 211.

Let A E C+I, A = V* 0 aiXa where ao E C ,a,a2,*.a,, (a) is the element w of V + I with

w - IrO (aO + a, X., + + amXcm ) E xm jt/IIc+1

for all m in Z+. There is at most one element w with this property. To see that there is exactly one choose elements Yn of X inductively so that

YO= 0(ao), llyYn + 07(an+1Xa) -yn+lxI < 2n 11x,,jV-n,




the choice being possible because xc, I is dense in WX. Put

Zm = E (Yn + 0(an+IXa)a m n>m

We then have

zo = 0'(ao + a,X., + * * * + a Xa: )ynx I + znXC Xl.

Hence irz0 is an element of .1'/I4a+1 with 7rzO - 7rO (ao + alXa + * +am,cXc) E x 21/lIa+ 1 for all m. We put iP(A) = rzo. is a map (a -1 II and is an algebra homomorphism because i(A) + i(B) and i(A) (B) satisfy the defining condition for i(A + B) and i(AB).

COROLLARY 5.6. Under the same hypotheses as 5.5, if (E is a subalgebra of

C% + containing Ca-, and X. and 0; (E >.I' is an extension of 00X with 0 (Xa) =Xa: and rO =0 on (E then, for all nonzero A in x I-# 0(A)x maps Ia+i one to one onto itself 0(A) is not a zero divisor in WQ.

PROOF. By Lemma 5.2 we have A = MB where M is a monomial and B is regular in Ci a:+ 1 . Let b E UI1 with ir(b) = (B). By Corollary 5.4 x i-? O(M)x maps I',+ , one to one onto itself as it is the composition of maps x F x,x and their inverses. We have 7r(0 (A) - 0(M)b) = (A) - P(M)t(B) = 0 so 0 (A) -

0(M)b E I+ I1- Let i E 1a +1 with 0(M)j= 0 (A) - 0(M)b. Then 7r(b +j)= /(B) and 0(A) = 0(M) (b + j). Because B is regular, i(B) is regular in ff /11a+ and, as 1a+ I C U = Rad VI1, b + j is regular in U 1. Thus x - (b + j)x maps ,a+ i one to one onto itself and hence so does x i- 0(A). If y E 11 with

0 (A)y = O then 0 (A)xa + 1 Y = 0 so x + 1 Y = 0 and hence y = O, because

xa + 1 is not a zero divisor and x. + 1 E IC, +

LEMMA 5.7. Let a be a countable ordinal and O00 an algebra isomorphism

a - 9t1 with Oa(X0)=x,, <3<a. Then 0a has an extension Oc, + +1

-+ l with O.+ l (Xp) = xp, ,B < a + 1. Oa+ l is an algebra isomorphism.

PROOF. Construct 4 and 0 as in Lemma 5.5. Y is an extension of 0 to the polynomials in (za+1 with Y(X,) = x, < a. Let -0: CiS. be a maximal extension of 0 with 7rO = i to a subalgebra X of G,+ 1 .,Let P, Q E Cy with PQ'1 E Ca+ and let a E?X 1with 7r(a) = P(PQ-1). Then 0(P) -0(Q)a E ker 1r = I<,+ 1 so, by Corollary 5.6, there is j E I. with 0 (Q) j = 0 (P) - 0(Q)a. Thus b = a + j satisfies 0(P) = O(Q)b and, since 0(Q) is not a zero divisor (Corollary 5.7), is the only element of l 1 which does. Defining O(PQ- 1) = b we extend 0 to ( a+ 1 n A where R is the subfield of Ra + l generated by (E.

Since 0 is already maximal, (i = fy c+l n R. If A E (ic+ 1 5 ?a+ 1 is transcendental over R then choose a e 211 with



54 B. E. JOHNSON

ira = iA and extend 0 to the subalgebra of (+ generated by (E and A by

defining 0 (A) = a. As 0 is already maximal we see that every element of i is algebraic over ?.

Let A E 15 . Then A is algebraic over R . Let P be the minimum polynomial of A over P. The degree of P is at least 2. By the remark at the end of Lemma 5.2, by multiplying P by a suitable monomial element of P.+ , and hence in P, we can assume that the coefficients of P are in (E and at least one is regular in 5 . Let Q be a polynomial over P with Q(A) = 0. Then Q = PR for some polynomial R with coefficients from !. If the coefficients of Q are in 5 then, by considering the coefficient of A'+' in Q where A' is the lowest degree term in P with regular coefficient and Ai is the lowest degree term in R with coefficient

ri such that rklrl E 9 for all coefficients rk of R (some r1 have this property, e.g. the coefficient of maximum absolute value), we see that the coefficients in R are all in (. Thus if we can find a' E l1 with 7ra' = tA, (OP)(a') = 0, where 0 (V,7- Ai Y) = n (0 (Ai) Y', then we can extend 0 to the algebra generated

by (E and A by defining 0 (A) = a'. This contradicts the maximality of 0 and shows $ = 0 is one to one by Corollary 5.6.

Let a C 2t1 with ira = 4A. We have 7r(OP)(a) = gP(A) = 0 so (OP)(a) C

I+ I As P(A) = 0 is the polynomial of minimal positive degree satisfied by A, P'(A) # 0. Thus P'(A) = MB where M is a monomial in ( ,+ , and hence in i,

and B is regular in (,+ 1 (Lemma 5.2). There exists b E 21 with (OP') (a) =

0 (M)b, irb = iB because if bo E 911 has irbo = iB then (OP')(a) - O(M)bo C ker r = Ia+ I so we put b = bo + i where j C I+ 1 satisfies 0 (M)j

(0 P')(a) -0 (M) bo (Corollary 5.6). b is invertible because irb = iB is invertible in U 1/I+ I and I.+ I C U = Rad I1. Again using Corollary 5.6 let k CIoa+I

with 0 (M)3k = (OP)(a). We have, by Taylor's theorem for polynomials, (OP) (a + 0 (M)2j) = 0 (M)3[k+bj+ 2 Cnin] wherek Ib1 C Em=

Rad 21', b is invertible, c2, . .. . cm W. By [6, Lemma 3.28] there is! e Ct with k + bi + Ym-2 c, jn = 0. As Et is the set of singular elements of W, b G

u and m-2 C in - 1 E U we see that b + Em-2 Cn in - = c is regular in 2t1

so =-c- 1lk EC Io,+ . By Lemma 5.4, 0(M)2j e Io+ I so if a' = a + 0(M)2j then 7ra' = A and (OP)(a') = 0.

Lemma 5.7 provides the inductive step to extend from ( - to ( +. Since

(20 = C, Lemma 5.7, together with ordinary induction, shows that (a can be normed for finite ordinals a. To complete the transfinite induction we need analogues of Lemmas 5.5 and 5.7 which deal with limit ordinals.

LEMMA 5.8. Let a be a countable ordinal and for all ,B < at let 09 be an al-

gebra isomorphism (L - o U 1 with 03(X7) =x-, 0g l I - = 0 for y< <j. Then




there is an isomorphism 4: (fi ot )t 11/4 with 4 'I 9, = nr0 for 3 < oz.

PROOF. If a is a successor ordinal then this is Lemma 5.5. If el is a limit ordinal let go < 1? <,B2 - - - be a sequence of ordinals less than a and such that if 3 < a then 1 < ,3 for some i. Let A E $a . We show that there is exactly one element w of WA 1/ with

W - 7TO+ (Paepn+1 A) AE x 1/IO ,

for all n. To do this we define elements Yn of 2t inductively so that

Yo= 00,po(PljoA))

Yn + Vn - Yn +1XnI < 2 nllxollV1 * l

IIx1 11-1,

where vn E t1 such that

00 3On - 1 On0n + 1 (PctOn + 1A)an(C! )

On + 1 (PcOn + 1 A - pXn A).

To see that vn exists consider A'= n +Pc *nA - pan A E n +I. We have

A' = MB where B is regular in , and M is a monomial in by Lemma

5.2. As A = 0 and P l B is regular in 9, we see M = O. Thus if M = X7 * * * X7kXs 1l* * * X61, in reduced form (that is none of the

Sj is a yi) with y1 y 2>* k and 51 >1 6 >1 * * >1 s then On+ 1 > 5 1 >

On and y1 > 8 1. By Lemma 5.3.

xyk p(BxI x+na? Xa Xy 1 'yk 0On+ 1(B) EW x1 C X6 1 n O * Xn-1 "6 x * I1

and hence for some vn E -

X6 ** X6I0On+1(A') =x1 x7'k 0nA(B)

0 Xp1 X61 **X6 Vw,

and so, since the x6 are not zero divisors, vn has the required property. Put

00

zo = (yn + Vn Yn+IXOn)Xo X)nX-p

00

Zm = (Jyn +m + Vn +m -Yn +m + 1 Xn +m)XOm + 1 pn +m-1 n=1



56 B. E. JOHNSON

Then zo = m+1 (Pom + 1 A) + x-o ..

xpm(Zm-Ym + 1i) so w = 7rzo is as

required. The remainder of the proof is the same as that of 5.5.

LEMMA 5.9. Let a be a countable ordinal and for all (3 < a let 09 be an algebra isomorphism (E - t1 with 0(X7) = x) and 0p I% = O fory <f,B Then there is an isomorphism Oa,: 5 Xx 2Il with 0 Ol I =% p for all B < a.

PROOF. We take fi -U,< (p< , 0: (i R'1 the map defined by I S,:

p and 4 is given by Lemma 5.8. The proof now is almost identical with that of Lemma 5.7.

THEOREM 5.10. There is an algebra isomorphism 0; i -- X11.

PROOF. Lemmas 5.7 and 5.9 show, by transfinite induction, the existence of a system {0a; a < co, } of algebraic isomorphisms O.; (a I1 with 0 a -0 for ,B < oz. As = U<r ( we can define 0 by 0 Ia = 0 afor all

a < co. 0 is then the required map.

COROLLARY 5.11. There is an algebra norm on '.

6. Embedding algebras of fornal power series in i O. In this section we show how 1 can be embedded faithfully in j the real algebra obtained by adjoining a unit to 30 (= jo(V) for some ultrafilter V on N). The process is similar to Allan's method for embedding in Zt1 [1] in that 11 is first embedded in a quotient algebra and this embedding is later lifted to an embedding in I O. This result follows from the isomorphisms constructed in Corollaries 4.8 and 4.11 if we assume the continuum hypothesis. The construction in this section does not depend on the continuum hypothesis and can be extended in a similar way to the extensions in ?5 of the results of Allan's paper.

LEMMA 6.1. Let x Ej0,x > ?,I= n 1 xn i . Then there is an isomorphism >': I - O/I with 4(X) = xrx where ir is the quotient map j i /L

PROOF. Put t = {h, h E1i , for all n E Z+ there is a polynomial Pn with real coefficients such that h - Pn(x) E x n j } is a subalgebra of i 0 and Pn + 1 (x) - Pn(x) E xn jo. Because j 0 C *RO, if Q is a nonzero polynomial of degree S n and a C i 3 then IQ(x)l > xna. Thus Pn and Pn + ,1 have the same terms of degree < n.

For h C e let #(h) = 2m=O amXm where Pn = E2o amXm for all n. t

is an algebra homomorphism 6 - ??l. We now show t is surjective. Let a =

z aiX' E ?,. Consider the sets



NORMING C(w) AND RELATED ALGEBRAS 57

A = Xn+I + ? anX';n E Z+[ B = xn+l + E anXn;n cEZ+ i=o i=o

in i 0. As jO is an l-set (the proof of this is the same as the proof of [4, Lem- ma 13.7], and A < B, with A, B countable, there is v E jG with A < v < B. Put Ym = x- m (v - %0 ai xi) E *R. Then-x < Ym < x so Ym E iand v E .

Clearly t(v) = a and t(x) = X. The kernel of t is I and we take 4 to be the inverse of the bijection t41/I

- 1 induced by t .

THEOREM 6.2. Let x E Io, x > 0. There is an algebra isomorphism 0; 8

ji 'O with 0(X)=x.

PROOF. Define 4 as in the lemma and let 0 be the homomorphism from the unital subalgebra of 01 containing X into i O with 0(X) = x. Let 0 be a maximal extension of 0 satisfying irO = . Denote the domain of 0 by N. The proof now follows that of 5.7 up to the point at which we seek a solution i E I to the equation k + bj + m2 cn jn = 0 where k E I, b is regular in io and C, .. 9C c E io1

To show that such a solution exists let b = XI + b', b' E )0 and choose preimages {Kp}, { Ip}, {ynp } of k, b', cn respectively in CR, the space of convergent sequences of real numbers. As k = xk' for some k' E iI we can assume

{Kp} Eco, we can also choose {I p} eco. PutRp = max(p-1, 3IK IIXXI1). Since X $ O, Rp is well defined and Rp - O as p oo. For each n, {ynp} is

a bounded sequence so there is K E R+ with mn=2 I'YnpI < K for all p. If we choose p so large that Rp < 1I lBpI < IX1/3,Rp K- 1XI/3 we have, for any complex number z with I z I = Rp

m

Kp + (XI + p)Z + E ynpZn -Z

n=2

<LXIIR p+ IXIRP +(E Ynpl R2<IXzI. 3 p3\n2 np/p

This shows that the variation of Arg(Kp + (I + t)Z + y 7npzn) as z de- scribes the circle rp centre 0 and radius Rp once in the positive direction is the same as the variation of Arg Xz, that is 27T. Hence for large values of p, Kp +

(X + P3)z + n? 'YnpZn = 0 has exactly one solution z inside rF. As rp is np2 p p p symmetric about the real axis and the coefficients are real, zp is real. If j is the element of j0 corresponding to {zp} E co then i satisfies the polynomial equation and j E I because j=-(b + m=2 Cn ijn 1 )- k with k E I, the inverse existing because b is invertible and ZmU2 en InE1 EO.



58 B. E. JOHNSON

REFERENCES

1. G. R. Allan, Embedding the algebra of formal power series in a Banach algebra, Proc. London Math. Soc. (3) 25 (1972), 329-340. MR 46 #4201.

2. W. G. Bade and P. C. Curtis,Jr., Homomorphisms of commutative Banach algebras, Amer. J. Math. 82 (1960), 589-608. MR 22 #8354.

3. W. F. Donoghue,Jr., The lattice of invariant subspaces of a completely continuous quasi-nilpotent transformation, Pacific J. Math. 7 (1957), 1031-1035. MR 19, 1066.

4. L. Gillman and M. Jerison, Rings of continuous functions, University Ser. in Higher Math., Van Nostrand, Princeton, N.J., 1960. MR 22 #6994.

5. L. Gillman, General topology and its relations to modern analysis and algebra. II, Proc. Second Prague Topological Sympos., 1966.

6. L. H6rmander, An introduction to complex analysis in several variables, Van Nos- trand, Princeton, N.J., 1966. MR 34 #2933.

7. N. Jacobson, Lectures in abstract algebra. Vol. III: Theory of fields and Galois theory, Van Nostrand, Princeton, N.J., 1964. MR 30 #3087.

8. B. E. Johnson and A. M. Sinclair, Continuity of linear operators commuting with continuous linear operators. II, Trans. Amer. Math. Soc. 146 (1969), 533-540. MR 40 #4791.

9. A. M. Sinclair, Homomorphisms from C*-algebras, Proc. London Math. Soc. (3) 29 (1974), 435-452.

SCHOOL OF MATHEMATICS, THE UNIVERSITY, NEWCASTLE-UPON-TYNE, ENGLAND



norming c(Ω) and related algebras

Documents