On the unit-1-stable rank of rings of analytic functions

Publicacions Matemátiques, Vol 36 (1992), 439-447 . ON THE UNIT-1-STABLE RANK OF RINGS OF ANALYTIC FUNCTIONS Abstract JOAN JOSEP CARMONA, JULIÁ CUFÍ ANDPERE MENAL In this paper we prove a general result for the ring H(U) of the analytic functions on an open set U in the complex plane which implies that H(U) has not unit-1-stable rank and that has some other interesting consequences . We prove also that in H(U) there is no totally reducible elements different from the zero function . Introduction Let A be a commutative ring with unity . A pair of elements (al, a2) CA 2 is said to be unimodular if there exists (b l , b2) E A 2 such that a l b, + a2b2 = 1 . We will denote by U2(A) the set of all unimodular pairs and by Uj(A) =A -1 the set of inversible elements of A . One says that the unimodular pair (al, a2) is reducible if it is possible to find x E A such that al + xa2 E A -1 . The ring A is said to have stable rank 1if each unimodular pair in A is reducible in A . This is a special case of the concept of stable rank n introduced by Bass [1] . This notion has been useful in treating some problems in K-theory . Moreover Vasershtein [12] has calculated the stable rank of rings of continuous functions and rings of differentiable functions in Rn and related it to the topological dimension of the domain space . Concerning to rings of holomorphic functions P . Jones, D . Marshall and T . Wolff [3] proved that the disc algebra has stable rank l . Previously L . A . Rubel [7] had observed that the same is true for the ring H(U) of holomorphic functions'on the open set U CC . Different proves of these results can be found in the paper of G . Corach and F . Suárez The first two authors dedicate this paper to the memory of Pere Menal, who introduced them to the subject of this paper during the last period of his life . Partially supported by DGICYT PB89-0311 and DGICYT PB89-0296 . 44 0  J . J . CARMONA, J . CUFÍ, P . MENAL [2] , where the case of rings of holomorphic functions of several complex variables is also considered . The more dificult problem to decide if the algebra H'(D) of bounded analytic functions in the unit disc D, has stable rank 1 has recently been answered positively by S . Treil [10] . In [4] P . Menal and J . Moncasi introduced the concept of unit-1-stable rank . A unimodular pair (al, a2) EU2 (A) is said to be totally reducible if there exists an element uE A -1 such that al +ua2 E A -1 . The ring A is said to have unit-1-stable rank if each unimodular pair in A is totally reducible . In [7] L . A . Rubel proved that H(C) has not the unit-1-stable rank property . The question to decide if the disc algebra has unit-l-stable rank arose and was studied by R . Mortini and R . Rupp and by ourselves . Mortini and Rupp,communicated to us the negative answer to this question and using some ideas of their proof we obtain a more general result that has some other interesting consequences . This is the content of the first part of the present paper . The result of Mortini and Rupp appears in [5] where the question to characterize the totally reducible elements of the disc algebra is also considered . They find a sufficient condition for an element to be totally reducible in this algebra . Given a ring A one says that an element a E A is totally reducible if for each b E A such that the pair (a, b) E U2 (A) then (a, b) is a totally reducible pair . In the second part of this paper we consider the totally reducible elements of the ring H(U) . In this case the situation is completely different from the dise algebra because we show that the zero function is the only . totally reducible element of H(U) . We are able to obtain these kind of algebraic properties of rings of analytic functions by using deep theorems of the function theory of one complex variable . Unit-1-stable rank Rom now on we deal with the ring H(U) . If a E H(U) \ {0}, then we denote by Z a , the discret closed set in U of the zeros a . Each zero is considered with the corresponding multiplicity . So when we write Z a = Zb we mean that a and b have the same zeros with the same multiplicity . We recall that a pair (a, b) with a, b E H (U) is unimodular if only if Z a fl Zb = 0 . We can prove the following general result . UNIT-1-STABLE RANK  44 1 Theorem 1 . Assume that (an), (bn), (en), (d n ,) ments of H(U) satisfying i) anbn + cndn = 1 for all n >_ 1 and bn, dn are invertible elements of H(U) . ii) The sequences (a n ), (c n ) are uniformly convergent on compact subsets of U to a, c E H(U) \ {0} respectively . Then either Z a n Z c = 0or Z a = Z, are sequences of eleProof : First we prove that (a nbn ) is anormal sequence in D \ Zaa . Fix a point zo E D \ Zae . Then there exists a closed disc A C U \ Zac with center zo and a number S > 0 such that lan(z)cn(z)1 >_ 5, for z E A, n > v , v large enough . Consider the sequence given by (a nb n ) if n >_ v . We will prove that this sequence is normal in the classical sense in A . Since bn is invertible and an has no zeros in A we see that anbn never takes the value 0, n > v . If an(z)bn(z) = 1 for some z E A, it follows from i) that cndn(z) = 0 . But dn is invertible and cn(z) :y~ 0 if n >_ v and this is a contradiction . By Montel's Theorem [8, p . 350] , (a,,,bn)n>  is a normal sequence in 0 and so is (anbn)n>1 . Since zo was an arbitrary point of the open set U \ Zae, it follows from [8, p . 51] that (anbn)n>1 is normal in U \ Zac . Assume Z a n Z, z,~ 0and let us fix a point a E Z a n Z, Take a closed disc O1 C U with center a and such that (O1 \ {a}) n Zac = 0 . The result will follow if we prove that any ~3 E Zac is a common zero of a and c with the same multiplicity in a than in c . For such a (~ let 02 be a closed disc with center ,P and such that (02 \ {,3}) n Zac = 0 . Let K = 01 U~2 . By ii) (cn) converges uniformly to c in K and (1/c,) converges uniformly on 8K . Since d, . = (1 - an b n )/c n we know that a partial sequence of (d,,) is either uniformly convergent or uniformly divergent on 8K . In the first case this partial sequence (dn) is uniformly bounded in 801 and, by the maximum modulus principle, also in O 1 . Then a partial sequence of (dn(a)) is convergent and the corresponding partial of ((cndn)(a)) tends to 0 . Since b,, = (1 - c,,,dn)/an and on 8K we get that same partial sequence of (bn) is uniformly a a bounded on K . Therefore (an(a)bn(a)) tends to 0 and this together with the fact that (cn(a)dn(a)) tends to 0 contradicts i) . Therefore (dn) is uniformly divergent on 8K . Since d n is invertible, by the minimum modulus principle we get that (dn) is uniformly divergent on K . Since dñ 1 = anbndn 1 + c n we get that (-a n bn dn 1 ) converges uniformly to c in K . Also (an) tends to a uniformly on K and since Za, n 80 2 = 0, by Hurwitz's Theorem [8, p . 158] the function a has the same zeros than c in 02 . 44 2  J . J . CARMONA, J . CUFÍ, P . MENAL The following simple result shows that the hypothesis about the invertibility of b,, and d n cannot be weakned . Proposition 1 . Let a, c E H(U) .  Then there exist sequences (an), (b n ), (cn), (d n ) such that (a n ) and (cn) are almost uniformly convergent to a and c respectively such that a n b n +c nd n = 1 and b n is invertible for all n > 1 . Proof .. For each \ > 0 let us consider the set AA= {z E Uja(z) + = 0} n Z, . Since >, 0Y implies AA n Ay= 0, the set of A such that Aa :,A 0 is at most countable . Therefore we can choose a sequence ( , \n) of positive numbers that converges to 0 such that Aa n = 0 . Put an(z) = a (z)+A n and c n = c, n >_ 1 . Cleary the sequences (a n ) and (c n ) converge uniformly to a and c and Z Qn n Z, is empty for all n _> 1 . Since H(U) has stable rank 1 this implies that there are b n invertible and d  such that anbn + c nd n = 1 . Proposition 1 says in particular that the topological stable rank of H(U), in the sense of Reiffel [6], is 2 . From the Theorem 1 we can deduce the result of Mortini and Rupp Corollary 1 . Let f be a nonzero element of H(U) . Then f has some zero in U if and only if there is a positive integer n such that the un¡- modular pair (f, 1 - nf 2 ) is not totally reducible in H(U) . Proof . Since (nf) f+(1-nf 2 ) = 1 it is clear that (f, l-nf 2 ) is totally reducible when f has no zeros . Conversely, assume there exist un E H(U) -1 such that v n = fu, + 1/n - f 2 E H(U) -1 for all n >_ 1 . Then 1 = fu n vñ 1 + (1/n - f 2)vñ 1 . It follows from Theorem 1 that f has no zeros . Corollary 2 . Let A be a subring of H(U) . If f E A is totally reducible in A, then f has no zeros in U or f is identically 0 . Let co : A - B be a ring homomorphism . We say that cW has stable rank 1 provided that for any x, yE A with xA+yA= A there exists c E S with cp(x) +W(y)c E B -1 . If c can be chosen to be invertible in B, then we say'that cp has unit-1-stable rank . Corollary 3 . Let U be an open set of C and let A be a ring . If ep : A -~ H(U) is a ring homomorphism with unit-1-stable rank, then W(A) C C . UNIT-1-STABLE RANK  443 Proof .. Corollary 1 implies that cp(a) E H(U) -1 when a E A, a 7¿ 0 . Let a :,1 : 0 and assume that W(a) is not constant . Then W(a)(U) contains an algebraic numberand so there is a nonzero polynomial P E Z[t] such that P(W(a)) has some zero in U . Since P(W(a)) = ~p(P(a)) we conclude that P(a) = 0 and so P(cp(a)) = 0 . This shows that W(a) takes only finitely many values and so it must be constant which is a contradiction . Corollary 4 . Let EC C[z] be the set of all polynomials without zeros in the closed unit disc . Then the ring R = C[z]r has stable rank 1 but not unit-1-stable rank . Proof .- Clearly we can view R as a subring of the disc algebra A(D) . If (a, b) is a unimodular pair in R , there exists an element f E A(D) such that a + f b is invertible [2] [3] . Since f can be approximated uniformly by polinomials we can assume that f itself is a polynomial . Then a + b fE R and R has stable rank 1 . It follows from Corollary 3 that R has not unit-1-stable rank . Another applications of Theorem 1 are some results that guarantee the existente of a fixed disc contained in the image of the unit disc for each element of some classes of functions . In this line we recall the classical results of Bloch [11, p . 262], Koebe [9, p . 197] and also the interesting one referred in [9, p . 502] . Here we consider the class of functions fg where f is a fixed function and g is a holomorphic function without fixed points in an open set and also the class of all the funcions f .g where fE S and g is as before . We write S for the set of all f E H(D) such that f is one to one and f (0) = 0, f'(0) = 1 . Proposition 2 .  There exists a universal constant r > 0 such that the dise D(0, r) is contained in the image of every function f g, where fE S and g E H(D) has no fixed points in D . Proof .. Assume our conclusion is false . Then for each positive integer n, there exist zn with 1zn1 < ñ, f n E S and 9n without fixed points such that fn9n - zn E H(D) -1 . Write 9n (z) = z - hn(z), where h n E H(D) -1 . We obtain 1 = fnun + (zfn - z n)vn I '  wlth  un, vn E HA-1But S is anormal class [9, p . 200] , so there exits a partial sequence of (fn) uniformly convergent to some f, f E S . Applying Theorem 1 we conclude that Zf = Z,f , a contradiction . 444  J . J . CARMONA, J . CUFÍ, P . MENAL Proposition 3 . Let U be an open set with 0 E U and let f be a function that is neither invertible nor zero . Then there exiss a constant r = r(f) > 0 such that the disc D(0, r) is contained in the image of every function fg , where gE H(U) has no fixed points in U . Proof . If the statement is not true, then for each n there would exist z,, 1, with Izn1 < ñ and gn without fixed points such that f .gn - z n E H(U) -1 . Proceeding as before we obtain 1 = fun + (z f - zn)vñ 1 ,  with  u n , vn E H(U) -1 . Now we aply Theorem 1 and the conclusion follows . The constant that appears in Proposition 2 is less or iqual than 1, by Koebe's Theorem . Considering the functions f (z) = z, g(z) = ez -1 one can see that r <_ é . It would be interesting to find a constructive proof of Proposition 2 and the best value of r . Totally reducible elements We are going now to consider the totally reducible elements of the ring of analytic functions in an open set . Let A be a Banach algebra . For this special case every element u E A -1 is totally reducible . In fact , for each f E A, the pair (u, f) is unimodular and u + e f E A -1 if e < 11fu 1 11 ' For the disc algebra A(D) formed by the functions which are continuous on D and holomorphic in D, Mortini and Rupp proved that each outer function in A(D) is totally reducible [5] . The situation is completely different for the ring H(U) as the following theorem shows . Theorem 2 . Let U be an open set in C . A function f E H(U) is totally reducible in H(U) if and only if f is the zero function in U . For the proof we consider separately the two different cases U= C and U 7~ C . Proof for the first case : Let f be totally reducible in H(C) . We know, using Corollary 2, that f = eh with h an entire function . If the un¡- modular pair (e h , z) would be totally reducible then there would exist k, l E H(C) such that ehe k +ze l = 1 . ÜNIT-1-STABLE RANK  44 5 The function ze l never takes the value 1 and takes the value 0 only once at the origen . By the great Picard Theorem [8, p . 353] ze l must be a polynomial and this contradicts the fact that it never takes the value 1 . To prove the Theorem 2 for U z,~= C first of all we remark that the existente of some fE H(U), f :,A 0 totally reducible implies that each function in H(U) -1 is also totally reducible and this existente is equivalent to an interpolation problem as the following lemma shows . Lemma . Let U be an open set of C . Then the following are equivalent . i) There exists a function fE H(U) -1 which is totally reducible in H(U) . ii) For each closed and discrete set {zn}, counting every z n with some multiplicity, there exists a function h E H(U) whose zeros are {z n }, with the corresponding multiplicity, and such that h never takes the value 1 in U . iii) Each function g E H(U) -1 is totally reducible in H(U) . Proof of the lemma : i) =~> ii) Let fE H(U) -1 be totally reducible . Given the set {zn} take k E H(U) with {zn} as its zero set . By i) there are a, b E H(U) -1 such that af + bk = 1 . So a = 11 fbk and h = bk satisfies the requirements of ii) . ii) ==> iii) Let g E H(U) -1 and let l E H(U) be arbitrary . The pair (g, l) is unimodular . Let {z,,} be the zero set of l . By ii) there is some hE H(U) that takes the value 1 on {z n } and h(z) :7É 0 . Now the functions b = 11 f hh E H(U) -l and a = s E H (U) -1 verify ag + bl = 1 . We need the following results [11, p . 215-204] . Theorem (Ahlfors) . Let fE H(D) and assume that lim sup  T (r)  = +oo . r-1  log 11 1 r Then f takes any value infinitely often in D with one possible exception . Here T(r) is the characteristic function of Nevalinna [11, p . 196] . Theorem . Let fE H(D) and let {zn} be its zero set . Assume that r-°° 1 ( 1I zn I )1+P = + oo, for all p > 0 . Then lim sup,1 iof = + oo . 44 6  J . J . CARMONA, J . CUFÍ, P . MENAL Proof for the second case : Assume first that U=D . Let (zñ) be a sequence such that lim n ~~ zñ = 1 and r_,' 1 (1 - jzñ¡)'+P = + oo for all p > 0 . Let {z  ,} be the sequence of the same points but doubling their multiplicity . We show that ii) of the lemma is not satisfied for {zn} . Let h E H(D) be any function that vanishes on {zn} . We have h = hó for some ho E H(D) vanishing at {zñ} . By Ahlfors Theorem ho may omit only one value and so its square h takes any complex value infinitely often . For a general open set U let A be any disc A C U such that there is a point a E ao n aU . l t suffices to take a sequence {zn}, z n E A as before with lim e -,,, zñ = a . For this sequence, condition ii) of the lemma is not satisfied in U, since it is not satisfied on A . References 1 . H . BASS, K-theory and stable algebra, Publi . Math IHES 22 (1964), 5---60 . 2 .  G . CORACH AND F . DANIEL SUÁREZ, Stable rank in holomorphic functions algebras, Illinois J . of Math . 29 (1985), 627-639 . 3 .  P . W . DONES, D . MARSHALL AND T . WOLFF, Stable rank of the disc algebra, Proc . Amer . Math . Soc . 96 (1986), 603-604 . 4 .  P . MENAL AND J . MONCASI, K1 of Von Neumann regular rings, Journal of Pure and Applied Algebra 33 (1984), 295-312 . 5 .  R . MORTINI AND R . RUPp, Totally reducible elements in rings of analytic functions, Comm . i n Algebra (to appear) . 6 .  M . A . RIEFFEL, Dimension and stable rank in the K-theory of C*-algebras, Proc . London Math . Soc . 46 (1983), 303--333 . 7 .  L . A . RUSEL, Linear compositions of two entire functions, Amer . Math . Monthly 85 (1978), 505-506 . 8 .  S . SAKS AND A . ZYGMUND, "Analytic functions," Elsevier Pub . Company, 1971 . 9 .  G . SANSONE AND J . GERRETSEN, "Lectures on the theory of functions of a complex variable," Wolters Noordhoff, 1969 . 10 . S . TREIL, The stable rank of the algebra H°° equals 1, Preprint Leningrad University (1991) . 11 . M . Tsu,Ii, "Potential theory in modere function theory," Chelsea Publishing Company, New York, 1975 . UNIT-1-STABLE RANK  44 7 12 .  L . N . VASERSTEIN, Stable rank of rings and dimensionality of topological spaces, Functional Anal . Appl . 5 (1971), 102-110 . Departament de Matemátiques Universitat Autónoma de Barcelona 08193 Bellaterra (Barcelona) SPAIN Rebut el 2de MarC de 1992