ring of entire functions

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The ring of entire functions

Jonas Bjermo

U.U.D.M. Project Report 2004:15

Examensarbete i matematik, 20 pong

Handledare och examinator: Karl-Heinz Fieseler

Juni 2004

Department of Mathematics

Uppsala University


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The ring of entire functions.

Examensarbete

Matematiska Institutionen

Uppsala Universitet

Jonas Bjermo

August 19, 2004

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Acknowledgment

I am most grateful to my supervisor Karl-Heinz Fieseler for presenting theidea of this thesis and for answering all my questions about this thesis andrelated subjects.

1 Introduction

In 1940, O. Helmer [5] proved that, in the ring of all entire functions, everyfinitely generated ideal is a principal ideal. In this text we show, based on [6,

chapter 6.3], that this is also valid in rings of functions holomorphic in arbi-trary open connected subsets G C denoted O(G). To prove this, severaltools from complex function theory are used, such as the concept of greatestcommon divisors which depends on Weierstra products, and Mittag-Lefflerseries which help us to prove the important lemma of Wedderburn.

Then, as in [3] and [4], we consider the ideals in the ring O(G) withthe difference that we use the concept of divisors to describe them. Weintroduce so called filters and deal with them instead of the distributionsof zeros for the functions in the ideal. Both maximal and prime ideals areconsidered. We will see that the factor ring O(G)/m, where m is maximalis isomorphic to C both when the functions in m have at least one common

zero and when they have no common zeros.

2 Holomorphic and meromorphic functions

2.1 Holomorphic functions

We will start to recall some results from complex analysis. A domain is anon-empty, open subset of C. From now on it will always be denoted D. Aconnected domain is called a region and will be denoted G in the rest of the

text.

Definition A function f: D C is called holomorphic in the domain Dif f is complex-differentiable at every point of D; f is called holomorphicat c D if there is an open neighborhood U of c lying in D such that therestriction f|U of f to U is complex differentiable in U.

Remark The set of all points at which a function is holomorphic, is al-ways open in C. A function which is holomorphic at c is always complex-differentiable at c, but the converse do not need to be true.

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Theorem 2.2 (Identity Theorem [7 p. 228]) The following statements about

a pair f, g of holomorphic functions in a region G C are equivalent:

1. f = g.

2. The coincidence set {w G|f(w) = g(w)} has a cluster point in G.

3. There is a point c G such that f(n)(c) = g(n)(c) for all n N.

Proof 1) 2) is trivial.2) 3) Set h := f g O(G). Then by the hypothesis the zero-

set M := {w G|h(w) = 0} has a cluster point c G. If there is anm N with h(m)(c) = 0 then we consider the smallest such m. From

above we see that we have the factorization h(z) = (z c)m

hm(z) withhm(z) =

m

h()(c)! (z c)

m O(B) for all open balls B centered atc in G and hm(c) = 0. Because of its continuity hm is zero free in someneighborhood U B of c. Then M (U \ {c}) = , that is c is not acluster point of M. This contradiction shows that there is no such m, thatis h(n)(c) = 0 for all n N, hence f(n) = g(n) for all n N.

3) 1) Set h := f g, and let Sk = Z(h(k). Each Sk is relatively closed

in G because h(k) O(G) is continuous. Then S := 0 Sk is relativelyclosed in G. However, if z1 S then the Taylor series of h in any openB C centered at z1 equals the zero series. Hence h

(k)|B = 0 for everyk N, that is B S. This implies that S is open in G. Since G is a region

(hence connected), G is the only non-empty subset of G which is both openand closed. Hence S = G, and therefore f = g.

A consequence of the Identity Theorem is

Theorem 2.3 The domainD C is connected if and only if the ring O(D)is an integral domain.

Proof ) Suppose f, g O(D), f is not the zero function but f g is, thatis f(z)g(z) = 0 for all z D. There is some c D where f(c) = 0 anda neighborhood U D of c in which f is zero-free. Then g(U) = {0} andsince D is a region (hence connected) this means that g(D) = {0} by theIdentity Theorem. That is, g = 0, the zero element of the ring O(D).

) If D is not a region then it would not be connected. It would bepossible to express D as the disjoint union of the non-empty open sets D1and D2. Let us define f, g in D by

f(z) := 0 for z D1 and f(z) := 1 for z D2,g(z) := 1 for z D1 and g(z) := 0 for z D2.These functions are holomorphic in D, and neither is the zero function

in O(D) but f g is. This contradicts the hypothesis that O(D) is an integraldomain.

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2.2 Meromorphic functions

The Taylor expansion does not apply to functions that fail to be holomorphicat some points. For such functions we have another expansion called theLaurent expansion. Let r1 0, r2 > r1, and c C, and consider the regionA = {z C: r1 < |z c| < r2}. Let us recall from complex analysis that if afunction f is holomorphic in A, it can be written f(z) =

n=0 an(z c)

n +n=1

bn(zc)n where both series on the right side of the equation converge

absolutely on A and uniformly on any set of the form B1,2 = {z C: 1 |z c| 2} where r1 < 1 < 2 < r2. The series is called the Laurentseries or the Laurent expansion around c in A. The series n=0 an(z c)n,respectively,

n=1 b

n(zc)n is called its regular, respectively, principal part.

When r1 = 0 in the Laurent expansion we have a special case. In thiscase, f is holomorphic in {z : 0 < |z c| < r2}, and we say that c is anisolated singularity of f.

Definition Let f be holomorphic in a domain D except for one point c D,i.e., f is holomorphic in D \ {c}, then c is called an isolated singularity of f.

As seen above, if f is holomorphic in {z : 0 < |z c| < r2}, then we canexpand f in a Laurent series: f(z) = ...+ bn(zc)n + ...+

b1(zc) +ao +a1(z c)+

a2(z c)2 + ... valid for 0 < |z c| < r2. There are three different types of

isolated singularities depending on the number of coefficients bn = 0. If allthe coefficients bn = 0 is zero then we say that c is a removable singularity,and if the number of bn are infinite c is called an essential singularity. Weare interested in the case when all but a finite number of the coefficients bnis zero.

Definition If c is an isolated singularity of f and if all but a finite numberof the bn in the Laurent expansion are zero, then c is called a pole of f. Thehighest integer k such that bk = 0, is called the order of the pole.

Definition A meromorphic function on D is a pair (f, P), where P D isdiscrete and f: D \ P C is holomorphic with a pole at each point of P.

The set P is called the pole-set of f. Usually we write f instead of (f, P)and P = P(f).

The set of all function meromorphic in a domain D is denoted M(D).

Remark Because the pole-set is allowed to be empty, the functions holo-morphic in D are also meromorphic in D.

The pole-set of each function f = 0 meromorphic in D is, just as for thezero-set of a holomorphic function, a discrete and relatively closed subset of

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D. Then it follows that the pole-set of each function meromorphic in D is

either empty, finite, or countably infinite [7 p. 316].From complex analysis it is known that poles (locally) arise via the

formation of reciprocals of holomorphic functions.

Proposition 2.4 Letf O(D\{c}) and letc D be an isolated singularityof f, then c is a pole of order m if and only if there exists a function g O(D), g(c) = 0, such that f(z) = g(z)(zc)m for z D \ {c}.

Proof ) By definition c is a pole iff f(z) = bk(zc)k

+ bk1(zc)k1

+ ... +b1

(zc) +

n=0 an(z c)n = 1

(zc)k(bk + bk1(z c) + ... + b1(z c)

k1 +

n=0 an(z c)

n+k), where bk = 0. This is valid in D \ {c}. Let g(z) =

bk + bk1(z c) + ... + b1(z c)k1

+

n=0 an(z c)n+k

, this implies thatg O(D) and g(c) = bk = 0.

) Let g O(D) where g(z) =

n=0 an(z c)n and set f = g(z)(zc)n .

Then f =

n=0 an(z c)nm, and by definition, it has a pole at c of order

m 1.

Theorem 2.5 Letf O(D \ {c}), thenf has a zero of order m at c if and

only if 1f(z) has a pole of order m at c. If f O(D) and f(c) = 0 then

f(z)f(z)

also has a pole of order m at c.

Proof The function f O(D \ {c}) has a zero of order m iff f(z) = (z

c)

m

h(z), where h(z) is holomorphic at c with h(c) = 0. Set g(z) :=

1

h(z) ,g O(D), g = 0. Then f(z) = (z c)m 1

g(z) , by prop 2.4 this is equivalent

to that 1f(z) has a pole of order m at c.

As we already know, a function f with a pole of order m at c can bewritten f(z) =

1n=m bn(z c)

n +

n=0 an(z c)n. We can then see that to

ask when f has a pole of order m at c is the same as asking when (z c)mfis bounded.

Theorem 2.6 Let m N, m 1 and let f O(D \ {c}). Then f hasa pole of order m at c if and only if there is a neighborhood U of c lyingin D and positive finite constants M, M

such that for all z U \ {c}

M|z c|m |f(z)| M|z c|m.

Proof ) The function f has a pole of order m at c iff h := 1f

has a

zero of order m at c, which can be written in the form (z c)mh for anh O(U), U D small enough. Let M = infzU{|h(z)|

1} > 0 andM = supzU{|h(z)|

1} < . The claim now follows from |f(z)| = |z

c|m|h(z)|1.) |(z c)mf(z)| M for z U\ {c} shows that (z c)mf is bounded

near c and |(z c)m1f(z)| M|z c|1 shows that (z c)m1f is notbounded near c. That is, c is a pole of order m.

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Let f O(D). We say that f increases uniformly to around c, written

limzc f(z) = , if for every finite M there is a neighborhood U of c in Dsuch that infzU\{c} |f(z)| M

Then a consequence of theorem 2.6 is the following

Corollary 2.7 The function f O(D \ {c}) has a pole at c if and only iflimzc f(z) = .

In view of this corollary, at each pole, we choose f(z) := for z P(f).So meromorphic functions are special mappings from D C to C {}.

Iff, g M(D) with pole-sets P(f), P(g) are given, then P(f) P(g) isalso discrete and relatively closed in G. In G \ P(f) P(g) both f and g areholomorphic, and hence f+g, f g are holomorphic. For each c P(f)P(g),

f and g can be written as Laurent expansions with finite principal part in anopen neighborhood U of c, with U (P(f) P(g)) = {c}. Then both f + gand f g can be written as Laurent expansions with finite principal part. Thepoint c is then either a pole or a removable singularity of f+g and f g. Thusthe pole-set of these functions are subsets of P(f) P(g), which are discreteand relatively closed in G. This implies that f + g, f g M(D). Fromthe rules of calculating with holomorphic functions it follows that M(D)is a commutative ring with unity with respect to pointwise addition andmultiplication. The ring O(D) is a subring of M(D).

In the ring O(D), division of an element f is possible only when thevalue of f is zero-free in D. But in the ring M(D), division of functions

which have zeros in D is possible.Now we define the zero-set of a meromorphic function.

Definition Let f M(D), then Z(f) is called the zero-set of f if Z(f) isthe zero-set of the holomorphic function f|(D \ P(f)) O(D \ P(f)).

Remark Z(f) is relatively closed in G and Z(f) P(f) = .

Now we define the order function for a meromorphic function. If a func-tion f = 0 is meromorphic, then it can be developed uniquely into a Laurentseries

m a(z c)

with a C, m Z and am = 0.

Definition If f = 0, f M(D) the number m in the Laurent series above

is called the order of f at c and is denoted oc(f).

From the definition it follows directly that for an f meromorphic at c:

1. f holomorphic at c oc(f) 0

2. Ifm = oc(f) < 0, then c is a pole of f of order m.

As for holomorphic functions the product rule 2.1 is also valid for mero-morphic functions.

Now we prove that the ring of all functions meromorphic in a region isa field.

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Theorem 2.8 Letu M(D), then u is invertible, that is uv = 1 for some

v M(D), if and only if the zero-set Z(u) is discrete in D.

Proof ) For c D, uv = 1 implies that u(c) = 0 i ff v(c) = andu(c) = iffv(c) = 0. This means that Z(u) = P(v) and P(u) = Z(v), thenby the definition of a meromorphic function the zero-set Z(u) is discrete.

) The set Z(u) P(u) is discrete and relatively closed in G. Choosev := 1/u, then v is holomorphic in G \ Z(u) P(u) and have a pole at everypoint of Z(u). Every point c P(u) is a removable singularity of v becauselimzc 1/u(z) = 0. Hence v M(D)

From this theorem we see that the quotient of two elements f, g M(D)exists in the ring M(D) if Z(g) is discrete in D.

A consequence of theorem 2.8 is

Corollary 2.9 LetG C be a region, then M(G) is a field.

Proof If f M(G), f = 0 and G is a region, then G \ P(f) is a regionand f|(G \ P(f)) is a holomorphic function which is not the zero elementof O(G). Therefore Z(f) is discrete in G and then f is invertible. Hence,every element of M(G) \ {0} is invertible and therefore M(G) is a field.

Because M(G) is a field it contains no proper ideals = {0}. The ringO(G) on the other hand, has an interesting ideal structure.

3 Preparation for the ideal theory in O(G)

To prove some important result about the ideal structure in the integraldomain O(G) we make use of some important tools. These are the lemmaof Wedderburn and the concept of greatest common divisors. In the proofof Wedderburns lemma we make use of Mittag-Leffler series which will alsobe considered. We start with ideals.

3.1 Ideals

Since we are going to consider the ideal structure in the ring of functionsholomorphic in open connected subsets of C we give some basic definitions.In this subsection R is a commutative ring with unity.

Definition Let R be a ring. A subset I R, I = is called an ideal in Rif I is an additive subgroup and ax I for all a I, x R.

Now, let M = be any subset of R and L be the set of all finite linearcombinations i.e. L = {u =

ni=1 rifi : ri R, fi M}. Then L is an ideal

in R and M is said to generate L.

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Definition An ideal I in a ring R is called finitely generated if there is a

finite set that generates I. A ring R is called Noetherian if every ideal in Ris finitely generated.

If we can choose M with only one element, we have a special case.

Definition An ideal I in a ring R is called principal if it is generated by anelement f, i.e. if I = {rf : r R} for some f R. A integral domain R iscalled a principal ideal domain if every ideal is a principal ideal.

Remark A finitely generated ideal generated by M = {f1,...,fr} is usuallydenoted I = Rf1 + ... + Rfn, and a principal ideal generated by f is usuallydenoted (f) or Rf.

Definition An ideal I in a ring R is called maximal provided that, I = Rand there does not exist an ideal J = R such that I J.

Definition An ideal I = R is called a prime ideal if ab I implies thateither a I or b I for a, b R.

3.2 Convergent series and sequences of complex functions

Let X be a metric space.

Definition A sequence of functions fn: X C is called uniformly conver-gent in A X to f: A C if for every > 0 there exists an n N suchthat |fn(x) f(x)| < for all n n and all x A.

A series

f of functions converges uniformly in A if the sequence sn =n f of partial sums converges uniformly in A.Remark The limit functions f and

f are uniquely determined.

We introduce the supremum semi-norm |f|A := sup{|f(x)|: x A} for subsets A X and functions f: X C. The set V := {f: X C: |f|A < } is a C-vector space and the mapping f |f|A fulfills:

1. |f|A = 0 f|A = 0.

2. |cf|A = |c||f|A.

3. |f + g|A |f|A + |g|A. for all f, g V, c C.

We see that the sequence fn converges uniformly in A to f exactly whenlim |fn f|A = 0.

Since convergence does not need to be uniform on the whole X, weintroduce the following

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Definition A sequence of functions fn: X C is called locally uniformly

convergent in X if every point x X lies in a neighborhood Ux in whichthe sequence fn converges uniformly.

A series

f is called locally uniformly convergent in C when its asso-ciated sequence of partial sums is locally uniformly convergent in X.

Uniform convergence implies locally uniform convergence. Let A1, ...Ambe subsets of X. If the sequence fn: X C converges uniformly in each ofthese subsets, then it also converges uniformly in the union A1 ... Am.A consequence of this is

Theorem 3.1 If the sequence fn converges locally uniformly in X, then itconverges uniformly on each compact subset K of X.

Proof Every point x K has an open neighborhood Ux in which fn isuniformly convergent. The open cover {Ux: x K} of the compact set Kadmits a finite subcover, say Ux1 ,...,Uxm . Then fn converges uniformly inUx1 ... Uxm , and therefore so in the subset K of this union.

Definition A sequence or series converges compactly in X if it convergesuniformly on every compact subset of X.

So we may reformulate 3.1:

Corollary 3.2 Local uniform convergence implies compact convergence.Let X be a metric space. We call X locally compact if each of its points

has at least one compact neighborhood. Then it is easy to see the following

Theorem 3.3 If X is locally compact, then every compactly convergent se-quence or series in X is locally uniformly convergent in X.

In locally compact spaces local uniform convergence and compact con-vergence are equivalent. This is also valid in domains in C since they arelocally compact.

To guarantee that every re-arrangement of a series will converge locally

uniformly we introduce the following concept.Definition A series

f of functions f: X C is called normally conver-

gent in X if each point of x has a neighborhood U which satisfies

|f|U 1and choose a function f: = gcd{f2,...,fn}. By the induction hypothesis weknow that f = a2f2 + ... + anfn where a2, ..., an O(G). Now, becausef = gcd{f1, f} we have 1 = gcd{f1/f, f /f}. Then by Wedderburns lemma

f1/f and f /f satisfy the equation 1 = (f1/f)a + (f /f)b with functionsa, b O(G), which implies f = f1a + f b = f1a + (f2a2 + ... + fnan)b. Leta1: = a and a: = ab, 2. Hence f = a1f1 + ... + anfn.

We are now able to prove our main theorem of this section.

Theorem 4.2 (Principal ideal theorem) An ideal I O(G) is finitely gen-erated if and only if it is a principal ideal.

Proof ) We reformulate the claim as follows: IfI is generated by f1,...,fn,then I = O(G)f, where f = gcd{f1,...,fn}. Every non-zero subset in O(G)

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has a gcd, then by proposition 4.1, O(G)f I. Since f divides all of

the f1,...,fn, we see that f1, ...fn O(G)f; hence I O(G)f thereforeI = O(G)f, which means that I is a principal ideal.

) This direction is trivial.

Since f divides every f1,...,fn the functions f1,...,fn vanish where fdoes, and since f = a1f1 + ... + anfn the function f vanish at the commonpoints where f1,...,fn vanish, we have Z(f) = i=1,...,nZ(fi).

However, the next example shows that there exists ideals in O(G) thatare not finitely generated.

Example Let G be a region and let A G be infinite, discrete and closed inG. The set a := {f O(G) : f(a) = 0 for all except a finite number of a A} is an ideal in O(G). Assume that a is finitely generated, i.e. a =O(G)f1 + ... + O(G)fn. Then there exists a kj N such that for all k kj ,fj(ak) = 0, j = 1, . . ,n. Choose k0 :=max{k1,...,kn}, then for h a, h =n

j=1 gjfj we have h(ak) = 0 for all k k0. For any discrete and closedsubset T G we can construct a holomorphic function f with zero set T.Let T := A\ {a}, then f(a) = 0 and f a, but f is not a linear combinationof f1,...,fn. Contradiction!

Thus we have proved

Proposition 4.3 No ring O(G) is Noetherian, and therefore never a prin-

cipal ideal domain.

We see that every ideal in O(G) either is principal or not finitely gener-ated.

Definition An integral domain R is called a unique factorization domainif the following conditions are valid:

1. Every non-unit element p = 0 can be factored into a product of a finitenumber of irreducibles.

2. Ifp1,...,pn and q1,...,qm are two factorizations of the same element of

R into irreducibles, then n = m and the qj can be renumbered so thatfor pi and qi we have pi = qiui, where ui are units in R.

The functions (z c), c G are, up to unit factors, precisely the primesof O(G). Functions f = 0 in O(G) with infinitely many zeros in G cannotbe written as the product of finitely many primes. From the fact that suchfunctions exists, by the general Weierstra product theorem, we get

Proposition 4.4 No ring O(G) is a unique factorization domain.

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4.2 Fixed and free ideals

An element of O(G) is a unit if and only if it vanishes nowhere, thereforeevery function in a proper ideal vanishes somewhere.

We now distinguish between two types of ideals.

Definition An ideal I in O(G) is called fixed if fIZ(f) = , otherwiseit is called free. A point c G is called a zero of an ideal I if f(c) = 0 forevery f O(G).

Remark A fixed ideal has at least one zero and a free ideal has no zeros.

From the principal ideal theorem 4.2, we have

Corollary 4.5 Any proper finitely generated ideal is fixed.

Definition An ideal is called closed if it contains the limit function of everysequence fn a that converges compactly in G.

We continue with a lemma.

Lemma 4.6 Let I O(G) be an ideal, where c G is not a zero, and letf, g O(G) where f vanishes only at c. If f g I, then g I

Proof Choose h I with h(c) = 1. Let n := oc(f). If n 1 then,

fzc g = fzc g(h(z) (h(z) 1)) = fzc gh h(z)1zc f g I because h(z)1zchas a removable singularity at c and gh I and f g I. Applying this ntimes gives ( f(zc)n ) g I. Since

f(zc)n is invertible, we have g I.

By Weierstra product theorem for arbitrary regions [5 p.93], everyf = 0 that is holomorphic in an arbitrary region G C can be written inthe form f = u

1 f where u is a unit in the ring O(G) and

1 f

converges normally in G and f has exactly one zero c in G of order 1.

Proposition 4.7 If I is a closed free ideal in O(G), then I = O(G).

Proof Let f I, f = 0 and let f =

f be a factorization off as describedabove. Then fn :=

n f O(G) converges compactly in G to 1 by

theorem 3.7. We write fn = fn

1 f = fnfn+1. Since f0 = f I and fnhas no zeros in G \ {c}, it follows by induction and lemma 4.6 that fn Ifor all n 0. Since I is closed, it follows that 1 I and hence I = O(G).

Theorem 4.8 An ideal I O(G) is a principal ideal if and only if it isclosed.

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Proof ) Let I = O(G)f, and let gn = anf O(G) converge compactly

to g on G. Then an = gn/f converges compactly on G \ Z(f), by lemma3.6, to a function a O(G). Hence g = af I, so I is closed.

) By 3.14, I has a gcd f in O(G). Then I := f1I is a free ideal inO(G). Now, I is closed if I is, therefore by proposition 4.7 I = O(G) whichimplies that I = O(G)f.

From this and proposition 4.7 we have

Corollary 4.9 A proper free ideal in O(G) is never finitely generated.

However, a fixed ideal is not necessarily finitely generated. Let I be afree ideal in O(G), then (z c)I with c G would be a fixed ideal that is

not finitely generated.

Proposition 4.10 LetI = O(G) be a proper free ideal in O(G), then everyfunction f I has infinitely many zeros.

Proof Assume f O(G) has only finitely many zeros. Then there aref1,...,fr I with Z(f, f1,...,fr) = , hence 1 = gcd(f, f1,...,fr) I. Con-tradiction!

Because a non-zero polynomial only has a finite number of zeros we get

Corollary 4.11 No non-zero polynomial belongs to a proper free ideal.

We now describe the ideals, especially the free ones, with help of divisors.As in section 3.5 we denote D+(G) the set of positive divisors D D(G),i.e. such that D(z) 0 for all z G and D(z) > 0 for at least one z G.

Definition A non-empty subset F D+(G) is called a filter in D+(G) ifit satisfies

D D F = D F and D, D F = min(D, D) F

Furthermore let |F| := {|D|; D F}.

Proposition 4.12 There is an inclusion preserving bijection between theset of proper ideals = {0} in the ring O(G) and the family of all filters inD+(G), namely O(G) a F := (a) := {(f); f a \ {0}}

with inverse

F a := I(F) := {f O(G); f = 0 (f) F}

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Proof Every function f in an ideal a determines a divisor (f) by the order

function and a contains all other functions that is divided by f. Therefore(a) fulfills, by theorem 3.12 and prop. 4.1, the conditions for a filter.

On the other hand, given a filter F, the set I(F) O(G) is an ideal, sincefor f, g I(F)\{0} we have f+g I(F) because of (f+g) min((f), (g)) F (when we assume f + g = 0) while f O(G), g I(F) f g I(F)is obvious. We have now to show I((a)) = a and (I(F)) = F for an ideala O(G) repr. a filter F D+(G). The inclusions a I((a)) and(I(F) F are obvious. Now, if f I((a)) \ {0}, then (f) = (g) with someg a, hence f = ug with a unit u O(G) resp. f = ug a. On the otherhand, take a divisor D F, write D = (f) with some function f O(G).But then f I(F) resp. D = (f) (I(F)).

4.3 Maximal ideals

Now we are going to describe maximal ideals of the ring O(G). The maximalideals are also described with help of divisors. We will see that the freemaximal ideals have a simpler structure than the fixed ones. Further weconsider the factor rings O(G)/m for both free and fixed maximal ideals.We start to recall some results from algebra.

Let N be an ideal of a ring R. Then the subsets a + N = N + a ={a + n|n N} of R called the additive cosets of N form a ring R/N with

the binary operations defined by (a + N) + (b + N) = (a + b) + N and(a + N)(b + N) = ab + N. This ring is called the factor ring of R moduloN.

The fundamental homomorphism theorem for rings says that if : R R is a ring homomorphism with kernel N, then the map : R/N (R)given by (x + N) = (x) is an isomorphism. If : R R/N is thehomomorphism given by (x) = x + N, then for each x R, we have(x) = (x). See [1 p.330].

An ideal m of R is maximal if and only if R/m is a field [1 p.336].We also recall that every integral domain D can be enlarged to a field F

such that every element of F can be expressed as a quotient of two elements

of D. F is called the field of fractions of D, see [1 section 5.4].

Definition Let F be a field. A F-algebra A is a ring A such that:

1. A is a vector space under addition.

2. k(ab) = (ka)b = a(kb) for all k F and a, b A.

A F-algebra A is called a division algebra if it as a ring is a division ring.

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Definition Let A and B be F-algebras. A function f: A B is an F-

algebra homomorphism if it is a ring homomorphism and if f(ka) = kf(a)for all a A and k F.

From the last definition we see that the elements in C stay fixed undera C-algebra homomorphism.

Theorem 4.13 Let m be an ideal in O(G). Then the following statementsare equivalent:

1. m is a closed maximal ideal in O(G).

2. There exists a point c G such that m = {f O(G) : f(c) = 0} =O(G)(z c).

3. There exists aC-algebra homomorphism : O(G) C with kernel m.

Proof 1) 2) Since m is closed, m = O(G)f and since m is maximal, mis prime. This implies that f is irreducible, which means that f only hasone zero of order 1.

2) 3) The evaluation map c : O(G) C defined by c(f) = f(c) isa homomorphism with kernel m.

3) 1) Let : O(G) C be a C-algebra homomorphism, then z O(G)implies (z) = c for some c C. Therefore (z c) = 0. Suppose c / G,then 1

zc O(G) and we would have 1 = (1) = (1

zc )(z c) = 0 .

Hence c has to belong to G. Now let f O(G) and write f = f(c) (z c) f(z)f(c)

zc . Since keep complex numbers fixed and (z c) = 0, it followsthat (f) = f(c). That is, = c where c is the evaluation map. By thefundamental homomorphism theorem, O(G)/m is isomorphic to C, wherem = {f O(G) : f(c) = 0} is the kernel of the map : O(G) C . Hencem is maximal and clearly closed.

Corollary 4.14 Ifm is a maximal fixed ideal, then the residue fieldO(G)/mis isomorphic to C.

Proof Since c is onto O(G), O(G)/m is isomorphic to C.

The maximal ideals that are not closed, i.e. free, have a more compli-cated structure than the closed and hence fixed ones. We will now charac-terize the maximal ideals, both fixed and free. From proposition 4.12 we seethat an ideal m O(G) is maximal if and only if M := (m) is a maximalfilter in D+(G).

Proposition 4.15 A filter M D+(G) is maximal iff

min(D, D) > 0, D M = D M.

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Proof ) Suppose M is maximal and the given condition is not valid,

then D would together with the elements in M generate a filter properlycontaining M. Hence M would not be maximal.

) Suppose M is not maximal, that is, there exists a filter N M.Then for any D N and D M we have min(D, D) > 0, hence the givencondition implies that D M, i.e. N M and N = M.

To a divisor D D+(G) we associate its reduced divisor red(D) definedby

red(D)(z) :=

1 if D(z) > 0 ;0 otherwise.

A filter F D+(G) is called reduced, ifD F = red(D) F; we definered(F) F to be the unique reduced filter in D+(G) with |red(F)| = |F|.

Proposition 4.16 The filter F D+(G) is maximal iff red(F) = F andA G discrete, A B = , B |F| = A |F|.

Proof ) Since red(F) F in any case, the maximality of F impliesred(F) = F. Assume then that A / |F| would be a discrete subset, such thatA B = for all B |F|. Then F = {D : |D| A B for some B |F|}would be a filter containing F as proper subfilter.

) Assume min(D, D) > 0 for all D F. Then |D| |D| = forall D F, hence |D| |F|, i.e. |D| = |D| for some D F. But thenD red(D) = red(D) red(F) = F, whence D F.

Note that since the support of a divisor is the zero-set of the associatedfunction a maximal ideal m is free if |D| = , D (m).

Theorem 4.17 If m is a maximal free ideal, then O(G)/m contains theset of all rational functions C(z) as a subfield.

Proof By corollary 4.11, m can not contain a non-zero polynomial. There-fore the composition C[z] O(G) O(G)/m is an injective map fromC[z] O(G)/m, that is C[z] is isomorphic to a subring of O(G)/m. Be-cause m is maximal O(G)/m is a field. Then the subfield C[z] can beenlarged to the field of fractions C(z).

Corollary 4.18 The field C is a subfield of O(G)/m. Hence the residuefieldO(G)/m of a maximal free ideal may be considered as a division algebracontaining C as a proper subfield.

In the remaining part of this section we investigate O(G)/m more closely.The next two lemmas are going to help us. From now on let G = C. Firstwe note that:

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1. If (f) = (g) then f = gh where h is a unit (a nowhere vanishing

function).

2. If {zn}n=1 is any closed, discrete subset of C and if {wn}

n=1 is any

sequence of complex numbers then there is an f O(C) such thatf(zi) = wi, i = 1, 2,....

3. A field F is called algebraically closed if every non-constant polynomialin F[T] has a zero in F.

Let CN be the set of all complex-valued sequences.

Lemma 4.19 The field O(G)/m is algebraically closed.

Proof Let h m be a function with (infinitely many) simple zeros, saya0, a1, a2,.... Consider the following factorization of the quotient map:

O(G) O(G)/(h) K := O(G)/m.

Now the map O(G) CN, f (f(an)) has kernel (h) and is, accordingto 2), onto, hence O(G)/(h) = CN and K = CN/m with a maximal idealm CN. Let p K[T] be a non-constant monic polynomial and lift it to amonic polynomial q(T) = (qn(T))nN (C

N)[T] (C[T])N. In particular,all the polynomials qn(T) C[T] are monic of the same degree > 0, so forany n there is a zero bn C of the polynomial qn. Then the image in Kof the sequence (bn)nN is a zero of p(T). So C

N/m and hence O(G)/m isalgebraically closed.

Lemma 4.20 The cardinality of O(C)/m is equal to the cardinality of thecontinuum.

Proof The dense subset A := Q(i)[z] of O(C) is countable. Considerthe map : AN O(C), where AN is the set of all sequences (fn)nNwith fn A and ((fn)nN) := 0 if fn is not compactly convergent and((fn)nN) := limn fn else. Since is surjective and |A

N| = |R| = thecardinality of the continuum, we have |O(C)| |R|. Hence O(C)/m has as

most |R| elements. But all complex numbers are incongruent (mod m), soO(C)/m has at least |R| elements. Therefore O(C)/m contains precisely|R| elements.

From algebra we recall that a field F is an extension field of K if K isa subfield of F. A subset S of F is said to be algebraically independentover K if for all n > 0, f K[x1,...,xn] and pairwise distinct s1,...,sn,f(s1,...,sn) = 0 implies f = 0.

A transcendence base of F over K is a subset S of F which is alge-braically independent over K and is maximal in the set of all algebraically

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independent subsets of F. Moreover, every transcendence base of F over

K has the same cardinality [8. p.315]. The cardinality |S| of any base S iscalled the transcendence degree of F over K and is denoted tr.d.F/K.

In [8. p.317] it is proved that if F1 and F2 are algebraically closed fieldextensions of the field K1 resp. K2 with tr.d.F1/K1=tr.d.F2/K2, then everyisomorphism of fields K1 = K2 extends to an isomorphism F1 = F2.

We are now able to state the final theorem of this section.

Theorem 4.21 Letm be a free maximal ideal, then O(G) is isomorphic asa ring to C.

Proof Since O(G)/m contains C it has tr.d over Q at least |R| and bylemma 4.20 it has tr.d at most |R| over Q. Hence O(G)/m has tr.d |R| overQ. The set of complex numbers also have tr.d |R| over Q, and both C andO(G)/m is algebraically closed over Q. Hence by the above O(G)/m = C.

If we replace the integral domain O(G) with any integral domain withgreatest common divisor in which Wedderburns lemma is true, the principalideal theorem is valid.

Moreover, for non-compact Riemann surfaces the theorems of Weierstraand Mittag-Leffler are also valid and hence are Wedderburns lemma and theexistence of greatest common divisors. Therefore, if we replace regions in Cwith non-compact Riemann surfaces, all the results in the previous sectionremain valid.

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5 Prime ideals

We are going to investigate the filters of prime ideals.

Definition A filter P D+(G) is called prime if D1 + D2 P = D1 P D2 P.

Note that a filter P is prime if and only if p := I(P) O(G) is a primeideal. We want to investigate that situation more in detail. For a divisorD D+(G) we set

s(D) := sup{D(z)|z G} N {}.

Lemma 5.1 Let P D+(G) be a prime filter. Then P is maximal if andonly if there is a divisor D0 P with s(D0) < .

Proof ) This is clear since for P maximal we have P = red(P).) Since P is prime, we may assume s(D0) = 1. Take now D

D+(G)with min(D, D) > 0 for all D P, in particular for D = D0. Write D0 =min(D, D0) + D1. Since P is prime, either min(D

, D0) P or D1 P. Inthe first case we obtain that also D P because of D min(D, D0) P,in the second case we would get 0 = min(D1, D0) P, a contradiction.

Corollary 5.2 If a prime ideal is fixed, then it is maximal.

Lemma 5.3 If P D+(G) is prime, then so is red(P); in particularM :=red(P) is maximal.

Proof Assume D1 + D2 red(P). Then |D| = |D1 + D2| with someD P, and we may even assume D D1 + D2 resp. D = D3 + D4,where |D3| = |D1|, |D4| = |D2|. Since P is prime, we obtain D3 P orD4 P, say D3 P. But then D1 red(P).

Let M D+(G) be a maximal filter. Denote F := |M|. We shallinvestigate the set of filters F M with |F| = F. Note first of all thatF P(G) is an ultrafilter of discrete subsets of G:

1. F;

2. A B with B G discrete and A F = B F;

3. A, B F = A B F

4. A G discrete, A B = , B F = A F.

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Furthermore note that M can be reconstructed from F, namely M con-

sists of all divisors D, such that there is a subset A F with D(z) > 0 forall z A.

We define on D+(G) a new relation:

D D : A F with D|A D|A.

That relation is obviously reflexive and transitive, it becomes even anti-symmetric if we replace = with the equivalence relation:

D D : A F with D|A = D|A.

But the most interesting fact here is that for any two divisors D, D D+(G) one of the relations D D

or D D holds. Assume that D D

does not hold, with other words that for A := {z G; D(z) < D(z)} onehas A B = for all B F. But F being an ultrafilter, we may concludethat A F resp. D D. Furthermore note that D D F impliesD F: Let A F with D|A D|A and A = |D0| with some D0 F.Then F min(D, D0) D

, hence D F.The set of divisors is totally ordered with respect to the relation .

Proposition 5.4 Let F, F D+(G) be filters with |red(F)| = |red(F)| =

F. Then either F F or F F.

Proof Suppose F F, then for all D F, D F, D D F does nothold. Then D D F holds, which implies that D F. Hence F F.

So we see that the set of all filters F with |red(F)| = F is totally orderedwith respect to inclusion. But when is such a filter F prime?

Proposition 5.5 Let F D+(G) be a filter with |F| = F. Then F isprime if and only if for all n N+ we have: nD F = D F.

Proof The given condition is obviously necessary, but also sufficient: As-sume D1 + D2 F, say D1 D2. But then 2D2 D1 + D2 F and hence

2D2 F resp. D2 F.

From this it follows that an ideal p m for a maximal ideal m O(G)is prime if and only if fn p implies f p. These prime ideals are alsototally ordered in m with respect to inclusion.

To a divisor D M we associate the filter FD := {D D+(G); D

D}.Furthermore we define the filter PD :=

n=1 nFD.

Remark Note here that nFD = FnD and that PD is never empty.

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Proposition 5.6 The filter PD := n=1 nFD is prime, and it is the largest

non-maximal prime filter contained in the filter FD.

Proof Assume mD PD. Then mD nD for all n N, in particular

mD kmD resp. D kD for all k N and thus D PD. The secondpart follows from lemma 5.1.

For example, if D = red(D) M is a reduced divisor, then PD consistsof all divisors D, such that for every n N there is a subset A F withD(z) n for all z A.

Let {Di|i I} be the set of all divisors D M \ P. Then, any non-maximal prime filter P D+(G) with |P| = F is now an intersection

P =

iI Pi

with prime filters Pi = PDi of the above type.

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6 References

1. J. B. Fraleigh, A first course in abstract algebra, Addison-Wesley Pub-lishing Company, 1999.

2. J. E. Marsden, M. J. Hoffman, Basic complex analysis, W. H. Freemanand Company, New York, 1997.

3. M. Henrikssen, On the ideal structure of the ring of entire functions,Pac. Journ. Math. 2(1952), 179-184.

4. M. Henrikssen, On the prime ideals of the ring of entire functions,Pac. Journ. Math. 3(1953), 711-720.

5. O. Helmer, Divisibility properties of integral functions, Duke Math. J.6(1940), 345-356.

6. R. Remmert, Classical topics in complex function theory, Springer-Verlag, New York, 1998.

7. R. Remmert, Theory of complex functions, Springer-Verlag, New York,1998.

8. T. W. Hungerford, Algebra, Springer-Verlag, New York, 1974.

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