57.3. THE EXISTENCE OF AN ANALYTIC FUNCTION WITH GIVEN VALUES1799

where g1, · · · ,gn are given functions in H (Ω). This ideal is also denoted as [g1, · · · ,gn] andis called the ideal generated by the functions, {g1, · · · ,gn}. Since there are finitely manyof these functions it is called a finitely generated ideal. A principal ideal is one which isgenerated by a single function. An example of such a thing is [1] = H (Ω) .

Then there is the following interesting theorem.

Theorem 57.3.4 Every finitely generated ideal in H (Ω) for Ω a connected open set (re-gion) is a principal ideal.

Proof: Let I = [g1, · · · ,gn] be a finitely generated ideal as described above. Then if anyof the functions has no zeros, this ideal would consist of H (Ω) because then g−1

i ∈ H (Ω)and so 1 ∈ I. It follows all the functions have zeros. If any of the functions has a zero ofinfinite order, then the function equals zero on Ω because Ω is connected and can be deletedfrom the list. Similarly, if the zeros of any of these functions have a limit point in Ω, thenthe function equals zero and can be deleted from the list. Thus, without loss of generality,all zeros are of finite order and there are no limit points of the zeros in Ω. Let m(gi,z)denote the order of the zero of gi at z. If gi has no zero at z, then m(gi,z) = 0.

I claim that if no point of Ω is a zero of all the gi, then the conclusion of the theoremis true and in fact [g1, · · · ,gn] = [1] = H (Ω) . The claim is obvious if n = 1 because thisassumption that no point is a zero of all the functions implies g ̸= 0 and so g−1 is analytic.Hence 1 ∈ [g1] . Suppose it is true for n− 1 and consider [g1, · · · ,gn] where no point ofΩ is a zero of all the gi. Even though this may be true of {g1, · · · ,gn} , it may not be trueof {g1, · · · ,gn−1} . By Corollary 57.1.7 there exists φ , a function analytic on Ω such thatm(φ ,z) = min{m(gi,z) , i = 1,2, · · · ,n−1} . Thus the functions {g1/φ , · · · ,gn−1/φ} .areall analytic. Could they all equal zero at some point, z? If so, pick i where m(φ ,z) =m(gi,z) . Thus gi/φ is not equal to zero at z after all and so these functions are analytic thereis no point of Ω which is a zero of all of them. By induction, [g1/φ , · · · ,gn−1/φ ] = H (Ω).(Also there are no new zeros obtained in this way.)

Now this means there exist functions fi ∈ H (Ω) such that

n

∑i=1

fi

(gi

φ

)= 1

and so φ = ∑ni=1 figi. Therefore, [φ ] ⊆ [g1, · · · ,gn−1] . On the other hand, if ∑

n−1k=1 hkgk ∈

[g1, · · · ,gn−1] you could define h≡∑n−1k=1 hk (gk/φ ) , an analytic function with the property

that hφ = ∑n−1k=1 hkgk which shows [φ ] = [g1, · · · ,gn−1]. Therefore,

[g1, · · · ,gn] = [φ ,gn]

Now φ has no zeros in common with gn because the zeros of φ are contained in the set ofzeros for g1, · · · ,gn−1. Now consider a zero, α of φ . It is not a zero of gn and so near α,these functions have the form

φ (z) =∞

∑k=m

ak (z−α)k , gn (z) =∞

∑k=0

bk (z−α)k , b0 ̸= 0.