1800 CHAPTER 57. INFINITE PRODUCTS

I want to determine coefficients for an analytic function, h such that

m(1−hgn,α)≥ m(φ ,α) . (57.3.20)

Let

h(z) =∞

∑k=0

ck (z−α)k

and the ck must be determined. Using Merten’s theorem, the power series for 1−hgn is ofthe form

1−b0c0−∞

∑j=1

(j

∑r=0

b j−rcr

)(z−α) j .

First determine c0 such that 1− c0b0 = 0. This is no problem because b0 ̸= 0. Next youneed to get the coefficients of (z−α) to equal zero. This requires

b1c0 +b0c1 = 0.

Again, there is no problem because b0 ̸= 0. In fact, c1 = (−b1c0/b0) . Next consider thesecond order terms if m≥ 2.

b2c0 +b1c1 +b0c2 = 0

Again there is no problem in solving, this time for c2 because b0 ̸= 0. Continuing thisway, you see that in every step, the ck which needs to be solved for is multiplied by b0 ̸=0. Therefore, by Corollary 57.1.7 there exists an analytic function, h satisfying 57.3.20.Therefore, (1−hgn)/φ has a removable singularity at every zero of φ and so may beconsidered an analytic function. Therefore,

1 =1−hgn

φφ +hgn ∈ [φ ,gn] = [g1 · · ·gn]

which shows [g1 · · ·gn] = H (Ω) = [1] . It follows the claim is established.Now suppose {g1 · · ·gn} are just elements of H (Ω) . As explained above, it can be

assumed they all have zeros of finite order and the zeros have no limit point in Ω sinceif these occur, you can delete the function from the list. By Corollary 57.1.7 there existsφ ∈ H (Ω) such that m(φ ,z) ≤ min{m(gi,z) : i = 1, · · · ,n} . Then gk/φ has a removablesingularity at each zero of gk and so can be regarded as an analytic function. Also, as before,there is no point which is a zero of each gk/φ and so by the first part of this argument,[g1/φ · · ·gn/φ ] = H (Ω) . As in the first part of the argument, this implies [g1 · · ·gn] = [φ ]which proves the theorem. [g1 · · ·gn] is a principal ideal as claimed.

The following corollary follows from the above theorem. You don’t need to assume Ω

is connected.

Corollary 57.3.5 Every finitely generated ideal in H (Ω) for Ω an open set is a principalideal.

Proof: Let [g1, · · · ,gn] be a finitely generated ideal in H (Ω) . Let {Uk} be the compo-nents of Ω. Then applying the above to each component, there exists hk ∈H (Uk) such thatrestricting each gi to Uk, [g1, · · · ,gn] = [hk] . Then let h(z) = hk (z) for z ∈Uk. This is ananalytic function which works.