WEIERSTRASS THEOREMS AND RINGS OF HOLOMORPHIC FUNCTIONS

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1 WEIERSTRASS THEOREMS AND RINGS OF HOLOMORPHIC FUNCTIONS YIFEI ZHAO Contents. The Weierstrass factorization theorem 2. The Weierstrass preparation theorem 6 3. The Weierstrass division theorem 8 References 9 We organize this set of notes around a few theorems of Weierstrass. Write O C n for the sheaf of holomorphic functions on C n. In the first three sections, we deduce the following algebraic results as consequences of the Weierstrass theorems: (i for each open set Ω C n, the ring O C n(ω is not Noetherian, (ii the local ring O Cn,0 is factorial, (iii the local ring O Cn,0 is Noetherian. The two later sections can be read independently of the first.. The Weierstrass factorization theorem The Weierstrass factorization theorem generalizes the fundamental theorem of algebra to transcendental functions. It asserts that every entire function f factors as f(z = e ( z g(z E pn z n n where g is some entire function, E pn are so-called elementary functions that generalize linear functions, and are the roots of f. The proofs of this section mostly follow [3, 5]... Statement. For any nonnegative integer p, define the pth elementary function E p on C as E p (z = ( z exp (z + z zp p Theorem. (Weierstrass factorization theorem. Suppose f is an entire function with f(0 0. Let z, z 2, be the zeroes of f (allowing repetition, and p, p 2, be any sequence of nonnegative integers such that ( +pn r r > 0 = < Then there exists an entire function g such that f(z = e g(z ( z E pn Note that (i The sequence p n = n satisfies this condition according to the root test. (ii If f has an order-k zero at z = 0, simply apply the theorem to f(z/z k.

2 2 YIFEI ZHAO The meat of the theorem is the convergence of the infinite product to an entire function ( z E(z = E pn with the property that (i E(z has exactly one zero at each, and (ii the order of the product does not matter. Indeed, assuming (i, we find that f(z/e(z is an entire function with no zeroes. Since C is simply connected, we have f(z/e(z = e g(z for some holomorphic function g(z. There are some standard examples of Weierstrass factorizations. For f(z = sin(πz/πz, the zeroes are precisely the nonzero integers. Hence p n = suffices. We find sin(πz πz = e ( g(z z ( z exp = e n n ( g(z z2 n 2 n 0 by multiplying together the factors associated to n and n. In fact, in this example, the function g(z can be taken to be zero..2. Infinite product. Recall that the infinite product of complex numbers a, a 2, is defined as a i = lim N a i N if the limit exists. The infinite product of a sequence of functions is defined pointwise. Proposition.2. Let Ω be an open subset of C. Given a sequence of holomorphic functions u, u 2, on Ω such that u n converges uniformly on each compact set K Ω, we have (i The infinite product ( + u n converges uniformly on each K, hence defines a holomorphic function P on Ω. (ii The order of zeroes of P at z satisfies ord z (P = ord z ( + u n (iii The function u is independent of the order of the infinite product, i.e. for any reordering {n k } of natural numbers, the infinite product k ( + u n k again converges uniformly on each K, and defines the same holomorphic function P. Proof. We leave to the reader verifications of the following inequalities (or see [3, Lemma 5.3]: Claim.3. Let a,, a N be complex numbers, and set Then we have N N p N = ( + a n, p N = ( + a n. N p N p N, p N exp a n. This requires a more refined version of the Weierstrass factorization theorem. I learned this argument from Bjorn Poonen s answer here: <

3 WEIERSTRASS THEOREMS AND RINGS OF HOLOMORPHIC FUNCTIONS 3 (i Fix a compact subset K. Let ɛ > 0. There exists an N 0 such that z K = u n (z ɛ. n=n 0 Let P N denote the Nth partial product of the ( + u n s. Then P N N N ( + u n exp u n < C (. where C is a constant (that only depends on K. We have used the claim in the second inequality. For M N N 0, we have M P M P N = P N ( + u n P N P N n=n+ ( M ( n=n+ exp ( + u n ( M n=n+ u n P N (e ɛ (by N N 0 Finally, using the inequality e ɛ < 2ɛ, we obtain (by claim (by claim P M P N < 2ɛ P N (.2 It follows from the bound on P N (. that P M (z is Cauchy for each z K. Hence exists. The inequality (.2 then shows that N N 0 P (z = lim N P N (z (.3 = P P N 2Cɛ so the convergence is uniform. (ii Note that the right-hand-side is a finite sum, because otherwise u n (z = for infinitely many n, and un would not converge. Suppose z is contained in the interior of a compact set K Ω. We first show that P (z 0 if + u n (z 0 for all n. Indeed, it follows from (.2 that for any ɛ > 0, there is some N 0 with N N 0 = P P N ( 2ɛ. If P N (z 0 for all N, then P (z 0. Now, say each + u n has an order-k n zero at z. We replace it by a holomorphic function + v n defined as (w z kn ( + v n (w = ( + u n (w. Note that we have only changed finitely many u n s. Hence the hypothesis is still satisfied by the functions v n, and + v n (z 0 for all n. Set d = ord z ( + u n Then we have n=0 ( + v n (w = P (w (w z d on K {z}, and the left hand side does not vanish at z by the argument above. It follows that P has a pole of order d at z.

4 4 YIFEI ZHAO (iii Again fix a compact set K Ω. Let ɛ > 0, and N 0 be taken as in (i. Let Q M be the partial product of the ( + u nk s. In other words, M Q M = ( + u nk For each N N 0, suppose M is so large that {, 2,, N} {n, n 2,, n M }. Then k= Q M P N 2ɛ P N 2Cɛ by a similar computation as in (.2. Thus the limit of Q M is identical to that of P N on K..3. Infinite product of elementary functions. To apply the Proposition, we first establish a bound for the elementary functions. Lemma.4. E p is an entire function with a simple zero at z =. Furthermore, for z, there holds E p (z z p+. Proof. Only the bound requires a proof. Indeed, E p (z = E p (0 E p (z = [0,z] The derivative may be computed and expanded into a power series: E p(wdw. E p(w = w p exp(w + w wp p = wp + a p+ w p+ + a p+2 w p+2 + with a p+, a p+2, 0. Hence E p (z = b p+ z p+ + b p+2 z b+2 + again with b p+, b p+2, 0. Therefore, the entire function satisfies ϕ(z ϕ( =. ϕ(z = E p(z z p+ Proposition.5. Let z, z 2, be a sequence of nonzero complex numbers (allowing repetition such that, and p, p 2, be any sequence of nonnegative integers such that ( pn+ r r > 0 = < Then (i The infinite product ( z E pn converges uniformly on each compact set K C, hence defines an entire function E. (ii E has a zero of order m at z C if and only if z occurs m times in the sequence z, z 2,. (iii The function E is independent of the order of the infinite product. This Proposition implies the Weierstrass factorization theorem, as we have discussed before. Proof. These are line-by-line restatements of the conclusions of Proposition.2, where we set Ω = C, and ( z u n (z = E pn Hence, we only need to verify the uniform convergence of ( z E p n z p n+ on each compact set K C. This is precisely given by the hypothesis.

5 WEIERSTRASS THEOREMS AND RINGS OF HOLOMORPHIC FUNCTIONS 5.4. Further consequences. Here we are concerned with the general problem of finding a holomorphic function with given data of zeroes. Theorem.6. Let Ω C be open. Suppose A Ω such that A has no limit point in Ω, and for each α A, there is a positive integer m(α. Then there exists a holomorphic function f on Ω which has a zero of order m(α for each α A, and no zero elsewhere. Proof. The problem is trivial if A is finite, where a polynomial f suffices. From now on we assume that A is infinite. Note that A is countable. Indeed, each α A is the center of an open ball B α such that {α} = B α A. Hence the cardinality of A is no greater than that of the rational points in C. We now let α, α 2, be an enumeration of elements of A, where each α occurs m(α times. The idea is to multiply elementary functions associated to each α n. But we cannot do so naïvely without running against the problem of convergence. So we first perform some reductions. Embed C in the Riemann sphere P = C { }. After a coordinate change, we may assume Ω and / A. This assumption implies the following: (i The complement S = C Ω is compact. Indeed, it is closed because Ω is open, and is bounded because Ω. (ii A is bounded. Indeed, since Ω is not a limit point of A, it admits a neighborhood that does not intersect A. Now, let β, β 2, be a sequence in S such that This construction is possible by (i. It follows that β n α n = inf β S β α n β n α n 0; otherwise, infinitely many α n would belong to a closed subset B Ω, which can be taken to be compact by (ii, so A would have a limit point in B. Now, we set ( αn β n f(z = E n z β n We claim that it converges uniformly on each compact K Ω. Indeed, compactness guarantees some γ > 0 such that z K = z β n γ However, for sufficiently large n, we have α n β n γ/2. Thus ( E αn β n n z β n α n β n z β n n+ < By Proposition.2, the infinite product converges and f(z has a zero precisely at each α n. Corollary.7. Let Ω C be open and nonempty. Then the ring O C (Ω is not Noetherian. Proof. There exists a sequence A of distinct complex numbers α, α 2, in Ω such that A has no limit point in Ω. Let f, f 2, be a sequence of holomorphic functions on Ω such that Then we have a proper chain of ideals f n has a simple zero at precisely α n, α n+, (f (f 2 (f n (f n+ since each f n /f n+ only has removable singularities in Ω, but if f n+ = f n g for some g O C (Ω, then which is a contradiction. 0 f n+ (α n = f n (α n g(α n = 0

6 6 YIFEI ZHAO Corollary.8. Let Ω C n be open and nonempty. Then the ring O C n(ω is not Noetherian. Proof. There is an embedding f : C C n such that f (Ω is nonempty in C. Hence we have a surjection of rings f : O C n(ω O C (f Ω. If O C n(ω were Noetherian, so would O C (f Ω be, but this is impossible by Corollary The Weierstrass preparation theorem The proofs in this section mostly follow [] and [2]. 2.. The theorem. In one variable, each nonzero holomorphic germ f O C,0 factors as f(z = z k u(z for some k and some unit u O C,0. In particular, the zeroes of f are discrete. The Weierstrass preparation theorem is a generalization of this statement to higher dimensions, where the linear factor z k is replaced by the Weierstrass polynomials. Definition 2.. A Weierstrass polynomial near 0 is a function P of n complex variables z,,, w, defined in a neighborhood of 0, of the form P (z, w = w d + a d (zw d + + a 0 (z where d is a nonnegative integer, and each a i (z is a holomorphic function defined in a neighborhood of 0 such that a i (0 = 0. It is better to think of a Weierstrass polynomial in terms of its germ in the local ring O C n,0. There are inclusion maps: P Cn,0 O C n,0[w] O Cn,0 where P C n,0 denotes the germs of Weierstrass polynomials. The first map embeds P C n,0 as a multiplicative submonoid of O C n,0[w], and the second map is a ring map. Theorem 2.2 (Weierstrass preparation theorem. Let f O Cn,0 and suppose f(0, w is not identically zero. Then f = gh for some g P C n,0 and h O C n,0. Furthermore, an expression of this form is unique. In particular, the degree of g is precisely the degree of vanishing of f(0, w at w = 0. Proof. Since f(0, w is not identically zero, we may find ɛ > 0 and r > 0 such that w < r, f(0, w = 0 = w = 0 (2. Hence f(z, w is nonzero and holomorphic in a neighborhood of the tube z < ɛ, w = r = f(z, w 0 (2.2 V = {(z, w : z < ɛ, w = r} For each z with z < ɛ, we may let b (z,, b d(z (z be the zeroes of f(z, w, allowing repetition. Then (2. shows that b i (0 = 0. Set d(z g(z, w = (w b i (z = w d(z + a d(z (zw d(z + + a 0 (z i=

7 WEIERSTRASS THEOREMS AND RINGS OF HOLOMORPHIC FUNCTIONS 7 where each a i (z is a symmetric polynomial in the b i (z s. Hence each a i (0 = 0 and a i (z can be expressed in terms of the functions d(z b q i (z = 2π w q f (z, wdw w =r f(z, w w i= and therefore are holomorphic. In particular, by taking q = 0 in the above equation, we find that d(z is holomorphic, so it must be a constant. Hence g(z, w defines a Weierstrass polynomial. Now, set f(z, w h(z, w = g(z, w and it remains to show that h(z, w extends to a holomorphic function on some neighborhood of zero. Indeed, for fixed z, the functions f and g (as functions of w have the same data of zeroes, so h(z, w is defined and holomorphic in w. It follows that h(z, w = 2π w =r h(z, ξ dξ (2.3 ξ w Since f and g have identical data of zeroes, g is also holomorphic and nonzero in a neighborhood of V. Thus h is holomorphic in a neighborhood of V. It now follows from (2.3 that if w is sufficiently close to 0, then h is holomorphic as well. To prove the uniqueness, note that given such an expression, f and g must have the same data of roots in a neighborhood of zero. On the other hand, g is the unique (germ of Weierstrass polynomial with this data. Hence h O Cn is also uniquely determined O C n,0 is factorial. The result will follow from a few easy lemmas. Lemma 2.3. Let g P C n,0 be irreducible as an element in O C n,0[w]. Then g is irreducible in O C n,0. Proof. Suppose g = f f 2 is a proper factorization in O C n,0. In other words, both f and f 2 vanish at zero. Since g(0, w is not identically zero, the same is true for f (0, w and f 2 (0, w. The Weierstrass preparation theorem applies, and we have f = g h, f 2 = g 2 h 2, where g, g 2 P Cn,0 are of nonzero degree and h, h 2 O C n,0. Hence g = (g g 2 (h h 2 The uniqueness in Weierstrass preparation theorem now shows g = g g 2, so we obtain a proper factorization in O C n,0[w]. Lemma 2.4. Let g P Cn,0, and suppose there exist g,, g n O C n,0[w] such that g = n g i. i= Then there exist g,, g n P C n,0 with deg( g i = deg(g i, such that n g = g i (2.4 i= Proof. Again, the fact that g(0, w is not identically zero implies that same thing for each g i (0, w. Let g i = g i hi be a factorization with g i P C n,0 and h i O C n,0. So (2.4 again follows from the uniqueness in Weierstrass preparation theorem. It remains to show that deg( g i = deg(g i. Indeed, we have deg( g i = deg( g i (0, w deg(g i (0, w deg(g i but n deg( g i = deg(g = i= so we must have equality deg( g i = deg(g i for all i. n deg(g i i=

8 8 YIFEI ZHAO Theorem 2.5. The local ring O Cn,0 is factorial. Proof. We first assume that O C n,0 is factorial. By Gauss lemma, the subring O C n,0[w] of O C n,0 is again factorial. Using Weierstrass preparation theorem, we may express each f O C n,0 uniquely as the product of a unit and a Weierstrass polynomial. Hence we only need to show that Weierstrass polynomials admit unique factorization into Weierstrass polynomials that are irreducible in O Cn,0. Indeed, each g P C n,0 admits a unique factorization n g = i= where each g i is an irreducible element of O C n,0[w]. If n =, then g is irreducible in O C n,0[w], hence irreducible in O C n,0 as well by Lemma 2.3. If n 2, then each g i has lower degree than g. By Lemma 2.4, we may change each g i into g i P C n,0 with degrees preserved. Hence, assuming that each g i has a unique factorization into Weierstrass polynomials that are irreducible in O Cn,0, the same holds for g itself. g i 3. The Weierstrass division theorem The proofs in this section follow [2]. 3.. The theorem. The Weierstrass polynomials behave like monomials in dimension-one in another way: one may divide by them and obtain a polynomial remainder of lesser degree. This is expressed by the Weierstrass division theorem. Theorem 3. (Weierstrass division theorem. Suppose f O Cn,0 and g P Cn,0 has degree d. Then there exists unique elements h O Cn,0 and r O C n,0[w] of degree < d such that f = gh + r Proof. Find ɛ > 0 and r > 0 such that z < ɛ, w = r = g(z, w 0. So g is nonzero and holomorphic in a neighborhood of the tube V = {(z, w : z < ɛ, w = r}. Set 2 h(z, w = 2π f(z, ξ ξ =r g(z, ξ(ξ w dξ Hence h is a holomorphic function in a neighborhood of zero. We need to show that r = f gh O C n,0[w] and has degree < d. This follows from a straightforward computation: r(z, w = = 2π 2π ξ =r ξ =r By definition of the Weierstrass polynomial, Hence g(z, ξ g(z, w ξ w 2 I find this expression of h a bit magical. f(z, ξ ξ w dξ 2π ξ =r ( f(z, ξ g(z, ξ g(z, w g(z, ξ ξ w g(z, w = w d + a d (zw d + + a 0 (z g(z, wf(z, ξ g(z, ξ(ξ w dξ dξ = (ξd w d + a d (z(ξ d w d + + a (z(ξ w ξ w

9 WEIERSTRASS THEOREMS AND RINGS OF HOLOMORPHIC FUNCTIONS 9 is a polynomial in w of degree d, with coefficients being holomorphic functions in z and ξ. It follows that r(z, w O C n,0[w] and has degree < d. For the uniqueness, it suffices to show that 0 = gh + r = h = 0, r = 0 Indeed, when z is sufficiently close to zero, g(z, w has d roots (as a function of w. However, deg(r < d, so h must be identically zero. Thus the same is true for r. Theorem 3.2. The local ring O C n,0 is Noetherian. Proof. Assume O C n,0 is Noetherian. By Hilbert s basis theorem, O C n,0[w] is again Noetherian. Let I be any nonzero ideal of O Cn,0, and let f O Cn,0 be a nonzero element. Then after a coordinate change if necessary, we may assume f(0, w is not identically zero. Weierstrass preparation theorem shows that f = gh for some g P C n,0, and h a unit. Thus g I. Let f be any element in I. Then Weierstrass division theorem shows that f = g h + r (3. for some r I O C n,0[w]. On the other hand, I O C n,0[w] is finitely generated by some g,, g n as an ideal of O C n,0[w]. Hence (3. expresses f as a O C n,0-combination of g, g,, g n. In other words, I is finitely generated. References [] Griffiths, Phillip, and Joseph Harris. Principles of algebraic geometry. John Wiley & Sons, 204. [2] Huybrechts, Daniel. Complex geometry: an introduction. Springer Science & Business Media, [3] Rudin, Walter. Real and complex analysis. Tata McGraw-Hill Education, 987.