University:
Rice University
Course:
MATH 221 | Honors Calculus III
Academic year:

2021
Views:

30

Pages:

4
Author:

MathMonk

Questions and Answers Sheet 5 Rational Functions Question #1 In Hartshorne's Algebraic Geometry the function field K(Y) of a variety Y is defined as the set of equivalence classes with f being a regular function on the open subset 𝑈 ≠ ∅. We have = if f and g agree on 𝑈 ∩ 𝑉. How does one evaluate a "rational function on Y'' f? Is it f(P) using the Evaluation map for polynomials? In 𝑃∈𝑈 this case, I don't see that this is well defined: Let and =, then g(P) might not even be 𝑉 defined! Answer: The point is this: on a variety Y, given open subsets U and V and regular functions 𝑓: 𝑢 → 𝑘 and 𝑔: 𝑉 → 𝑘 , if 𝑓|𝑈∩𝑉 = 𝑔|𝑈∩𝑉 , there is a unique regular function ℎ: 𝑈 ∪ 𝑉 → 𝑘 such that 𝑓 = ℎ|𝑈 and 𝑔 = ℎ|𝑉 . (There's only one thing it can be – you just have to check that it works!) Therefore, in any equivalence class of regular functions, there is a unique maximal representative. In particular, you can evaluate a rational function at a point P so long as there is some representative defined at P. They are called rational functions as a matter of tradition, not because they are actually functions. (It is tempting to think of rational functions on Y as being regular maps 𝑌 → 𝑃𝑃1 , but that doesn't work in general: for example, 𝑥 𝑦 defines a regular function 𝐴2 {(0,0)} 1 → 𝑘, hence a rational function 𝐴2 → 𝑘, but there is no way of extending it to a regular map 𝐴2 → 𝑃 .) Question #2 Is a rational function or a rational equation or none of these? 1. 𝑦 = 5𝑥 3 − 2𝑥 + 1 2. 𝑔(𝑥) = 7𝑥 3 − 4√𝑥 + 1√𝑥 2 + 3 Answer: A rational function is defined as 𝑓(𝑥) = 𝑃(𝑥) 𝑄(𝑥) for any polynomials P and Q (constant polynomials as well). Thus, we see that (1) is a rational function with 𝑃(𝑥) = 5𝑥 3 − 2𝑥 + 1 and Q(x)=1. We also see that (2) is not a rational function because it contains square roots. Question #3 Is the section df associated to a rational function f on a curve X a global section of the canonical sheaf 𝜔𝑋 ? I know its zeroes are the ramification points, but does it have poles? Answer: Yes, df is a rational section of 𝜔𝑋 , which one could write rigorously as 𝑑𝑓𝜖Γ(𝑋, 𝜔𝑋 ⊗𝑂𝑋 𝐾𝑋 ). Beware however that not all rational sections of 𝜔𝑋 are of this form: the simplest example is (or on 𝑃𝐶1 ) which is not the differential of any rational function. 𝑑𝑧 𝑧 on CC Question #4 Lagrange's rational function theorem states that if one has two rational functions in multiple variables 𝑓(𝑥1 , 𝑥2 , … 𝑥𝑛 ) and 𝑔(𝑥1 , 𝑥2 , … 𝑥𝑛 ) then one can express f as a rational function in g if and only if the set of permutations that keep g unchanged is a subset of the set of permutations that preserve f. Is anyone familiar with the proof of this theorem? While it is fairly clear that if f can be expressed in terms of g the set of permutations that keep g unchanged has to be the subset of those that keep f unchanged, the converse is far from obvious. Answer: Consider the field 𝐾 = 𝑄(𝑥1 , 𝑥2 , … 𝑥𝑛 ), and consider 𝐾𝑔 ⊂ 𝐾 to be the subfield generated by g. Define 𝐾 𝐻𝑔 = 𝐴𝑢𝑡 ( ) = {𝜎𝜖𝐴𝑢𝑡(𝐾): 𝜎(𝛼) = 𝛼∀𝛼𝜖𝐾𝑔 } 𝐾𝑔 Similarly, define 𝐾𝑓 and 𝐻𝑓 . Then you want to show that 𝐾𝑓 ⊂ 𝐾𝑔 ⇔ 𝐻𝑔 ⊂ 𝐻𝑓 If all the hypotheses are satisfied, this is merely the Fundamental Theorem of Galois Theory Question #5 Let f be a rational function on a compact connected Riemann surface X. The rational function f induces a holomorphic map 𝑓: 𝑋 → 𝑅1 (𝐶). Let x be a point on the Riemann sphere 𝑃1 (𝐶). ow can I check that if b is a branch point of 𝑓 by looking at the derivative of f? How does this work when 𝑋 = 𝑃1 (𝐶) ? Answer: If 𝑏𝜖𝑋 and 𝑓(𝑏) ∞ then b is a branch point iff f'(b)=0 (derivative wrt. an arbitrary local coordinate; the ramification index is the maximal k s.t. 𝑓 (𝑘) (𝑏) = 0 (the number of branches meeting at b is k+1)). If 1 𝑓(𝑏) − ∞, replace f with . 𝑓 Question #6 Let f be holomorphic in all of C except for poles at z=0 and z=1. Further f satisfies: lim 𝑓(𝑧) = 0 |𝑧|→∞ I've to prove that f is a rational function. Approach: I considered the function ℎ(𝑧) = 𝑧 𝑛 (𝑧 − 1)𝑚 𝑓(𝑧), with n, m the orders of the poles 0 and 1. Clearly h is holomorf in 𝐶 {0,1} . But because the limit of h for 𝑧 → 0 or 𝑧 → 1 is finite (and non zero), by Riemanss theorem it follows that h can be extended to an entire function. If h has a removable singularity at ∞, we know that h must be constant, so that f is for sure a rational function. If I now can exlude the case that ∞ can be an essential singularity of h, we are done because then it follows that h is polynomial and thus f is a rational function. But how to approach this latter case? Answer: 𝑖 if you only had pole at z=0, you could do the following. Suppose 𝑓(𝑧) = ∑∞ 𝑖=−𝑛 𝑎𝑖 𝑧 near 0. Then 𝑔(𝑧) = −1 𝑖 𝑓(𝑧) − ∑𝑖=−𝑛 𝑎𝑖 𝑧 is holomorphic near 0, and it is fairly easy to see that it is also holomorphic everywhere 𝑖 else, and limit at ∞ is 0. By Liouville, g=0, whence 𝑓 = ∑−1 𝑖=−𝑛 𝑎𝑖 𝑧 , which is a rational function. Basically, the same thing can be done for 2 of poles. Question #7 𝑘[𝑥,𝑦] Let k be a field, 𝐴 = (𝑦2 ,𝑥𝑦) . I know that the only associated points of Spec A are [(x,y)] and [(y)]. I want to 𝑥−2 show that (𝑥−1)𝑥 is not a rational function. Intuitively it makes sense to me that this function is not defined at the origin [(x,y)], hence not a rational function. But I am having trouble arguing it precisely. I would appreciate any comments. Answer: The problem comes from the fact that this function should be defined on the open set 𝑈 = Spec𝐴𝑥,(𝑥−1) (inverting x and x-1) or on a smaller open set so that x and x-1 can be invertible. However such U could not contain all the associated points of the scheme (which is necessary in Vakil's definition of rational section) Question #8 Let 𝑓: 𝑋 → 𝑃1 be a rational function of degree 𝑑 ≥ 2 on a curve X. Let 𝑛 ≥ 2 be a divisor of d. Does there exist a curve Y with a rational function 𝑔: 𝑌 → 𝑃1 of degree n such that f factors through g Can we factor f in some sense? Note that by de Franchis' theorem the curve Y, if it exists, will almost always be isomorphic to X. 𝑑 Example. The rational function 𝑥 ↦ 𝑥 𝑑 clearly factors as 𝑎 ↦ 𝑥 𝑛 and 𝑦 ↦ 𝑦 𝑛 . Answer: Construct a cover of degree 4 such that over 0𝜖𝑃1 there is an unramified point and a point of ramification index 3. Then this cover can't factorize through of subcover of degree 2 (edit because it would force the ramification indexes to be 1 or even). eg: k(X) given by 𝑦 3 (𝑦 + 1) + 𝑥 = 0 and f given by (𝑥, 𝑦) ↦ 𝑥 . In this case X is even isomorphic to 𝑃1 ! Question #9 Let 𝑅(𝑡1 , … , 𝑡𝑛 )𝜖𝑅(𝑡1 , … , 𝑡𝑛 ) be a rational function in n variables. Suppose 𝑅𝑖 ≔ 𝑅(1,1, … ,1, 𝑡𝑖 , 1,1, … ,1) is a polynomial in 𝑡𝑖 for 1 ≤ 𝑖 ≤ 𝑛 . Is it true that 𝑅(𝑡1 , … , 𝑡𝑛 ) is a polynomial? Answer: 𝑅(𝑥, 𝑦) = 𝑥𝑦 + 1 𝑥+𝑦 Question #10 Let X be an integral and Noetherian scheme. We consider an open subset U of X and 𝑓𝜖Γ(𝑈, 𝑂𝑋 ) . Hence we may regard (U,f) as a rational function in X. I would like to prove that if 𝑝𝜖𝑋 − 𝑈 (and there is no rational function defined at p that is equivalent to (U,f)), there exists an open subset V of X containing p such that we may write the inverse rational function of (U,f) as (V,g) for certain 𝑔𝜖Γ(𝑉, 𝑂𝑋 ). The idea behind this is that if a rational function has a pole at a point, then its inverse rational function has a zero, and in particular it is defined at that point. Answer: This is not true. Morally, the point is that, in dimension at least 2, rational maps can fail to be defined at a point in ways that are more complicated than just "having a pole". 𝑦 𝑥 𝑥 𝑦 Counterexample: let 𝑋 = 𝐴2 , let p=(0,0), and let 𝑓 = . Then the inverse rational function is 𝑔 = but (every representative of) that function is not defined at p either. A bit more conceptually, you can view a rational function on X as a rational map 𝑓: 𝑋 → 𝑃1 . If f has a pole at p, then f is a regular map in a neighbourhood of p, and it maps p to ∞𝜖𝑃1 . But not every rational map to 𝑃1 is regular everywhere --- in fact, if X is a projective variety say, "most" rational maps to 𝑃1 will fail to be regular at some points.