Get Expert Help for Your Rational Functions Questions

Let $X$ be a locally Noetherian scheme and let $f$ be a rational function on $X$ (i.e. the equivalence class of a pair $(U,f)$, where $f\in <!--\; \in \; -->{\mathcal{O}}_X(U)$ and $U$ contains the associated points of $X$, under obvious equivalence relation).
While reading Vakil's notes I wondered how could we define poles of such a rational function. After some thought I came up with the following definition: I'd say that a regular codimension one point $p$ is a pole if it's not in the domain of definition of $f$. If $X$ is also an integral scheme (or at least if all the stalks of ${\mathcal{O}}_X$ are integral domains, in which case we can cover $X$ with integral schemes), then this definition would coincide with the usual one, namely using the discrete valuation at $p$.
But there is something unnatural about my definition, since I was not able to relate the rational function with the discrete valuation on ${\mathcal{O}}_{X,p}$ and consequently was not able to determine the order of the pole. So I'd like to know if it's possible to define a meaningful notion of poles for rational functions on locally Noetherian schemes and how would it relate to my definition. By extension, consider the same question about zeros.

Example
$\frac{6x+2}{x^2-<!--\; -\; -->9}=\frac{6x+2}{(x+3)(x-<!--\; -\; -->3)}$
I know how to find the vertical and horizontal asypmtotes and everything, I just don't know how to find end behavior for a RATIONAL function without plugging in a bunch of numbers.

Let $I\subseteq <!--\; \subseteq \; -->\mathbb{R}$ be an interval and $f:I\to <!--\; \backslash rightarrow\; -->\mathbb{R}$ a continuous function. We’ll say that f is totally rational if the following propositions are true for any $x\in <!--\; \in \; -->I$:
1. If $x\in <!--\; \in \; -->\mathbb{Q}$ then $f(x)\in <!--\; \in \; -->\mathbb{Q}$
2. If $f(x)\in <!--\; \in \; -->\mathbb{Q}$ then $x\in <!--\; \in \; -->\mathbb{Q}$
A simple example of such a function is the identity function f(x)=x. More generally any function of the form f(x)=ax+b with $a,b\in <!--\; \in \; -->\mathbb{Q}$ will do. Another class of functions that are totally rational are those of the form
$f(x)=\frac{ax+b}{cx+d}\text{with}a,b,c,d\in <!--\; \in \; -->\mathbb{Q}\text{and}x\ne <!--\; \ne \; -->-<!--\; -\; -->\frac{d}{c}.$
Besides functions of these kinds (and piecewise combinations thereof) I cannot find any other examples of such functions. It is easy to see, for instance, that any higher-order polynomial or rational function will fail condition (2).
But do other totally rational functions exist?

Question: Give an example of a rational function that has vertical asymptote $x=3$ now give an example of one that has vertical asymptote $x=3$ and horizontal asymptote $y=2$. Now give an example of a rational function with vertical asymptotes $x=1$ and $x=-<!--\; -\; -->1$, horizontal asymptote $y=0$ and x-intercept 4.
My solution: $(a)\frac{1}{(x-<!--\; -\; -->3)}$. This is because when we find vertical asymptote(s) of a function, we find out the value where the denominator is 0 because then the equation will be of a vertical line for its slope will be undefined.
$(b)\frac{2x}{(x-<!--\; -\; -->3)}$. Same reasoning for vertical asymptote, but for horizontal asymptote, when the degree of the denominator and the numerator is the same, we divide the coefficient of the leading term in the numerator with that in the denominator, in this case $\frac{2}{1}=2$
$(c)\frac{(x-<!--\; -\; -->4)}{(x-<!--\; -\; -->1)(x+1)}$. Same reasoning for vertical asymptote. Horizontal asymptote will be $y=0$ as the degree of the numerator is less than that of the denominator and x-intercept will be 4 as to get intercept, we have to make $y$, that is, $f(x)=0$ and hence, make the numerator 0. So, in this case; to get x-intercept 4, we use $(x-<!--\; -\; -->4)$ in the numerator so that $(x-<!--\; -\; -->4)=0⟹<!--\; ⟹\; -->x=4$.
Are my solutions correct of have I missed anything, concept-wise or even with the calculations?

A graph has a $y$-intercept at −5, no $x$-intercepts, and discontinuous points at (−1,−5) and (3,−5). I want to form an equation for this graph, but I don't know how the $y$-intercept relates to the graph of a rational function. Here's all I have so far:
$y=\frac{(x+1)(x-<!--\; -\; -->3)}{(x+1)(x-<!--\; -\; -->3)}$
I realize this isn't even close to being complete; I need some direction on how to finish the equation. Any help would be appreciated. Thanks.

Let $p_1,...,p_n$ be $n$ distinct points in $\mathbb{C}$, and let $q_1,...,q_n$ be $n$ distinct points in $\mathbb{C}$. Is there a rational function $f\in <!--\; \in \; -->\mathbb{C}(x)$ such that $f(p_i)=q_i$ for each $1\le <!--\; \le \; -->i\le <!--\; \le \; -->n$? Thinking of $\mathbb{C}(x)$ as a monoid under composition, this question asks whether $\mathbb{C}(x)$ acts $n$-transitively on $\mathbb{C}$.
More importantly, if this is false, then is there a nice description of the set of pairs $((a_1,...,a_n),(b_1,...,b_n))\in <!--\; \in \; -->{\mathbb{C}}^n\times <!--\; \times \; -->{\mathbb{C}}^n$ for which there is a rational function $f\in <!--\; \in \; -->\mathbb{C}(x)$ such that $f(a_i)=b_i$?
A rational function is determined by its preimages counting multiplicities at 3 distinct points, so I suspect this is false.

Let X,Y be affine varieties, we know that the coordinate ring of the product variety $X\times <!--\; \times \; -->Y$ satisfies $k[X\times <!--\; \times \; -->Y]\cong <!--\; \cong \; -->k[X]{\otimes <!--\; \otimes \; -->}_{k}k[Y]$.
My question is is it true that for rational function field, we also have $k(X\times <!--\; \times \; -->Y)\cong <!--\; \cong \; -->k(X){\otimes <!--\; \otimes \; -->}_{k}k(Y)$? If not, is there a way to relate $k(X\times <!--\; \times \; -->Y)$ with $k(X)$, and $k(Y)$?

Let $r_1,r_2,\dots <!--\; \dots \; -->,r_n$ be distinct complex numbers. Show that a rational function of the form
$f(z)=\frac{b_0+b_{1}z+\cdots <!--\; \cdots \; -->+b_{n-<!--\; -\; -->2}{z}^{n-<!--\; -\; -->2}+b_{n-<!--\; -\; -->1}{z}^{n-<!--\; -\; -->1}}{(z-<!--\; -\; -->r_1)(z-<!--\; -\; -->r_2)\cdots <!--\; \cdots \; -->(z-<!--\; -\; -->r_n)}$
can be written as a sum
$f(z)=\frac{A_1}{z-<!--\; -\; -->r_1}+\frac{A_2}{z-<!--\; -\; -->r_2}+\cdots <!--\; \cdots \; -->+\frac{A_n}{z-<!--\; -\; -->r_n}$
for some constant $A_1,\dots <!--\; \dots \; -->A_n$.
Solution:
Defining
$g(z)=f(z)-<!--\; -\; -->\frac{A_1}{z-<!--\; -\; -->r_1}+\frac{A_2}{z-<!--\; -\; -->r_2}+\cdots <!--\; \cdots \; -->+\frac{A_n}{z-<!--\; -\; -->r_n}$ where $A_1,\dots <!--\; \dots \; -->A_n$ are the residues of $f$ at each of the points $r_1,\dots <!--\; \dots \; -->r_n$.
I suppose I want to show that $g(z)$ is 0, but not sure how to do this. I know that $g(z)$ tends to 0 as $|z|\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}$....

I have been thinking of this problem recently and am wondering if anyone is able to prove/disprove it:
Let $R$ be a rational function, that is let $R(x)=p(x)/q(x)$ where $p$ and $q$ are polynomials of degree > 0. Assume that at some point x=a, the function $R(x)$ is undefined (e.g. by division by zero): That is $R(a)$undefined. Then
$\underset{x\to <!--\; \to \; -->a}{lim}R(x)=\underset{x\to <!--\; \to \; -->a}{lim}{R}^{\ast <!--\; \ast \; -->}(x),$
where $R^{\ast <!--\; \ast \; -->}(x)=p^{\prime }(x)/q^{\prime }(x)$, where $p^{\prime }(x)$ and $q^{\prime }(x)$ are the derivatives of the polynomials of $p$ and $q$ respectively.
Is this equation always true?

What is the general form of a rational function which has absolute value 1 on the circle $|z|=1$? In particular, how are the zeros and poles related to each other?
So, write $R(z)={\displaystyle \frac{P(z)}{Q(z)}}$, where $P,Q$ are polynomials in $z$. The condition specifies that $|R(z)|=1$ for all $z$ such that $|z|=1$. In other words, $|P(z)|=|Q(z)|$ for all $z$ such that $|z|=1$. What can we say about $P$ and $Q$?

I 'm trying to solve this exercise of Fulton Algebraic Curves:
Let $D=\sum <!--\; \sum \; -->n_{P}P$ be an effective divisor, $S=\{P\in <!--\; \in \; -->X:n_P>0\}$, $U=X\setminus <!--\; \setminus \; -->S$. Show that $L(rD)\subset <!--\; \subset \; -->\mathrm{\Gamma <!--\; \Gamma \; -->}(U,{\mathcal{O}}_X)$. Where $L(E)=\{f\in <!--\; \in \; -->K:{\text{ord}}_P(f)>-<!--\; -\; -->n_P,\mathrm{\forall <!--\; \forall \; -->}P\in <!--\; \in \; -->X\}$ (The linear vectorial space of the rational functions with pole multiplicity worst at $n_P$).
But I have no idea

Evaluate:
$\int <!--\; \int \; -->\frac{x}{x^3-<!--\; -\; -->x^2+1}dx$
None of the regular methods to integrate rational functions (substitution, partial fractions, etc.) seems to work.

Q is a rational function with $x\cdot <!--\; \cdot \; -->Q(x+2018)=(x-<!--\; -\; -->2018)Q(x),\mathrm{\forall <!--\; \forall \; -->}x\notin <!--\; \notin \; -->2018\text{ and }0$. If $Q(1)=1$. what is Q(2017)?
I tried substituting values in for Q(x) but there is no way to have both Q(1) and Q(2017) in the same equation, perhaps im missing a property of a rational function since i haven't used it yet.

If I have two algebraic numbers $\mathrm{\alpha <!--\; \alpha \; -->}$ and $\mathrm{\beta <!--\; \beta \; -->}$ and a rational function $w$ with rational coefficients (a function that's the ratio of two rational polynomials) that relates the two $\mathrm{\alpha <!--\; \alpha \; -->}=w(\mathrm{\beta <!--\; \beta \; -->})$ if I were to substitute $\mathrm{\beta <!--\; \beta \; -->}$ with one of it's algebraic cojugates ${\mathrm{\beta <!--\; \beta \; -->}}^{{}^{\prime }}$ in the rational function will I get a conjugate of $\mathrm{\alpha <!--\; \alpha \; -->}$ or rather is $w({\mathrm{\beta <!--\; \beta \; -->}}^{{}^{\prime }})$ an algebraic conjugate of alpha? I feel like there is a very obvious counter example but I've been struggling to find one.

If $K$ is an infinite field, and $K(x)$ is the field of rational function (in the indeterminate $x$), is it true that the extension $K(x)/K$ is a Galois extension?

Let $f$ be a meromorphic (equivalently, rational) function on the Riemann sphere with $k$ prescribed poles. What is the number of critical points (or critical values) of such function $f$? I believe the answer is $2k-<!--\; -\; -->2$, but why?

Let $L=\mathbb{C}(X,Y,Z)$ be the rational function field over the complex field and σ be automorphism of $L$ over $C$,
$\mathrm{\sigma <!--\; \sigma \; -->}(X)=Y,\mathrm{\sigma <!--\; \sigma \; -->}(Y)=Z,\mathrm{\sigma <!--\; \sigma \; -->}(Z)=X$
Moreover let $M$ be the intermediate field of the extension $L/\mathbb{C}$ fixed by the group $<\mathrm{\sigma <!--\; \sigma \; -->}>$. I think the degree of filed extension $L/M$ is 3 since the order of group $<\mathrm{\sigma <!--\; \sigma \; -->}>$ is 3. But is it true? I know this is true if the extension $L/C$ is finite degree Galois extension. But now the extension $L/C$ is infinite degree extension. So I don't know it is true.
Please give me some advice.

Let
$f(z)={\displaystyle \frac{z-<!--\; -\; -->a}{z-<!--\; -\; -->b}},z\ne b\ne a$
be a complex valued rational function.
How can I show that, if $|a|,|b|<<!--\; <\; -->1,$, then there is a complex number $z_0$ satisfying $|z_0|=1$ and $f(z_0)\in <!--\; \in \; -->\mathbb{R}$ ?
I have tried in many ways, but on success. Basically I tried to show that there is a unimodular complex number such that
${\displaystyle \frac{a-<!--\; -\; -->b}{z-<!--\; -\; -->b}}={\displaystyle \frac{\stackrel{¯<!--\; ¯\; -->}{a}-<!--\; -\; -->\stackrel{¯<!--\; ¯\; -->}{b}}{\stackrel{¯<!--\; ¯\; -->}{z}-<!--\; -\; -->\stackrel{¯<!--\; ¯\; -->}{b}}}.$
I could make a quadratic equation by using the fact that $\stackrel{¯<!--\; ¯\; -->}{z}={\displaystyle \frac{1}{z}},\mathrm{\forall <!--\; \forall \; -->}z\in <!--\; \in \; -->\mathrm{\partial <!--\; \partial \; -->}\mathbb{D}.$ Unfortunately I could not solve this question using that. So, I would like to see different (and somewhat general) approach.
Also, I would like to know that what happen if (both or atleast one) $a,b\notin \mathbb{D}.$Any comment or hint will be welcome. Thank you.

Let $f:\mathbb{C}-<!--\; -\; -->A\to <!--\; \backslash rightarrow\; -->\mathbb{C}$ be analytic and proper, where A is a finite point set in $\mathbb{C}$.
I.e. $f^{-<!--\; -\; -->1}(C)$ is compact for all compact sets $C\subset <!--\; \subset \; -->\mathbb{C}$.
Is it true that f is a rational function, i.e. $f(z)=\frac{p(z)}{q(z)}$ for complex polynomials p(z) and q(z), such that f has poles at A and $\mathrm{\infty <!--\; \infty \; -->}$?
Does the converse hold as well? I.e. if f is a rational function with poles at A and ∞, then f is proper.
I am pretty sure this is true, but am new with the concept of poles and am not sure where to start. I would appreciate any hints/help if possible.

Let $f(x)=p(x)/q(x)$ be a rational function. Let q have a root at some x=a. It is clear that if $q(x)$ divides $p(x)$ then the limit of $f$ as $x$ goes to a exists. But is the converse true as well?