# Why is this fact about homogeneous polynomials true?

$$\newcommand{\ord}{\operatorname{ord}}$$Let $$V(P)$$ denote a curve, i.e. a smooth projective variety of dimension one, over $$\mathbb{C}$$, and let $$\mathscr{K}$$ denote the corresponding field of rational functions on $$V(P)$$. To simplify notation, for any place (point) $$p \in V(P)$$ and divisor $$D = \sum n_p p$$, we define $$\ord_p(D) = n_p$$. (Paraphrased from here.)

... Then $$\ord_p(f+g) \ge \min( \ord_p(f), \ord_p(g))$$ for all $$f,g \in \mathscr{K}^{\times}$$.

Why does this inequality hold? It seems like a simple fact about homogeneous polynomials, but I have already spent more than a day trying to figure out why this is true and have failed.

Version 2 of Question (Unnecessary to Read) $$\newcommand{\div}{\operatorname{div}_{V(P)} }$$This question is based on Exercise 3.5.15 in Algebraic Geometry: A Problem Solving Approach by Garrity, Belshof, et al.

Let $$L(D) := \{ F \in \mathscr{K}(V(P)): F = 0\ \text{ or }\div(F) + D \ge 0 \}$$ (see here, for example). If I have that $$F, G \in L(D)$$, how do I show that $$F+G \in L(D)$$?

This is necessary to show that $$L(D)$$ is a vector space and thus that the statement of the Riemann-Roch theorem makes sense.

Another way to phrase this:

How can one show that, where $$I_p(\dots)$$ denotes the intersection multiplicity at $$p \in \mathbb{CP}^2$$, $$I_p(V(f_2 g_1 + f_1 g_2) \cap V(P)) - I_p(V(f_2 g_2) \cap V(P)) \ge \\ \min\{ I_p(V(f_1) \cap V(P))-I_p(V(f_2) \cap V(P)),\quad I_p(V(g_1) \cap V(P)) - I_p(V(g_2)\cap V(P)) \}?$$

Probably it is a corollary of Bezout's theorem somehow?

Page 2 of this document [Proposition 1.7 (i)] says that closure under addition follows from the fact that $$v_p (F+G) \ge \min\{ v_p(F), v_p(G) \}$$ for any $$F,G \in \mathscr{K}(V(P))$$ and any $$p \in V(P)$$. However I do not understand either why this is true or why it "should" be true. $$v_p$$ is supposed to denote something about "valuation" or "discrete valuation", I am not sure, since I have not taken a course in commutative algebra yet, it seems to be related to this, but there it is given as an axiom -- I am not sure how to show that the order of a zero/pole of a rational function is a "discrete valuation", although that seems to be what this comes down to. This page also makes the same claim (that the order of poles/zeros is a discrete valuation on a rational function field).

Version 1 of Question (Unnecessary to Read)

I get that the Riemann-Roch space is a subset of the complex vector space of rational functions (see my previous question), therefore all I need to show is that it is closed under scalar multiplication and addition. The proof that it is closed under scalar multiplication is easy: $$0 \in L(D)$$ by definition, and for any $$\lambda \in \mathbb{C} \setminus \{0 \}$$ one has that $$\div(\lambda F) = \div(F)$$ where $$F \in \mathscr{K}(V(P))$$.

Question: How can one show that $$L(D)$$ is closed under addition?

However, I do not understand at all how to show that it is closed under addition. I know that $$\div(FG)=\div(F)+\div(G)$$ (and that $$\div(F^{-1})=-\div(F)$$), but this doesn't tell me anything I need to know about $$\div(F + G)$$. Specifically, if $$F= \frac{f_1}{f_2}, G= \frac{g_1}{g_2}$$, then $$F+G = \frac{g_2 f_1 + f_2 g_1 }{f_2 g_2}.$$

What can we say about the zeros of the sum of two distinct polynomials?

This is apparently a non-trivial issue, see e.g. https://mathoverflow.net/questions/30072/roots-of-sum-of-two-polynomials, http://www.sciencedirect.com/science/article/pii/S0022247X02000458

1. As far as I can tell (and I am probably incorrect), the zero set of $$g_2f_1 + f_2g_1$$, denoted $$V(g_2 f_1 + f_2 g_1)$$, is equal to $$V(g_2 f_1) \cap V(f_2 g_1).$$ Note that I am using intersection in the sense of generalized multisets, where an element can not only occur multiple times (a zero with multiplicity), but also appear negatively many times (a pole).

I can not figure out a simple way to translate intersection into the arithmetic of divisors.

For instance, if $$D_1$$ corresponded to $$V(a)\cap V(P)$$ and $$D_2$$ corresponded to $$V(b) \cap V(P)$$, then $$D_1 + D_2$$ would correspond to $$(V(a)\cap V(P))\cup(V(b) \cap V(P))= (V(a) \cup V(b)) \cap V(P)$$, i.e. the addition of divisors corresponds to the union of zero sets and pole sets and thus to the multiplication of rational functions.

Since multiplication between divisors is undefined, nothing seems to correspond to the intersection of zero and pole sets, and thus to the addition of rational functions. (My thinking in terms of indicator functions due to exposure to probability theory may be hampering me here; however, it is supposedly a natural way to understand multisets.)

2. This webpage here says that closure under addition should follow immediately from the fact $$\div(F +G) \ge \min \{ \div(F) + \div(G) \}$$ (specifically they use a term by term comparison for each point $$p$$). However, I don't see at all how this fact is true, or why it should be true. In fact, I would expect an inequality in the other direction to hold, since $$V(a) \cap V(b) \subset V(a), V(a) \cap V(b) \subset V(b)$$.

Defintions: $$V(P) \subset \mathbb{CP}^2$$ means the zero set of a homogeneous polynomial. I would suppose that in particular the $$V$$ stands for "variety". $$F$$ and $$G$$ are (equivalence classes of) rational functions in the field of rational functions $$\mathscr{K}(V(P))$$ corresponding to $$V(P)$$. $$D$$ denotes a divisor. $$\div (F)$$ denotes the divisor generated by the function $$F \in \mathscr{K}(V(P))$$ with respect to $$V(P)$$; specifically if $$F =\frac{f_1}{f_2}$$, where $$f_1$$ and $$f_2$$ are homogeneous polynomials of the same degree, then $$\div (F) = \sum_{p \in V(f_1) \cap V(P)} n_p p -\sum_{p \in V(f_2) \cap V(P)} n_p p$$ where $$n_p$$ is the order/multiplicity of the zero of $$f_1$$, $$f_2$$ respectively -- note of course that zeroes of $$f_2$$ correspond in particular to poles of $$F$$. Finally, $$L(D)$$, which I think is sometimes called a "Riemann-Roch space", is the set of functions in $$\mathscr{K}(V(P))$$ defined to be $$L(D) := \{ F \in \mathscr{K}(V(P)): F = 0 \quad or \quad \div(F) + D \ge 0 \}$$

• $V(f_2g_1+f_1g_2)$ is certainly not $V(f_2g_1) \cap V(f_1g_2)$ Commented Sep 24, 2016 at 15:23

At least in the scheme theory foundations, this reduces to the following:

Let $(A,m)$ be a Noetherian regular local ring of dimension one - equivalently a DVR. These are the local rings for (closed) points on your curve.

$m = (t)$, and $t$ is said to be a uniformizer. In particular, this means that every element of $A$ can be written as $u t^n$ , where $u$ is a unit in $A$ $n \geq 0$, and similarly every element in $K(A)$ (the fraction field) can be written as $u t^n$, where now $n \in \mathbb{Z}$. The valuation of $ut^n$ is set to be $n$.

Suppose $n \geq n'$.

Then $v(ut^n + u'^{n'}) = v((u + u't^{n' - n})t^n) = n$.

This is because $(u + u't^{n' - n})$ is a unit in $A$. If it is not, then it lives in $(t)$, so $u$ lives in $(t)$, which is a contradiction since $u$ is a unit.

Along the lines of AreaMan's answer, I am essentially copying part of the proof from p. 6 of Stichtenoth's Algebraic Function Fields and Codes (which I don't really understand) which shows that some function called $$v_P$$ defined in terms of the $$t^n$$ mentioned in their answer is a discrete valuation -- the only non-trivial part to show is that the triangle equality holds, which is also the only part of the definition of discrete valuation which I am interested in anyway:

In order to prove the triangle inequality consider $$x,y \in F$$ such with $$v_P(x) = n, v_P(y)=m$$. We can assume that $$n \le m < \infty$$, thus $$x = t^n u_1$$ and $$y=t^m u_2$$ with $$u_1, u_2 \in \mathcal{O}_P^{\times}$$. Then $$x+y = t^n(u_1 + t^{m-n}u_2)=t^n z$$ with $$z \in \mathcal{O}_P$$. If $$z=0$$, we have $$v_P(x+y)=\infty > \min\{n,m\}$$, otherwise $$z=t^k u$$ with $$k \ge 0$$ and $$u \in \mathcal{O}_P^{\times}$$. Therefore $$v_P(x+y) = v_P(t^{n+k}u)=n+k \ge n =\min\{ v_P(x), v_P(y) \}.$$

Perhaps I will understand this later once I fully internalize the definitions (if ever) -- so I will try to copy as many as possible, again from Stichtenoth's book, which is listed as the reference for the MIT link I mentioned.

To a place $$P \in \mathbb{P}_F$$ we associate a function $$v_P: F \to \mathbb{Z} \cup \{\infty\}$$ as follows: choose a prime element $$t$$ for $$P$$. Then every $$0 \not=z \in F$$ has a unique representation $$z = t^nu$$ with $$u \in \mathcal{O}_P^{\times}$$ and $$n \in \mathbb{Z}$$. Define $$v_P(z):= n$$ and $$v_P(0):= \infty$$.

A place $$P$$ of the function field $$F/K$$ is the maximal ideal of some valuation ring $$\mathcal{O}$$ of $$F/K$$. Every element $$t \in P$$ such that $$P = t\mathcal{O}$$ is called a prime element for $$P$$.

$$\mathbb{P}_F:=\{ P| P\text{ is a place of }F/K \}$$.

If $$\mathcal{O}$$ is a valuation ring of $$F/K$$ and $$P$$ is its maximal ideal, then $$\mathcal{O}$$ is uniquely determined by $$P$$, namely $$\mathcal{O}=\{z \in F | z^{-1}\not\in P \}$$ compare proposition 1.1.5(b). Hence $$\mathcal{O}_P:= \mathcal{O}$$ is called the valuation ring of the place $$P$$.

Proposition 1.1.5(b) -- Let $$\mathcal{O}$$ be a valuation ring of the function field $$F/K$$. Let $$0\not=x \in F$$. Then $$x \in P \iff x^{-1} \not\in \mathcal{O}$$. Proof: is obvious (according to Stichtenoth).

A valuation ring of the function field $$F/K$$ is a ring $$\mathcal{O} \subseteq F$$ with the following properties:

(1) $$K \subsetneq \mathcal{O} \subsetneq F$$, and (2) for every $$z \in F$$ we have that $$z \in \mathcal{O}$$ or $$z^{-1} \in \mathcal{O}$$ (or both).

An algebraic function field $$F/K$$ of one variable over $$K$$ is an extension field $$F \supseteq K$$ such that $$F$$ is a finite algebraic extension of $$K(x)$$ for some elment $$x \in F$$ which is transcendental over $$K$$.

$$K$$ is a field. I am not sure what $$K(x)$$ denotes.

The elements of $$F$$ which are transcendental over $$K$$ can be characterized as follows: $$z \in F$$ is transcendental over $$K$$ if and only if the extension $$F/K(z)$$ is of finite degree. The proof is trivial (according to Stichtenoth).

The simplest example of an algebraic function field is the rational function field; $$F/K$$ is called rational if $$F=K(x)$$ for some $$x \in F$$ which is transcendental over $$K$$. Each element $$0 \not=z \in K(x)$$ has a unique representation $$z = a \cdot \prod_i p_i(x)^{n_i}$$ in which $$0\not=a \in K$$, the polynomials $$p_i(x) \in K[x]$$ are monic, pairwise distinct, and irreducible, and $$n_i \in \mathbb{Z}$$. (Thus I don't know what $$K(x)$$ is, because $$K[x]$$ is the ring of polynomials over $$K$$.)

This definition (of valuation ring) is motivated by the following observation in the case of a rational function field $$K(x)$$: given an irreducible monic polynomial $$p(x) \in K[x]$$, we consider the set $$\mathcal{O}_{p(x)}:= \left\{ \frac{f(x)}{g(x)} \mid f(x), g(x) \in K[x], p(x) \not\mid g(x) \right\}.$$ (This is the same as the definition of function field over a curve $$V(p(x))$$ in Garrity et al I think. EDIT: No, the definition in Garrity et al is this quotiented by its maximal ideal -- note that an ideal is maximal if and only if the quotient ring is a field.) It is easily verified that $$\mathcal{O}_{p(x)}$$ is a valuation ring of $$K(x)/K$$. Note that if $$q(x)$$ is another irreducible monic polynomial, then $$\mathcal{O}_{p(x)} \not= \mathcal{O}_{q(x)}$$.

The maximal ideal of $$\mathcal{O}_{p(x)}$$ is $$P_{p(x)}:= \left\{ \frac{f(x)}{g(x)} \mid f(x),g(x) \in K[x], p(x) \mid f(x), p(x) \not\mid g(x) \right\}$$

Then there is a whole section starting on p.8 about how these definitions apply to the simple case of a rational function field, which I think is the case I am interested in.

I think I have reduced the problem to showing the following lemma:

Let $$V(P(x,y,z))$$ be a smooth projective curve/one-dimensional variety, and let $$f(x,y,z)$$ and $$g(x,y,z)$$ be two homogeneous polynomials in $$\mathbb{C}[x,y,z]$$. If $$\deg f = m$$ and $$\deg g =n$$, then let $$a,b \in \mathbb{C}[x,y,z]$$ be two other homogeneous polynomials such that $$\deg a =n$$ and $$\deg b =m$$.

Assume that $$I_p(f,P)=d_1$$ and $$I_p(g,P)=d_2$$. Then show that $$I_p(af + bg) \ge \min(d_1,d_2)$$.

This is fairly easy to show in $$\mathbb{CP}^1$$ when everything can factorize nicely, but I am not quite sure how to show it for $$\mathbb{CP}^2$$, when even when projecting to an affine patch we are not guaranteed that the resultant will factor nicely because then it need not be homogeneous.

When $$f$$ and $$g$$ are "relatively coprime", i.e. $$\min(d_1,d_2)=0$$, then it is also easy to show/follows immediately. Maybe I could use induction -- although that seems like that should be unnecessarily complicated for this problem.

Anyway the key insight is that how $$D$$ interacts at all with the numerators of $$F$$ and $$G$$ is completely irrelevant for showing whether or not they are in $$L(D)$$; all $$D$$ needs to do is to cancel out the "poles" of the denominator. (Why? Because the part of the divisor coming from the numerator added to the zero divisor will always be greater than or equal to the zero divisor, i.e. the only possible obstacle to a function being in $$L(D)$$ comes from the negative coefficients in its denominator possibly being too large.) If the denominators $$f$$ and $$g$$ are "relatively coprime", then the fact that $$D$$ can cancel out both $$div(f)$$ and $$div(g)$$ automatically means that $$D$$ can cancel out both "at the same time", i.e. that it can cancel out $$div(fg) =div(f)+div(g)$$.

If they aren't relatively coprime, then this doesn't follow, so we need to hope that one can "factor out" "$$(x-p)^{\min(d_1,d_2)}$$" from both $$f$$ and $$g$$ such that one can factor that out from $$af+bg$$ in the numerator (where $$a$$ is the numerator of the rational function $$G$$ and $$b$$ is the numerator of the rational function $$F$$), so that it can cancel/combine with the "$$(x-p)^{d_1+d_2}$$" in the denominator, so that $$D$$ only has to cancel out either "$$(x-p)^{d_1}$$" or "$$(x-p)^{d_2}$$", which we know it can do a priori since it can cancel out both $$f$$ and $$g$$, but so $$D$$ doesn't have to cancel out "$$(x-p)^{d_1+d_2}$$", which we have no reason to believe that it can. Again, the proof should be easier in $$\mathbb{CP}^1$$ than in $$\mathbb{CP}^2$$, because a lot of these expressions in quotation marks should be much closer to being literally true in the former than the latter. Thus I will still need to think some more about how to prove the result.

EDIT: I was wrong when I said that projecting to an affine patch wouldn't guarantee that the polynomials are homogeneous -- they are, this is Exercise 3.3.58 in Garrity et al. and/or Lemma 5 from Section 7 of Chapter 8 of Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms.

Then by 3.3.62 in Garrity et al. and/or Lemma 6 from Section 7 of Chapter 8 of Cox, Little, O'Shea, Ideals, Varieties, and Algorithms, we can factor the projection of the resultant onto an affine patch. In general I suppose this assumes that an appropriate choice of projective change of coordinates has been made beforehand. Then if $$(1:\frac{r}{s})$$ is the projection of the point $$p$$ onto the affine patch, and $$d=\min(d_1,d_2)$$, then $$(sx-ry)^d$$ should divide $$Res(f,P;z)$$ and $$Res(g,P;z)$$ (I think, I am not quite sure how to argue this more rigorously), so then (hopefully/plausibly) it will also divide $$Res(af+bg,P;z)$$ which then would complete the proof of the lemma.

Obviously again I would like to be able to argue this rigorously, but I think that most likely won't be possible until I learn more about how to factorize multivariate polynomials.