Category Archives: course notes

ABC-theorem for Curves

Here we give the promised proof of the ABC-conjecture for function fields.

As always, $k$ is a perfect (e.g. finite) field and $K=k(X)$ is the function field of a smooth projective curve $X$ defined over $k$. We take elements $u,v \in K^*$ satisfying $u+v=1$ and consider the cover $u : X \mapsto \mathbb{P}^1_k$ corresponding to the embedding $k(u) \hookrightarrow K$. We want to determine the (schematic) zero- and pole-divisors of $u$ and $v$ and call them $A=div_0(u), B=div_0(v)$ and $C=div_{\infty}(u)=div_{\infty}(v)$.

Let $R$ be the integral closure of $k[u]$ in $K$, then in $R$ we can write the ideals $u$ and $(v)=(1-u)$ as products of prime-ideals (which correspond to schematic points of $X$)

$(u) = P_1^{e_u(P_1)} \cdots P_r^{e_u(P_r)}$
$(v) = Q_1^{e_u(Q_1)} \cdots Q_s^{e_u(Q_s)}$

and so $A = \sum_i e_u(P_i) [P_i]$ and $B = \sum_j e_u(Q_j) [Q_j]$. If $S$ is the integral closure of $k[\frac{1}{u}]$ in $K$, then we have in $S$ a decomposition

$(\frac{1}{u}) = R_1^{e_u(R_1)} \cdots R_t^{e_u(R_t)}$

and therefore $C = \sum_l e_u(R_l)[R_l]$. We already know that $deg(A)=deg(B)=deg(C)=n=[K : k(u)]$.

Case 1 : Let us assume that the field extension $K/k(u)$ is separable. Then, by the Riemann-Hurwitz formula (or rather, the scheme-version of it) we get the inequality (use that the genus of $\mathbb{P}^1_k$ is zero) :

$2 g_K – 2 \geq -2n + \sum^{scheme}_{P \in C} (e_u(P)-1) deg(P)$

Because for all points $e_u(P)-1 \geq 0$, the inequality only becomes better if we restrict the sum to a subset of points, say to the support of $A+B+C$ (that is to ${ P_1,\cdots,P_r,Q_1,\cdots,Q_s,R_1,\cdots,R_t }$). Then we get

$2 g_K -2 \geq -2n + \sum_{P \in Supp(A+B+C)} e_u(P)deg(P) – \sum_{P \in Supp(A+B+C)}deg(P)$
$~\qquad = -2n+3n-\sum_{P \in Supp(A+B+C)} deg(P)$

which gives us the required form of the ABC-conjecture for curves

$n=deg(u)=deg_s(u) \leq 2g_K – 2 + \sum_{P \in Supp(A+B+C)} deg(P)$

Case 2 : If $K/k(u)$ is not separable, take a maximal separable subfield $k(u) \subset M \subset K$, then by definition of $deg_s(u)$ and case 1 we have

$deg_s(u) \leq 2g_M – 2 + \sum_{P’ \in Supp(A’+B’+C’)} deg(P’)$

where $A’$ (resp. $B’$,$C’$) are the schematic fibers of the cover $Y \mapsto \mathbb{P}^1_k$ over the $k$-rational points $0$ (resp. $1$, $\infty$) and where $Y$ is the curve with function field $M$. We are done if we can show that $g_K=g_M$ and that in the cover $X \mapsto Y$ there is a unique point $P$ lying over each point $P’$ with $deg(P)=deg(P’)$.

As $K/M$ is purely inseparable, we have a tower of subfields

$M=M_0 \subset M_1 \subset \cdots \subset M_z=K$

such that $M_i / M_{i-1}$ is purely inseparable of degree $p$ for all $i$. That is, raising to the $p$-th power gives a field-isomorphism $M_i \simeq M_{i-1}$. The genus is a field-invariant, so $g_{M_i}=g_{M_{i-1}}$ and there is a bijection between the dvr’s in $M_i$ and $M_{i-1}$. That is, a bijection between points $P_i \leftrightarrow P_{i-1}$ of the corresponding curves $Y_i \mapsto Y_{i-1}$. Finally, because $t_{P_i}^p = t_{P_{i-1}}$ it follows that $deg(P_i)=deg(P_{i-1})$, and we are done by induction on $i$.

What is the ‘curve’ $\mathsf{Spec}(\mathbb{Z})$?

The mantra recited by $\mathbb{F}_1$-followers is that $\mathsf{Spec}(\mathbb{Z})$ is far too large to serve as the terminal object in the category of schemes, and, one should view it as a ‘geometric’ object over ‘something’ living ‘under $\mathbb{Z}$’ called $\mathbb{F}_1$ : the field with one element.

In this seminar we will encounter a fair number of proposals as to what this elusive object $\mathsf{Spec}(\mathbb{Z})$ viewed over $\mathsf{Spec}(\mathbb{F}_1)$ might be. Let’s start with the simplest and earliest proposal.

Smirnov’s proposal is that the smooth projective curve $\mathsf{Spec}(\mathbb{Z})$ should have as its schematic points the set $\{ 2,3,5,7,11,13,17,\cdots \} \cup \{ \infty \}$, that is, the set of all prime numbers together with $\infty$, and, that the degree of the ‘point’ $p$ should be equal to $log(p)$ whereas the degree of $\infty$ is equal to $1$.

Attempted explanation : We have seen before that a schematic point $P$ of a curve $C$ defined over $k$ corresponds to a discrete valuation ring in the function field $k(C)$ and that its degree $deg(P)$ equals $[\mathcal{O}_P/\mathfrak{m}_P : k]$.

By analogy, the schematic points of the ‘projective curve’ $\mathsf{Spec}(\mathbb{Z})$ should correspond to all discrete valuations on $\mathbb{Q}$, which by Ostrovski’s theorem are either the $p$-adic valuations $v_p(q)=n$ if $q=p^n \frac{r}{s}$ and $(r,p)=(s,p)=1$ or the real valuation $v_{\infty}(q) = -log |q|$ (minus sign because of the convention that the value of $0$ should be $\infty$).

To motivate the non-sensical definition of the degrees, recall that the degree of the divisor $div(f) = \sum_{P \in C} ord_P(f) [P]$ equals zero for all $f \in \overline{k}(C)$.

Now, if $f$ is in the function field $k(C)$, then its divisor must be invariant under the action of the Galois group $Gal(\overline{k}/k)$ (that is, $ord_{\sigma(P)}(f) = ord_P(f)$ for all Galois-automorphisms $\sigma$). But then, we can write $div(f)$ as a sum over the schematic points (which are the orbits of the geometric points under the action of the Galois group) and hence its degree is

$deg(div(P)) = \sum^{scheme}_{P \in C}~ord_P(f) deg(P) = 0$

where now the sum is taken over all schematic points of $C$. Once again, by analogy, if $f = \pm \frac{p_1^{e_1} \cdots p_r^{e_r}}{q_1^{f_1} \cdots q_s^{f_s}} \in \mathbb{Q}$, then its ‘divisor’ is

$div(f) = \sum_i e_i [p_i] – \sum_j f_j [q_j] – log |f| [\infty]$

and Smirnov’s proposal for the degrees of the scheme points of $\mathsf{Spec}(\mathbb{Z})$ is (up to a common multiple) the only one assuring that the degree of all such divisors is zero.

What is the field of constants?

In this proposal $\mathsf{Spec}(\mathbb{Z})$ is a smooth projective curve with function field $\mathbb{Q}$. To determine the ‘field’ over which it is defined we have (in analogy with the functionfield case where the field of constants is $K \cap \overline{k}$) to determine

$\mathbb{Q} \cap \overline{\mathbb{F}_1} = \mathbb{Q} \cap \pmb{\mu} = \{ +1,-1 \}$

So, $\mathsf{Spec}(\mathbb{Z})$ is not really a curve over $\mathbb{F}_1$, but rather over $\mathbb{F}_{1^2}$.

Smirnov’s surface

In a letter to Manin Smirnov explained that, as a first step towards the intersection theory on $\mathsf{Spec}(\mathbb{Z}) \times \mathsf{Spec}(\mathbb{Z})$ (which might lead to a proof of the Riemann hypothesis by mimicking Weil’s proof in the function field case) he embarked on the intersection theory in the somewhat easier product of two curves


Combining the above with the description of the projective line over $\mathbb{F}_1$, we can now depict this Smirnov surface

Recall that the schematic points of $\mathbb{P}^1~/~\mathbb{F}_1$ are $\{ 0,\infty \} \cup \{ [1],[2],[3],\cdots \}$ where the point $[n]$ represents all primitive $n$-th roots of unity and so has degree $\phi(n)$.

0-geometry: Hurwitz

The genus formula of Bernhard Riemann (left) and Adolf Hurwitz (right) asserts that if $\phi: C_1 \rightarrow C_2$ is a separable cover of curves, we have an inequality relating their genera:

$$2 g_{C_1} – 2 \geq \deg(\phi)(2 g_{C_2} -2) + \sum_{P \in C_1} (e_{\phi}(P)-1)$$

Let’s first understand all these terms. The cover $\phi: C_1 \rightarrow C_2$ is separable if the induced field-extension $\phi^{\ast}(\overline{k}(C_2)) \subset \overline{k}(C_1)$ is finite and separable. The dimension $[\overline{k}(C_1) : \phi^{\ast}(\overline{k}(C_2))]$ is called the degree of $\phi$. By using a discriminant argument as in these notes we know that for all but finitely many points $Q \in C_2$ there are exactly $\deg(\phi)$ points of $C_1$ lying over it.

In general, let $P \in C_1$ with corresponding discrete valuation ring $\mathcal{O}_P$ in $\overline{k}(C_1)$, then $\mathcal{O}_P \cap \phi^*(\overline{k}(C_2))$ is a discrete valuation ring in $\overline{k}(C_2) \simeq \phi^*(\overline{k}(C_2))$ and thus of the form $\mathcal{O}_Q$ for some $Q \in C_2$. Naturally we have $\phi(P)=Q$.

If $R$ is the integral closure of $\mathcal{O}_Q$ in $\overline{k}(C_1)$, then $R$ is a semi-local Dedekind domain and a PID. If $t_Q$ is a uniformizer of $\mathcal{O}_Q$ we have

$$(t_Q) = P_1^{e_1} \cdots P_r^{e_r}$$

where the $P_i$ are the maximal ideals of $R$ which corresponds to points $P_i \in C_1$. The integer $e_i$ is called the ramification index of $\phi$ in $P_i$ and will be denoted $e_{\phi}(P_i)$. Clearly we have that $\deg(\phi) = \sum_i e_{\phi}(P_i)$.

Further, for almost all $P \in C_1$ we will have $e_{\phi}(P)=1$. With these notations we can now begin the

Proof of the Riemann-Hurwitz inequality: Because $\phi$ is separable, we have an inclusion

$$\phi^*~:~\Omega_{C_2} \hookrightarrow \Omega_{C_1} \qquad \phi^*(f\,\mathrm{d}x)=\phi^*(f)\,\mathrm{d} \phi^*(x)$$

Take a point $Q \in C_2$ with uniformizer $t_Q \in \mathcal{O}_Q$ and write $\omega = f d t_Q \in \Omega_{C_2}$. For the finitely many $P_i \in C_1$ lying over $Q$ we have (as before) that
$\phi^*(t_Q) = u t_{P_i}^{e_i}$ with $e_i = e_{\phi}(P_i)$ and $u$ a unit in the discrete valuation ring $\mathcal{O}_{P_i}$. But then,

$$\phi^*(\omega) = \phi^*(f)\,\mathrm{d} \phi^*(t_Q) = \phi^*(f)\,\mathrm{d}(u t_{P_i}^{e_i}) = \phi^*(f)\left(e_i u t_{P_i}^{e_i-1} + \frac{\mathrm{d}u}{\mathrm{d}t_{P_i}} t_{P_i}^{e_i}\right)\,\mathrm{d}t_{P_i}$$

The valuation $\operatorname{ord}_{P_i}$ of $e_i u t_{P_i}^{e_i-1}$ is $e_i-1$ (unless $e_i=0$ in $k$, that is $char(k) | e_i$) whereas the valuation of $(\tfrac{\mathrm{d}u}{\mathrm{d}t_{P_i}}) t_{P_i}^{e_i}$ is $\geq e_i$. But then,

$\operatorname{ord}_{P_i}(\phi^* \omega) \geq \operatorname{ord}_{P_i}(\phi^*(f)) + e_i -1 = \operatorname{ord}_Q(f) e_i + e_i – 1$
$= \operatorname{ord}_Q(\omega) e_{\phi}(P_i) + e_{\phi}(P_i) -1$

Summing these inequalities over all $P \in C_1$ we get for the degree of the divisor

$\deg(\operatorname{div}(\phi^*(\omega))) \geq \sum_{P \in C_1} (e_{\phi}(P) \operatorname{ord}_{\phi(P)}(\omega)+e_{\phi}(P) -1)$
$=\sum_{Q \in C_2} (\sum_{P \in \phi^{-1}(Q)} (e_{\phi}(P) \operatorname{ord}_Q(\omega) + e_{\phi}(P) -1))$

and because we already know that $\deg(\phi) = \sum_{P \in \phi^{-1}(Q)} e_{\phi}(P)$ for all $Q \in C_2$, this is equal to

$=(\sum_{Q \in C_2} \deg(\phi) \operatorname{ord}_Q(\omega))+(\sum_{P \in C_1} (e_{\phi}(P)-1))$
$=(\deg(\phi))(\deg(\operatorname{div}(\omega))) + \sum_{P \in C_1} (e_{\phi}(P)-1)$

Plugging in the relation between the genus and the degree of the divisor of a nonzero differential form, we have here the Riemann-Hurwitz inequality!

The ABC-conjecture

In 1985 Joseph Oesterle (left) and David Masser (right) formulated the conjecture that for three relative prime integers satisfying $A+B=C$, the product of the prime divisors of $ABC$ is rarely much smaller than $C$.

More precisely, if $A,B,C \in \mathbb{Z}$ are such that $A+B=C$ and $\gcd(A,B,C)=1$, then their conjecture states that for each $\epsilon > 0$ there is a constant $M_{\epsilon}$ such that for all triples $(A,B,C)$ satisfying the conditions we have

$$\max( |A|, |B|, |C|) \leq M_{\epsilon}\left(\underset{p | ABC}{\prod} p\right)^{1+\epsilon}$$

The ABC-conjecture has several consequences, some obvious ones such as proving Fermat’s last theorem for large exponents, some less obvious such as Falting’s theorem. However, many people consider a proof the ABC-conjecture to be beyond the range of the available methods.

Since 2006 the ABC@Home project tries to find triples $(A,B,C)$ of large ‘quality’ meaning that the ratio

$$\operatorname{q}(A,B,C) = \frac{\log(C)}{\log(\operatorname{rad}(ABC))}$$

is as large as possible. To date, the champion-triple is $2+3^{10}109 = 23^5$ (discovered by Eric Reyssat) with a quality of $1.6299$.

If we write $u = \tfrac{A}{C}$ and $v=\tfrac{B}{C}$ then the ABC-conjecture can be recast as the statement that there is a constant $M_{\epsilon}$ such that when $u,v \in \mathbb{Q}^*$ satisfy $u+v=1$ we have

$$\max\left(\operatorname{ht}(u),\operatorname{ht}(v)\right) \leq M_{\epsilon} + (1+\epsilon)\left(\sum_{p | ABC} \log(p)\right)$$

where $A$ and $B$ are the numerators of $u$ and $v$ and $C$ is their common denominator, and where the ‘height’ $\operatorname{ht}(u)$ of a rational number $u=\tfrac{A}{C}$ with $(A,B)=1$ is $\max\left(\log|A|, \log|C|\right)$.

The latter formulation can be extended to the case of function fields of curves. So, let $K \in \mathsf{1Fields}$ with a perfect field of constants $k$ and suppose $u,v \in K^*$ are non-constants satisfying $u+v=1$. We need a substitute for the notion of height.

If $L$ is the maximal separable extension of $k(u)$ in $K$, then we call the dimension $[L : k(u)]$ the separability degree of $u$ and denote it with $\deg_s(u)$. Clearly, $\deg_s(u) \leq \deg(u) = [K : k(u)]$.

If $R$ is the integral closure of $k[u]$ in $K$, then there are maximal ideals $P_i$ in the Dedekind domain $R$ such that

$$(u) = P_1^{e_1} \cdots P_r^{e_r}$$

Because the local ring in $P_i$ is a discrete valuation ring in $K$ it determines a point in the curve $C$ with $K=k(C)$ (see here) also denoted $P_i$. But then, the zero-divisor of $u$ is $\operatorname{div}_0(u) = A = \sum_i e_i [P_i]$ with degree $\deg(A) = \sum_i e_i \deg(P_i)$.

Similarly, in the integral closure $S$ of $k[\tfrac{1}{u}]$ we have a decomposition

$$(\tfrac{1}{u}) = Q_1^{f_1} \cdots Q_s^{f_s}$$

and the pole-divisor of $u$ is $\operatorname{div}_{\infty}(u) = C = \sum_j f_j [Q_j]$ with degree $\deg(C) = \sum_j f_j \deg(Q_j)$. With these conventions, the ABC-conjecture for function fields can now be formulated as the following claim:

Let $K \in \mathsf{1Fields}/k$ and $u,v \in K^*$ with $u+v=1$, then

$$\deg_s(u) = \deg_s(v) \leq 2 g_K – 2 + \sum_{P \in \operatorname{Supp}(A+B+C)} \deg(P)$$

where $A=\operatorname{div}_0(u)$, $B=\operatorname{div}_0(v)$, $C=\operatorname{div}_{\infty}(u)=\operatorname{div}_{\infty}(v)$ and $g_K$ is the genus of $C$. Observe that there is no $\epsilon$ in this function field ABC-conjecture.

Perhaps surprisingly, the function-field ABC-conjecture can be proved fairly easily from the Riemann-Hurwitz genus formula. Details are in the book Number Theory in Function Fields by Michael Rosen (theorem 7.17) or in an upcoming prep-notes post.

0-geometry: Genus

In these rough prep-notes, we are working towards the proof of the Riemann-Hurwitz genus formula. “0-geometry” means we want to use only fields and their discrete valuations so that we can port some of this later to number fields.

Before we have seen that any field $K$ of transcendence degree $1$ over $k$ with $K \cap \overline{k} = k$ is really the function field $K = k(C)$ of a smooth projective curve $C$ defined over $k$.

A geometric point $P \in C$ is a discrete valuation ring $\mathcal{O}_P$ in the extended field $K^e = K \otimes \overline{k} = \overline{k}(C)$.

Aim: To determine the genus of $C$ from $K^e$ and the discrete valuation rings $\mathcal{O}_P$.

Divisors: For $f \in K^e$ and $P \in C$ we denote the valuation of $f$ in the discrete valuation ring $\mathcal{O}_P$ by $\operatorname{ord}_P(f)$ (that is, $f = u t^{\operatorname{ord}_P(f)}$ for $t$ is a uniformizer and $u$ a unit in $\mathcal{O}_P$).

We claim that there are only finitely many $P \in C$ such that $\operatorname{ord}_P(f) \not= 0$ and that $\sum_{P \in C} \operatorname{ord}_P(f) = 0$.

We can assume that $f \notin \overline{k}$ and so the subring $\overline{k}[f] \subset K^e$ is a polynomial ring. Let $R$ be the integral closure of $\overline{k}[f]$ in $K^e$ (which is a finite field extension of $\overline{k}(f)$ say of dimension $r$). Then $R$ is a Dedekind domain, projective of rank $r$ over $\overline{k}[f]$ and there are maximal ideals $\mathcal{P}_i$ in $R$ such that

$$(f) = \mathcal{P}_1^{e_1} \cdots \mathcal{P}_s^{e_s}$$

Because the localization of $R$ at $\mathcal{P}_i$ is a discrete valuation ring with residue field $\overline{k}$, each $\mathcal{P}_i$ defines a point $P_i \in C$ and we have $\sum_i e_i = r$.

Similarly, let $S$ be the integral closure of the polynomial algebra $\overline{k}[\frac{1}{f}]$ in $K^e$, then there are maximal ideals $\mathcal{Q}_j$ (corresponding to points $Q_j \in C$) such that

$$\left(\frac{1}{f}\right) = \mathcal{Q}_1^{f_1} \cdots \mathcal{Q}_t^{f_t}$$

and $\sum_j f_j = r$. But then the divisor of $f$ satisfies the claims

$$\operatorname{div}(f) = \sum_{P \in C} \operatorname{ord}_P(f) [P] = \sum_{i=1}^s e_i [P_i] – \sum_{j=1}^t f_j [Q_j]$$

Differentials forms: Consider the $K^e$-vectorspace $\Omega_C$ spanned by all ‘differential forms’ $\mathrm{d}f$ where $f \in K^e$, subject to the usual rules:

  • $\mathrm{d}(f+g)=\mathrm{d}f+\mathrm{d}g$ for all $f,g \in K^e$.
  • $\mathrm{d}(fg) = f\,\mathrm{d}g + g\,\mathrm{d}f$ for all $f,g \in K^e$.
  • $\mathrm{d}a = 0$ for all $a \in \overline{k}$.

We claim that $\Omega_C$ has dimension one. More precisely, if $x \in K^e$ is transcendental over $\overline{k}$ such that $K^e$ is a finite separable field extension of the subfield $\overline{k}(x)$, then $\Omega_C = K^e \mathrm{d}x$.

The proof is a computation. Let $g \in K^e$ have a minimal polynomial over $\overline{k}(x)$ of the form

$$G(Y) = Y^n + f_1 Y^{n-1} + \cdots + f_{n-1} Y + f_n$$

with all $f_i \in \overline{k}(x)$. Now consider these two polynomials in $\overline{k}(x)[Y]$ :

$G_1(Y) = n Y^{n-1} + (n-1) f_1 Y^{n-2} + \cdots + f_{n-1}$, and

$G_2(Y) = Y^n + \frac{\partial f_{1}}{\partial x} Y^{n-1} + \cdots + \frac{\partial f_{n-1}}{\partial x} Y + \frac{\partial f_n}{\partial x}$.

By the above equations among differential forms we get

$$0 = \mathrm{d} G(g) = G_2(g)\,\mathrm{d}x + G_1(g)\, \mathrm{d}g$$

Because $G_1(g) \not= 0$ by separability, it follows that $\mathrm{d}g \in K^e \mathrm{d}x$. Done!

Genus: In particular, if $t$ is a uniformizing parameter of the discrete valuation ring $\mathcal{O}_P$, then for any differential form $\omega \in \Omega_C$ there is a unique $f \in K^e$ such that $\omega = f\,\mathrm{d}t$. We define $\operatorname{ord}_P(\omega) =\operatorname{ord}_P(f)$. Clearly, this number depends only on $\omega$ (and $P$), but not on the choice of uniformizer (check!).

Slightly more involved is the claim that $\operatorname{ord}_P(\omega) \not= 0$ for finitely many $P \in C$. Here’s the idea:

Take $x \in K^e$ such that $K^e$ is a finite separable extension of $\overline{k}(x)$ of dimension $r$, write $\omega = f\,\mathrm{d}x$ and consider the corresponding cover $x\colon C \rightarrow \mathbb{P}^1$. As before, there are at most $r$ points of $C$ lying over a point $Q \in \mathbb{P}^1$.

Now, write $K^e=\overline{k}(x)(\alpha)$ and let $D \in \overline{k}(x)$ be the discriminant of the minimal polynomial of $\alpha$ over $\overline{k}(x)$. Then, away from the finite number of poles and zeroes of $D$, there are precisely $r$ points of $C$ lying over any point $Q \in \mathbb{P}^1$.

So, removing a finite number of points from $C$, in the remaining $P \in C$ we have $f(P) \not= 0,\infty$, $x(P) \not= \infty$ and $x-x(P)$ is a uniformizer of $\mathcal{O}_P$. But in such points we have $\operatorname{ord}_P(\omega) =\operatorname{ord}_P(f\,\mathrm{d}(x-x(P))) = 0$.

The number $\sum_{P \in C} \operatorname{ord}_P(\omega)$ is thus well-defined and we claim that it doesn’t depend on the choice of differential form. For, any other form can be written as $\omega’ = f \omega$ for some $f \in K^e$ and then we have

$$\sum_{P \in C} \operatorname{ord}_P(\omega’) = \sum_{P \in C} (\operatorname{ord}_P(f) +\operatorname{ord}_P(\omega))$$

and we know already that $\sum_{P \in C} \operatorname{ord}_P(f)=0$. The genus $g_C$ of the curve $C$ is then determined from that number by $2g_C – 2 = \sum_{P \in C}\operatorname{ord}_P(\omega)$.

Example: Take the projective line $\mathbb{P}^1$ corresponding to the purely transcendental field $\overline{k}(x)$ and consider $\omega = \mathrm{d}x$. In a point $\alpha \not= \infty$ we know that $x-\alpha$ is a uniformizer, so

$$\operatorname{ord}_{\alpha}(\omega) = \operatorname{ord}_{\alpha}(\mathrm{d}x) = \operatorname{ord}_{\alpha}(\mathrm{d}(x-\alpha)) = 0$$

In $\infty$ the uniformizer is $\frac{1}{x}$, whence

$$\operatorname{ord}_{\infty}(\omega) =\operatorname{ord}_{\infty}(\mathrm{d}x) =\operatorname{ord}_{\infty}\left(-x^2\,\mathrm{d}\left(\frac{1}{x}\right)\right) = -2$$

Thus, $\sum_{P \in \mathbb{P}^1} \operatorname{ord}_P(\omega) = -2$ and so the genus of the projective line $g_{\mathbb{P}^1} = 0$.