# 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

$\mathbb{P}^1~/~\mathbb{F}_1~\times~\mathsf{Spec}(\mathbb{Z})$

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 $g_{\mathbb{P}^1} = 0$.