3.4 Curves

def: Curve

$$\begin{align*} &\sphericalangle \\ &(X:=\mathbb P^n, T := \mathcal Z(\mathbb P^n)) \in \mathcal T \\ &C \in T^{c} \cap X^- \\ &\dim C = 1 \\ \hline \\ &C \in \mathfrak C \end{align*}$$

Function order at point

Proposition 3.4.1: Regular local ring at a smooth point of a curve is a discrete valuation ring

$$\begin{align*} &\sphericalangle \\ &(X:=\mathbb P^n, T := \mathcal Z(\mathbb P^n)) \in \mathcal T \\ &C \in \mathfrak C \\ &P \in \cancel S(C) \\ \hline \\ &\mathcal O^r_{C,P} \in \mathcal R^{\mathcal {DVR}} \end{align*}$$

Proof

Recall that $\mathcal O^r_{C,P}:=\{\frac{f}{g}, f, g, \in S(C)_d, g(P)\ne 0\}$ so it's a fraction of two polynomials with number of variables not greater than $n$ and thus $\mathcal O^r_{C,P} \in \mathcal R^{\mathcal N}$.

Next since $C \in T^{c} \cap X^-$ we know that $I(C) \in \mathfrak P_I(\overline F[x_1, \ldots, x_n])$ so $\mathcal O^r_{C,P} \in \mathcal R^{\mathcal {ID}}$.

Then $\mathcal O^r_{C,P} \in \mathcal R^{\mathcal {L}}$ since it's a localization by maximal ideal $\mathfrak m_P$.

By $(3.3.9)$ $\dim \mathcal O^r_{C,P}=\dim C = 1$. So we have

$$\mathcal O^r_{C,P} \in \mathcal R^{\mathcal N} \cap \mathcal R^{\mathcal {L}} \cap \mathcal R^{\mathcal {ID}}, \dim \mathcal O^r_{C,P}=1$$

Next, since $P \in \cancel S(C)$ by $(3.3.10) \dim_{\mathcal O^r_{C,P}/\mathfrak m_P}\mathfrak m_P / \mathfrak m_P^2=1$.

So by $(2.7.21)$ we have $\mathcal O^r_{C,P} \in \mathcal R^{\mathcal {DVR}}$

$\square$

def: Function order at point p

$$\begin{align*} &\sphericalangle \\ &(X:=\mathbb P^n, T := \mathcal Z(\mathbb P^n)) \in \mathcal T \\ &C \in \mathfrak C \\ &P \in \cancel S(C) \implies \mathcal O^r_{C,P} \in \mathcal R^{\mathcal {DVR}} \\ &v - \text{discrete valuation} \\ &f \in \mathfrak F(\mathcal O^r_{C,P})=F(C)\\ \hline \\ &\text{ord}_Pf:=v(f) \end{align*}$$

note

Recall that maximal ideal in $\mathcal O^r_{C,P}$ is a set of functions that vanish on $P$. Thus $\text{ord}_Pf$ is a multiplicity of root of function $f$ at point $P$.

Obviously poles (roots of denominator) counts as order with negative sign

note

From $(2.7.21)$ we know that there's a uniformizer $t$ with $\text{ord}_Pt=1$ such that $\lang f \rang_I=\lang t^{\text{ord}_P(f)} \rang_I \iff f = u t^{\text{ord}_P(f)}, u \in O^{r*}_{C,P}=\{f/g, f, g \in S(C)_d, f(P) \ne 0,g(P) \ne 0\}$

Proposition 3.4.2: Number of points non-zero and negative order

$$\begin{align*} &\sphericalangle \\ &(X:=\mathbb P^n, T := \mathcal Z(\mathbb P^n)) \in \mathcal T \\ &C \in \mathfrak C \cap \cancel S \\ &f \in F(C), f \ne 0 \\ \hline \\ &\begin{align*} &|P: \text{ord}_Pf\ne 0| < \infty \hspace{0.5cm} \tag{a}\\ &|P: \text{ord}_Pf< 0| =0 \implies f \in \overline F\hspace{0.5cm} \tag{b}\\ \end{align*} \end{align*}$$

Example: Order of function for ellpitic curve

Consider a curve over $\mathbb P^2$:

$$C: y^2z = x^3+xz^2,$$

We want to explore the order of a function

$$f(x,y,z):=\frac{y-x}{z}$$

at different points.

First let's check if this curve have some singular points:

$$J:= \begin{pmatrix} -2x^2-z^2 & 2yz & y^2-2zx \end{pmatrix} \\$$

The singular point would be some projective point $[x, y, z]$ that is not $[0,0,0]$ (which is prohibited in projective space) such that $\text{rank } J<2-1=1$ that is $J=(0\, 0\, 0)$. So if $2yz = 0$ it means that either $y = 0$ or $z=0$. If $z=0$ then $-2x^2-z^2=0 \implies x = 0, y^2 - 2zx = 0 \implies y=0$ so we got the point $[0,0,0]$ which is not in projective space. So $z \ne 0$ but then $y=0$ and we can use similar statements to figure out that it's impossible as well. So we established that $C$ is non-singular in every point. In particular every local ring of regular functions is a discrete valuation ring.

Now back to the function $f$. First note, that this function has the same level of homogeneity in numerator and denominator and thus is well-defined on the curve.

Let's figure out at which points $P$ we have $\text{ord}_Pf \ne 0$. Obviously this happens if either $y-x=0$ or $z=0$. If $z=0$ then from the equation of the curve $x=0$ and so the only point that works is $P_4 = [0,1,0]$. On the other hand if $z \ne 0$ we can assume $z=1$ and $y=x$ so $x^2=x^3+x \implies x = w_1, w_2, w_3, w_1 = 0, w_2=e^{i2\pi/3}, w_3=e^{-i2\pi/3}$. So we have 4 points with non-zero order:

$$P_1 = [w_1,w_1,1] \\ P_2 = [w_2,w_2,1] \\ P_3 = [w_3,w_3,1] \\ P_4 = [0,1,0] \\$$

Consider some point $P_i=[w_i, w_i, 1]$. We want to find maximal ideal in $\mathcal O^r_{P_i}$. Recall this is $\mathfrak m_{P_i}=\{f/g: f, g \in \overline F[x, y, z]/I(C)_d, f(P)=0, g(P) \ne 0\}$.

According to $(3.1.8)$ we have maximal ideal $M_{P_i}=\lang x-w_i, y -w_i, z-1 \rang_I, M_{P_i} \subseteq \overline F[x, y, x], M_{P_i} = \{f \in \overline F[x, y, z]: f(P)=0\}$.

Note that this is still not $\mathfrak m_{P_i}$ for three reasons: the polynomials are not homogeneous and the level of homogeneity in the numerator and denominator are different and the polynomials are in $\overline F[x,y,z]$ rather than in $\overline F[x,y,z]/I(C)$. We can fix this by homogeneization similar to $(3.3.2)$. That is we map $x \to x/z, y \to y/z, z \to z/z=1$ So we have:

$$\mathfrak m_{P_i} = \lang x/z-w_i, y/z-w_i, z/z - 1 \rang_I = \lang \frac{x-w_iz}{z}, \frac{y-w_iz}{z} \rang_I$$

But the curve $C$ is non-singular in every point that means that $\mathfrak m_{P_i}$ is generated by a single element - uniformizer. Which of $\frac{x-w_iz}{z}$ and $\frac{y-w_iz}{z}$ we should eliminate? Consider equation for $C$ in the form:

$$\frac{y^2}{z^2}=\frac{x^3}{z^3}+\frac{x}{z}=\frac{x}{z} \frac{z^2 + x^2}{z^2} \implies \\ \frac{x}{z}=\frac{y^2}{z^2}\cdot \frac{z^2}{z^2+x^2}$$

Note that $\frac{z^2}{z^2+x^2} \in \mathcal O^r_{P_i}$ and it's actually a unit in this ring. So for $w_1=0$ we know that $\mathfrak m_{P_1} = \lang \frac{y}{z} \rang_I$. The same holds for for $w_2, w_3$ (requires some calculations that we'll skip) so we have

$$\text{ord}_{P_i}(y-w_iz)/z=1 \\ \text{ord}_{P_i}(x - w_iz)/z=2 \\$$

Next

$$\text{ord}_{P_i}f=\text{ord}_{P_i}(\frac{y-w_iz}{z}-\frac{x-w_iz}{z})= \\ \text{ord}_{P_i}(\frac{y-w_iz}{z})+\text{ord}_{P_i}(1-\frac{x-w_iz}{y-w_iz}) \\$$

Consider

$$g(x,y,z):=\frac{x-w_iz}{y-w_iz}=\frac{x/z-w_i}{y/z-w_i}= \\ \frac{(x/z-w_i)(y/z+w_i)}{(y/z)^2-w_i^2}=\frac{(x/z-w_i)(y/z+w_i)}{(x/z)^3+x/z-w_i^2}=\\ \frac{(x/z-w_i)(y/z+w_i)}{(x/z)^3+x/z-w_i^3-w_i}=\\ \frac{(x/z-w_i)(y/z+w_i)}{(x/z-w_i)((x/z)^2+w_ix/z+w_i^2)+x/z-w_i} = \\ \frac{(y/z+w_i)}{(x/z)^2+w_ix/z+w_i^2 +1}$$

By plugging $w_i$ we can see that the denominator doesn't go to $0$ so

$$g(w_i,w_i,1)=0 \implies \text{ord}_{P_i}(1-\frac{x-w_iz}{y-w_iz}) = 0 \implies \\ \text{ord}_{P_i}f = 1$$

Now consider point $P_4 = [0,1,0]$. $M_{P_4}=\lang x, y-1, z \rang_I$ and $\mathfrak m_{P_4}=\lang x/y, z/y \rang_I$. We can eliminate $z/y$:

$$z/y=(x/y)^3+(x/y)(z^2/y^2)=(x/y)(x^2/y^2+z^2/y^2) \overset{\frac{x^2+z^2}{y^2} \in \mathcal O^r_{P_4}}\implies \\ \mathfrak m_{P_4}=\lang x/y \rang_I$$

Unlike the previous cases we cannot use the relation $z/y=(x/y)(x^2/y^2+z^2/y^2)$ to determine the order of $z/y$ because $(x^2/y^2+z^2/y^2)^{-1}\notin \mathcal O^r_{P_4}$. So let's find the order in a generic way:

$$z/y = g(x,y,z)x^a/y^a \implies g(x, y, z)=\frac{y^{a-1}z}{x^a}$$

If we consider $a=1, 2$ we'll have $g \notin \mathcal O^r_{P_4}$. For $a=3$:

$$g(x, y, z)=\frac{y^2z}{x^3}=\frac{y^2z}{y^2z-xz^2}=\frac{y^2}{y^2-xz} \in \mathcal O^r_{P_4}$$ $$\text{ord }_{P_4} z/y=3 \\ \text{ord }_{P_4} x/y=1 \\ \text{ord }_{P_4} x/z=1-3=-2 \\ \text{ord }_{P_4} y/z=-3 \\ \text{ord }_{P_4}f=\text{ord }_{P_4}\frac{x-y}{z} = \text{ord }_{P_4}\frac{y}{z}(x/y-1)=\\ \text{ord }_{P_4}(y/z)=-3$$

Finally we have

$$\text{ord }_{P_1} f= 1 \\ \text{ord }_{P_2} f= 1 \\ \text{ord }_{P_3} f= 1 \\ \text{ord }_{P_4} f= -3 \\$$

Notice that orders sum to 0. As we'll see later it's no coincidence.

Example: Singular curve

Consider a curve

$$C_2:y^2z = x^3+x^2z,$$ $$J_2:= \begin{pmatrix} -2x^2-2xz & 2yz & y^2-x^2 \end{pmatrix} \\ J_2([0,0,1])=\begin{pmatrix} 0 & 0 & 0 \end{pmatrix} \\$$

So this curve has a singular point $[0,0,1]$ and so it's not a non-singular curve and $\mathcal O^r_{[0,0,1]}$ is not a discrete valuation ring.

def: Ramification index

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C_1, C_2 \in \mathfrak C/F \\ &C_1 = \cancel S(C_1), C_2 = \cancel S(C_2) \\ &\phi: C_1 \rightsquigarrow_{Rat} C_2, \phi \ne \text{const} \\ &P \in C_1 \\ &t_{\phi(P)} - \text{uniformizer in } F(C_2)_F \\ \hline \\ &e_{\phi}(P)=\text{ord}_P(\phi^*t_{\phi(P)}), \end{align*}$$

If $e_{\phi}(P)=1$ we say that $\phi$ is unramified at P, otherwise it's ramified.

Example: Ramification index

Consider

$$\phi: \mathbb P^1/\R \to \mathbb P^1/\R, \phi([x,y]) \mapsto [x^3(x-y)^2, y^5]$$

We'd like to consider point $[0,1]$ and it's preimage $\phi^{-1}[0,1]=\{[0,1], [1,1]\}$

For point $P=[0,1]$ we'll have

$$\phi(P)= [0,1] \\ t_{[0,1]}=x/y \\ \phi^*t_{[0,1]}=\frac{x^3(x-y)^2}{y^5} \\ e_\phi(P)=\text{ord}_{P}\phi^*t_{\phi(P)}=3$$

For point $Q=[1,1]$ we'll have

$$\phi(Q)= [0,1] \\ t_{[0,1]}=x/y \\ \phi^*t_{[0,1]}=\frac{x^3(x-y)^2}{y^5} \\ t_{[1,1]}=\frac{x-y}{y} \\ e_\phi(Q)=\text{ord}_{P}\phi^*t_{\phi(P)}=\text{ord}_{[1,1]}\frac{x^3(x-y)^2}{y^5}=2$$

Thus we have

$$\sum_{P \in \phi^{-1}([0,1]) } e_\phi(P) = 5$$

Notice that it matches the degree of polynomials in $\phi$. We'll see later that it's no coincidence.

Morphisms and rational maps

Proposition 3.4.3: Smooth curve rational map is a morphism

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &C = \cancel S(C) \\ &(X, T^c, \mathcal O_X) \in \mathcal V \\ &\phi: C \rightsquigarrow_{Rat}X \\ \hline \\ &\phi: C \rightsquigarrow_{Rsp}X \end{align*}$$

Proof

Assume $\phi = [\phi_0, \ldots, \phi_k]$ Consider some point $P$ and define $m:=\min_i \text{ord}_P\phi_i$. Then $[t^{-m}\phi_0(P), \ldots, t^{-m}\phi_k(P)]$ has at least one component of order $0$ which is thus not equal to 0 and point is well defined.

Since point $P$ is arbitrary we proved that $\phi$ is a morphism.

$\square$

Example: Rational map to $\mathbb P^1$

Consider $C/F$ and $f = g/h \in F(C)_F$ then $\phi:=[f, 1]$ is a rational map on the set $U:=C \setminus Z(h)$. We can extend this map to points of $Z(h)$ by defining $\phi(P):=[h(P)\cdot f(P), 1\cdot h(P)]=[1,0]$. So

$$\phi(P) = \begin{cases} [f(P), 1], f \in \mathcal O^r_{C,P} \\ [1,0], f \notin \mathcal O^r_{C,P} \end{cases}$$

def: Rational map degree

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C_1, C_2 \in \mathfrak C/F \\ &\phi: C_1 \rightsquigarrow_{Rat} C_2 \\ \hline \\ &\deg\phi:=\begin{cases} [F(C_1)_F:\phi^*F(C_2)_F], \text{if } \phi \ne \text{const} \\ 0, \text{if } \phi = \text{const} \end{cases} \end{align*}$$

note

We call $\phi$ separable and purely inseparable iff the field extension has the same property

Example: Rational map degree

Consider

$$\phi: \mathbb P^1/\R \to \mathbb P^1/\R, \phi([x,y]) \mapsto [x^3(x-y)^2, y^5]$$

Then $F(\mathbb P^1)=F(x/y)$ so $\phi^*F(\mathbb P^1)=F(\frac{x^3(x-y)^2}{y^5})$ and thus

$$t:= x/y \\ \frac{x^3(x-y)^2}{y^5}=t^3(t^2-2t+1)=t^5-2t^4+t^3 \implies \\ \deg \phi=[F(x/y):F(\frac{x^3(x-y)^2}{y^5})]=5$$

$\square$

Proposition 3.4.4: Rational map of degree one is an isomopshism of smooth curves

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C_1, C_2 \in \mathfrak C \\ &C_i = \cancel S(C_i) \\ &\phi: C_1 \rightsquigarrow_{Rat} C_2 \\ &\deg \phi = 1 \\ \hline \\ &\phi: C_1 \cong C_2 \end{align*}$$

Proposition 3.4.5: Morphism between curves

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C_1, C_2 \in \mathfrak C \\ &\phi: C_1 \rightsquigarrow_{Rsp} C_2 \\ \hline \\ &(\phi = \text{const}) \vee (\phi(C_1)=C_2) \end{align*}$$

Proposition 3.4.6: Degree and ramification index

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C_1, C_2 \in \mathfrak C/F \\ &C_i = \cancel S(C_i)\\ &\phi: C_1 \rightsquigarrow_{Rat} C_2 \\ &\deg \phi > 0 (\phi \ne \text{const}) \\ &Q \in C_2 \\ \hline \\ &\sum_{P \in \phi^{-1}(Q)} e_{\phi}(P)=\deg \phi \end{align*}$$

Example: Degree and ramification index

Consider

$$\phi: \mathbb P^1/\R \to \mathbb P^1/\R, \phi([x,y]) \mapsto [x^3(x-y)^2, y^5]$$

From examples above we know that $\deg \phi = 5$ and $\sum_{P \in \phi^{-1}([0,1]) } e_\phi(P) = 5$.

$\square$

We already defined a pullback on ringed spaces and so on affine varieties. But we'll further need the inverse of pullback. The problem with the inverse is that $\phi_*: F(C_2) \to F(C_1)$ is not surjective so we cannot take $\phi_*^{-1}$ rightaway. To overcome this problem we use so-called norm map:

$$N_{F(C_1)/\phi^*F(C_2)}: F(C_1) \to \phi^*F(C_2), f \mapsto \prod_{\sigma \in \text{Gal}(F(C_1)/\phi^*F(C_2))}\sigma(f)$$

Thus image of $f$ under the norm map will be Galois invariant and will belong to $\phi^*F(C_2)$ so we can make the following definition:

def: Pushforward

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C_1, C_2 \in \mathfrak C\\ &\phi: C_1 \rightsquigarrow_{Rat} C_2 \\ \hline \\ &\phi_*:=(\phi^*)^{-1} \circ N_{F(C_1)/\phi^*F(C_2)} \end{align*}$$

Proposition 3.4.7: Function field is a finite separable extension of uniformizer

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n, F \in \mathcal F^{\mathcal P}, \text{char }F = p \in \mathfrak P \\ &C \in \mathfrak C/F \\ &P \in \cancel S(C) \\ &t_P - \text{ uniformizer in }F \\ \hline \\ &\begin{align*} &F(C)_F/_AF(t_P)_F \hspace{0.5cm} \tag{a}\\ &[F(C)_F:F(t_P)_F] < \infty \hspace{0.5cm} \tag{b}\\ &F(C)_F/_{\boxminus}F(t_P)_F \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

a.

By $(3.3.4)$ $\text{trdeg }F(C)_F/F=1$. On the other hand it's obvious that $t_P$ is transcendental over $F$ so $\text{trdeg }F(t_P)_F/F=1$, finally $t_P \subseteq F(C)_F$ so $F(C)_F/_AF(t_P)_F$

b.

Since $F(C)_F/_AF(t_P)_F$ and $F(C)_F$ is finitely generated over $F$ (and hence over $F(t_P)_F$) then $[F(C)_F:F(t_P)_F] < \infty$

c.

Consider some $z \in F(C)_F$, we know that $f \in \overline {F(t_P)_F}_{F(C)_F}$, that is $z$ is algebraic over $F(t_P)_F$ so there is $q(x):=\mathfrak{pol}_{F(t_P)_F}(a)=\sum_{i,j}a_{ij}t_P^ix^j$ (note that we can clear denominator in $t_P$ just by multiplying by least common factor of all denominators). Consider polynomial

$$q(t,x)=\sum_{i,j}a_{ij}t^ix^j$$

If there's a term $\sum_{i,j}a_{ij}t^ix^j$ with $j \ne 0 \pmod p$ then $\partial q(t,x)/\partial x \ne 0$ and so by $(2.9.9.b)$ $q(x) \in F(t_P)_{F,\boxminus}[x] \implies z \in F(t_P)_{F,\boxminus}$.

Assume otherwise that each term has $j = 0 \pmod p$ that is $q(t,x)=r(t,x^p)$. Note that since $F \in \mathcal F^{\mathcal P}$ by $(2.9.12.c)$every $a_ij=b_{ij}^p$ then by $(2.10.3)$ we have $r(t^p,x^p)=s(t,x)^p$ for some $s$. So we can write $q$ in terms of mod power $p$ of $t$ as follows:

$$q(t,x)=r(t,x^p)=\sum_{k=0}^{p-1}s_k(t,x)^pt^k$$

Consider $q(t_p, z)$ and note that

$$\text{ord}_Ps_k(t_P,z)^pt_P^k=p \cdot \text{ord}_Ps_k(t_P,z) + k = k \pmod p$$

Since $q(t_P,z)=0$ and each term $s_k(t_P,z)^pz^k$ has different order it's impossible to have anything but

$$s_0(t_P,z) = \ldots = s_{p-1}(t_P,z) = 0$$

(i.e. terms can't cancel out since the have different order of vanishing and can be represented as $ut^k$).

Obviously $\exists k$ such that $s_k(t,x)$ depends on $x$ (otherwise $q(t,x)$ is independent of $x$). Now $\deg_x s_k(t,x) \le 1/p\cdot q(t,x)$ and $s_k(t_P,z)=0$ which contradicts to $q(t,x)$ being minimal.

$\square$

def: Frobenius morphism

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F^{\mathcal P} \\ &\text{char }F = p \in \mathfrak P \\ &\mathbb P^n \\ &C \in \mathfrak C/F \\ &C^{\rho}:=Z({f^{\rho}: f \in I(C})) \\ \hline \\ &\rho_{\rightsquigarrow}: C \to C^{\rho}, [x_0, \ldots, x_n] \mapsto [x_0^{p}, \ldots, x_n^p] \end{align*}$$

note

Recall that we denote $f^{\phi}:=\phi(a_0)+\phi(a_1)x + \ldots$ and $\rho(a):=a^p$ thus $f^{\rho}:=a_0^p+a_1^px+ \ldots$.

note

By $(2.3.10)$ indeed $\rho: C \to C^{p}$ if we map $[x_0, \ldots, x_n] \mapsto [x_0^{p}, \ldots, x_n^p]$

note

We can apply Frobenius several times to get:

$$\rho_{\rightsquigarrow}^k: C \to C^{p^k}, [x_0, \ldots, x_n] \mapsto [x_0^{p^k}, \ldots, x_n^{p^k}]$$

Proposition 3.4.8: Degree of Frobenius morphism

$$\begin{align*} &\sphericalangle \\ &F \in \mathcal F^{\mathcal P} \\ &\text{char }F = p \in \mathfrak P \\ &\mathbb P^n \\ &C \in \mathfrak C/F \\ \hline \\ &\begin{align*} &(\rho_{\rightsquigarrow}^{k})^*F(C^{\rho^k})_F=F(C)^{p^k}_F := \{f^{p^k}: f \in F(C)_F\} \hspace{0.5cm} \tag{a}\\ &\rho_{\rightsquigarrow}^k - \text{purely inseparable} \hspace{0.5cm} \tag{b}\\ &\deg \rho_{\rightsquigarrow}^k = p^k \hspace{0.5cm} \tag{c}\\ \end{align*} \end{align*}$$

Proof

a.

$$(\rho_{\rightsquigarrow}^{k})^*F(C^{\rho^k})_F=\frac{f(x_0^{p^k}, \ldots, x_n^{p^k})}{g(x_0^{p^k}, \ldots, x_n^{p^k})} \overset{F \in \mathcal F^{\mathcal P}} = \\ (\frac{f(x_0, \ldots, x_n)}{g(x_0, \ldots, x_n)})^{p^k} = F(C)^{p^k}_F$$

b.

From $(a)$ it follows that we consider extension $F(C)/(\rho_{\rightsquigarrow}^{k})^*F(C^{\rho^k})_F=F(C) / F(C)^{p^k}_F$. Note that $\forall \alpha \in F(C): \alpha^{p^k} \in F(C)^{p^k}_F$ so by $(2.9.10.a)$ the extension is purely inseparable

c.

Take some non-singular point $P \in C$ and uniformizer $t_P$. Note that $F(C)/_{\boxminus}F(t_P)$ by $(3.4.7)$ and $F(C)/_{\Box}F(C)^{p^k}_F$ by $(b)$. Then we'll have the following tower field:

So the extension $F(C)_F/F(C)^{p^k}_F(t_P)$ is both separable and purely inseparable so by $(2.9.11)$ we have $F(C)_F=F(C)^{p^k}_F(t_P)$. Thus from $(a)$

$$\deg \rho^k=[F(C)^{p^k}_F(t_P):F(C)^{p^k}_F]$$

Obviously $t_P \in F(C) \implies t_P^{p^k} \in F(C)^{p^k}_F$. So all we need to show is $t_P^{p^{k}-1} \notin F(C)^{p^k}_F$. Assume otherwise $\exists f \in F(C): f^{p^k}=t_P^{p^{k}-1}$ then

$$p^{k}-1=\text{ord}_Pt_P^{p^{k}-1}=p^k\text{ord}_Pf \implies \\ \text{ord}_Pf = (p^{k}-1)/p=p^{k-1}-1/p$$

But that's impossible because order is integer.

$\square$

Divisors

def: Divisor

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ \hline \\ &D: = \sum_{P \in Y} n_P(P), n_P \in \Z, |P \in Y: n_P \ne 0| \lt \infty \\ &D \in \text{Div}(C) \end{align*}$$

note

Divisor if a formal sum of $n_P(P)$ that is it's just a mapping $C \to \Z$ where each point is assigned some integer. Additional property is the number of non-zero points is finite (i.e. the sum is finite)

note

$\text{Div}(C)$ is a actually a free abelian group where we define operation $+$ in the natural way ($n_P$ at each point is summed)

The Galois group $\text{Gal}(\overline F/F)$ acts on $\text{Div}(C)$ in the following way:

$$D^\sigma=\sum n_P(P^\sigma)$$

We say that divisor $D$ is defined over $F$ ($D/F$) if $\forall \sigma \in \text{Gal}(\overline F/F): D^{\sigma}=D$. We use the notation $\text{Div}_F(C)$ for a group of divisors defined over $F$.

note

If $D = n_1(P_1)+ \ldots +n_k(P_k)$ and we say that $D/F$ it doesn't necessary mean that $P_i^\sigma=P_i$, we just require that they're permuted in the right way. For example divisor $2((1+i, 1-i))+2((1-i, 1+i))$ is defined over $\R$.

def: Degree of divisor

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &D = \sum_{P \in C} n_P(P) \\ \hline \\ &\deg D:=\sum_{P \in C} n_P \end{align*}$$

def: Divisors of degree 0 subgroup

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ \hline \\ &\text{Div}^0(C) :=\{D \in \text{Div}^0(C): \deg D = 0\}\end{align*}$$

def: Divisor of a function

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &C = \cancel S (C) \\ &f \in F(C)^*\\ \hline \\ &\text{div}(f):=\sum_{P \in C}\text{ord}_P(f)(P) \end{align*}$$

Example: Divisors on elliptic curve

Assume $\text{char}F \ne 2$, let $e_1, e_2, e_3 \in \overline F$ and consider:

$$C: y^2 = (x-e_1)(x-e_2)(x-e_3)$$

If we consider a projective equation and assume $z=0$ then we'll get one point in $C: P_{\infty}=[0,1,0]$.

Define $P_i=(e_i, 0)$ and consider $f_i(x)=x-e_i$. It can have two points of non-zero order: $P_i$ and $P_\infty$ (for the latter note that projective version is $\frac{x-ze_i}{z}$). For the point $P_i$ we'll have $\mathfrak m_{P_i}=\lang (x-ze_i)/z, y/z \rang_I$. We already established in previous examples that in this case $\text{ord}_{P_i}f_i=2$.

For ${P_\infty}$ we estblished in examples above:

$$\text{ord}_{P_{\infty}}x/y=1 \\ \text{ord}_{P_{\infty}}z/y=3 \\ \text{ord}_{P_{\infty}}z/x=\text{ord}_{P_{\infty}}z/y + \text{ord}_{P_{\infty}}y/x = 3-1=2 \\ \text{ord}_{P_{\infty}}f_i=\text{ord}_{P_{\infty}} \frac{x-e_iz}{z}=\text{ord}_{P_{\infty}} \frac{x-e_iz}{y}+\text{ord}_{P_{\infty}}y/z= \\ \text{ord}_{P_{\infty}} x/y+\text{ord}_{P_{\infty}} 1-\frac{e_iz}{x}+\text{ord}_{P_{\infty}}y/z=1+0-3=-2$$

So we have:

$$\text{div}(x-e_i)=2(P_i)-2(P_\infty)$$

Now for function $y$ we have three zeros $P_i$ and some unknown order at $P_\infty$. We established in the examples above that order of zeros are 1 and just above that $\text{ord}_{P_{\infty}}z/y=3$ so we have:

$$\text{div}(y)=(P_1)+(P_2)+(P_3)-3(P_\infty)$$

$\square$

It's easy to check that

$$\text{div}(f^{\sigma})=(\text{div}(f))^{\sigma}$$

In particular $f \in F(C)_F \implies \text{div}(f) \in \text{Div}_F(C)$

Also note that $\text{ord}_P$ is a discrete valuation so

$$\text{div}: F(C)^* \rightsquigarrow_G \text{Div}(C)$$

def: Principal divisor

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &C = \cancel S (C) \\ &D \in \text{Div}(C) \\ &\exists f \in F(C)^*: D = \text{div}(f)\\ \hline \\ &D - \text{prinicipal} \\ &D \in \text{Div}_{\lang \rang}(C) \end{align*}$$

def: Equivalent divisors

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &C = \cancel S (C) \\ &D_1, D_2 \in \text{Div}(C) \\ &D_1 - D_2 \in \text{Div}_{\lang \rang}(C) \\ \hline \\ &D_1 \sim D_2 \end{align*}$$

def: Picard group

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &C = \cancel S (C) \\ \hline \\ &\text{Pic}(C):=\text{Div}(C)/\text{Div}_{\lang \rang}(C) \end{align*}$$

We denote $\text{Pic}_F(C)$ the divisors in $\text{Pic}_F(C)$ that are fixed by the Galois group. In general $\text{Pic}_F(C) \ne \text{Div}_F(C)/\text{Div}_{F,\lang \rang}(C)$.

Proposition 3.4.9: Zero divisor and constant funtion

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &C = \cancel S (C) \\ &f \in F(C)^* \\ \hline \\ &\text{div}(f)=0 \iff f \in \overline F^* \end{align*}$$

Proof

$\implies$

$\text{div}(f)=0$ implies $f$ doesn't have poles. Consider a rational map $C \to \mathbb P^1$:

$$\phi(P) = \begin{cases} [f(P), 1], f \in \mathcal O^r_{C,P} \\ [1,0], f \notin \mathcal O^r_{C,P} \end{cases}$$

By $(3.4.3)$ it's a morphism and it's not surjective since $f$ doesn't have poles so by $(3.4.5)$ it's constant.

$\impliedby$

Obvious.

$\square$

In the previous sections we discussed pullbacks and pushforwards of function fields. Now we define it for divisors. Assume $\phi: C_1 \rightsquigarrow_{Rat}C_2$:

$$\phi^*:\text{Div}(C_2) \to \text{Div}(C_1), (Q) \mapsto \sum_{P \in \phi^{-1}(Q)}e_\phi(P)(P) \\ \phi_*:\text{Div}(C_1) \to \text{Div}(C_2), (P) \mapsto \phi(P)(P)$$

Proposition 3.4.10: Pullbacks, pushforwards and divisors

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C_1, C_2, C_3 \in \mathfrak C \\ &C_i = \cancel S (C_i) \\ &\phi: C_1 \rightsquigarrow_{Rat} C_2, \phi \ne \text{const} \\ &f \in F(C_1)^* \\ &g \in F(C_2)^* \\ &D \in \text{Div}(C_2) \\ &\psi: C_2 \rightsquigarrow_{Rat} C_3, \phi \ne \text{const} \\ \hline \\ &\begin{align*} &\deg(\phi^*D)=(\deg \phi)(\deg D) \hspace{0.5cm} \tag{a}\\ & \phi^*(\text{div}(g))=\text{div}(\phi^*g), \hspace{0.5cm} \tag{b}\\ & \deg(\phi_*D) = \deg D , \forall D \in \text{Div}(C_2)\hspace{0.5cm} \tag{c}\\ & \phi_*(\text{div}(f))=\text{div}(\phi_*f)\hspace{0.5cm} \tag{d}\\ & \phi_*\circ \phi^* \text{acts as multiplication by } \deg \phi\hspace{0.5cm} \tag{e}\\ & (\psi \circ \phi)^*=\psi^* \circ \phi^*, (\psi \circ \phi)_*=\psi_* \circ \phi_*\hspace{0.5cm} \tag{f}\\ \end{align*} \end{align*}$$

Proof

a.

Easy to check using defintions and $(3.4.6)$

b.

$$\text{ord}_P(\phi^*f)=\text{ord}_P(\phi^*t_{\phi(P)}^{\text{ord}_{\phi(P)}f})=e_{\phi}(P)\cdot \text{ord}_{\phi(P)}f$$

c.

Obvious from definition

d, e, f

Left as an exercise

$\square$

Proposition 3.4.11: Degree of a principal divisor

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &C = \cancel S (C) \\ &f \in F(C)^* \\ \hline \\ &\deg(\text{div}(f))=0 \end{align*}$$

Proof

Consider a morphism $\phi: C \to \mathbb P^1$

$$\phi(P) = \begin{cases} [f(P), 1], f \in \mathcal O^r_{C,P} \\ [1,0], f \notin \mathcal O^r_{C,P} \end{cases}$$

Then from definitions $\text{div}(f)=\phi^*((0) - (\infty))$ (that is we are summing orders where $f$ is 0 (zeros) and substracting orders where it's a pole).

$$\deg\text{div}(f)=\deg\phi^*((0) - (\infty))=\deg \phi \cdot \deg((0) - (\infty)) = 0$$

$\square$

Differentials

def: Space of differential forms

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ \hline \\ & \Omega_C:=\{df: f \in F(C)\} \\ & 1.\ d(f+g) = df + dg \\ & 2.\ d(fg) = gdf + fdg \\ & 3.\ da = 0, \forall a \in \overline{F} \\ \end{align*}$$

note

$\Omega_C$ is a $\overline F$-vector space

Eventually we want to arrive at the definition of divisor of a differential. So let's start with pullback

def: Pull-back differential

$$\phi^*: \sum f_idx_i \mapsto \sum (\phi^*f_i)d(\phi^*x_i)$$

Proposition 3.4.12: Differentials properties

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &P \in C \\ &t_P - \text{uniformizer} \\ &p:= \text{char }F \\ \hline \\ &\begin{align*} &\Omega_C \in \mathcal M_{F(C)}, \dim_{F(C)} \Omega_C = 1 \hspace{0.5cm} \tag{a}\\ & \omega \in \Omega_C \implies \exists! g \in F(C): \omega=gdt_P, g:=\omega/dt_P\hspace{0.5cm} \tag{b}\\ &\text{ord}_P(\omega):=\text{ord}_P(\omega/dt_P) \text{ is independent of } t_P \hspace{0.5cm} \tag{c}\\ &\text{ord}_P(\omega) \ne 0 \text{ for finitely many } P \hspace{0.5cm} \tag{d}\\ &(p = 0) \vee (p \nmid \text{ord}_P(x)) \implies \text{ord}_P(f\cdot dx) = \text{ord}_P(f)+\text{ord}_P(x)-1 \hspace{0.5cm} \tag{e}\\ &\text{ord}_Pf \ge 0 \implies \text{ord}_Pdf/dt \ge 0 \hspace{0.5cm} \tag{f}\\ \end{align*} \end{align*}$$

def: Differential divisor

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &\omega \in \Omega_C \\ \hline \\ &\text{div}(\omega) := \sum_{P \in C}\text{ord}_P \omega(P) \end{align*}$$

Since by $(3.4.12.a)$ for every differentials $\omega_1, \omega_2$ we have $\omega_1= f\omega_2, f \in F(C)$ then:

$$\text{div}(\omega_1)-\text{div}(\omega_2) = \text{div}(f)$$

So each differential is in the same coset of the Picard group and we can make the following definition:

def: Canonical divisor

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ & \\ \hline \\ &\text{Div}_\Omega(C) - \text{the coset of Picard group for differential divisors} \\ &D \in \text{Div}_\Omega(C) - \text{canonical divisor} \end{align*}$$

Rieman-Roch theorem

We're almost ready to formulate the Rieman-Roch theorem. But before that couple more definitions.

def: Divisors partial order

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &D_1 = \sum{n_{p_1}}(P), D_2 = \sum{n_{p_2}}(P) \\ &\forall P \in Y: n_{p_1} \ge n_{p_2} \\ \hline \\ &D_1 \ge D_2 \end{align*}$$

def: L(D)

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ & D \in \text{Div}(C) \\ \hline \\ &L(D) := \{f \in F(C)^*: \text{div}(f) + D \ge 0\} \cup \{0\} \end{align*}$$

It can be proved that $L(D)$ is a finite dimenional $\overline F$-vector space.

We finally can state Rieman-Roch theorem. We'll use it extensively further for proving that there exist functions for some predefined divisor.

Theorem 3.4.13: Rieman-Roch

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ & D_\omega \in \text{Div}_\Omega(C) \\ \hline \\ &\exists g: \forall D \in \text{Div}(C): \dim_{\overline F} L(D)-\dim_{\overline F} L(D_\omega-D) = \deg D - g + 1 \end{align*}$$

def: Genus of a curve

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ \hline \\ &\text{gen}(C) \text{ is a } g \text{ from Rieman-Roch} \end{align*}$$

Corrolary 3.4.14: Divisors properties from Rieman-Roch

$$\begin{align*} &\sphericalangle \\ &\mathbb P^n \\ &C \in \mathfrak C \\ &D_\omega \in \text{Div}_\Omega(C) \\ &g:=\text{gen}(C) \\ \hline \\ &\begin{align*} & \dim L(D_\omega)=g\hspace{0.5cm} \tag{a}\\ & \deg D_\omega = 2g-2\hspace{0.5cm} \tag{b}\\ & \deg D > 2g-2 \implies \dim L(D)=\deg D - g+1\hspace{0.5cm} \tag{c}\\ & \text{gen}(\mathbb P^1)=0 \hspace{0.5cm} \tag{d}\\ \end{align*} \end{align*}$$

Proof

a.

Note that $L(0)$ is a set of functions with no poles so by $(3.4.2.b)$ we have $L(0)=\overline F \implies \dim L(0)=1$. Thus taking Rieman-Roch with $D=0$:

$$\dim L(0)-\dim L(D_\omega)=\deg 0 -g + 1 \implies \\ \dim L(D_\omega) = g$$

b.

Take $D=D_\omega$:

$$\dim L(D_\omega)-\dim L(0)=\deg D_\omega -g + 1 \implies \\ \deg D_\omega = 2g-2$$

c.

First consider some arbitrary $D'$ with $\deg D'<0$ and some $f \in L(D'), f \ne 0$. Then by $(3.4.11)$ we have $0=\deg(\text{div}(f))\ge \deg(-D')=-\deg D' \implies \deg D' \ge 0$. That means that $L(D')=\{0\}$. So we established:

$$\deg D'<0 \implies \dim L(D')=0$$

Since $\deg D_\omega=2g-2$ and $\deg D > 2g-2$ we have $\deg (D_\omega - D) <0 \implies \dim L(D_\omega - D)=0$ and so

$$\dim L(D)=\deg D - g+1$$

d.

Consider a differential $dx \in \Omega_{\mathbb P^1}$:

$$\forall \alpha \in \overline F: \text{ord}_\alpha dx=\text{ord}_\alpha d(x-\alpha)\overset{(3.4.12.e)}=\text{ord}_\alpha (x-\alpha)-1=0 \\ \text{ord}_{\infty} dx = \text{ord}_{\infty} x-1 = -2$$

So $\text{div}(dx)=-2(\infty)$. Next for any $\omega \in \Omega_{\mathbb P^1}$ by $(3.4.12.b)$ $\deg(\text{div}(\omega))=\deg(\text{div}(dx))=-2$.

In particular there cannot be any differential with $\text{div}(\omega)\ge 0$ and so $\dim L(D_\omega)=0$. From $(a)$ we get $g=0$.

$\square$