Algebraic Geometry · Study tools

InteractiveHartshorne, Ch. I §1: Affine Varieties

Affine varieties and the Zariski topology

An affine variety is the set of common zeros of some polynomials. The subject starts from a dictionary between such sets and ideals of the polynomial ring. The Nullstellensatz says exactly how faithful that dictionary is, and the explorer shows what the ideal knows that the picture does not.

The idea

Fix an algebraically closed field $k$ , and write $A=k[x_1,\dots,x_n]$ and $\mathbb{A}^n=k^n$ .

Algebraic sets. For

T\subseteq A

let

Z(T)=\{P\in\mathbb{A}^n : f(P)=0\text{ for all }f\in T\}

. For

Y\subseteq\mathbb{A}^n

let

I(Y)=\{f\in A : f(P)=0\text{ for all }P\in Y\}

. The sets

Z(T)

are the closed sets of the Zariski topology, because

Z(\mathfrak a)\cup Z(\mathfrak b)=Z(\mathfrak a\mathfrak b)

and

\bigcap_i Z(\mathfrak a_i)=Z\big(\sum_i\mathfrak a_i\big)

.

Affine variety. An irreducible closed subset

Y

of

\mathbb{A}^n

, meaning one that is not the union of two proper closed subsets. Its coordinate ring is

A(Y)=A/I(Y)

, and the dimension of

Y

as a topological space equals the Krull dimension of

A(Y)

.

Nullstellensatz. For every ideal

\mathfrak a\subseteq A

,

\ I(Z(\mathfrak a))=\sqrt{\mathfrak a}

. So

Z

and

I

are inverse, inclusion-reversing bijections between radical ideals and closed sets. Prime ideals correspond to varieties and maximal ideals to points.

A coarse topology. On

\mathbb{A}^1

the closed sets are the finite sets and

\mathbb{A}^1

itself, so any two nonempty open sets meet. The Zariski topology is not Hausdorff.

In the explorer $k=\C$ , but only the real points can be drawn. Slide the line and compare the ideal $(f,g)$ with what you see.

Explore

Z(f)

Z(g)

points of

Z(f,g)

The curves

Show

Position of the line, a

Worked example

Let $Y=Z(xy,\ xz,\ yz)\subset\mathbb{A}^3$ . Find its irreducible components, their prime ideals, and $I(Y)$ .

Which points satisfy all three equations?

The products of every pair of coordinates vanish, so at least two of $x,y,z$ are zero. That is the union of the three coordinate axes: $Y=Z(y,z)\cup Z(x,z)\cup Z(x,y)$ .
Why is each axis irreducible, and what is its ideal?

The $x$ -axis is $Z(y,z)$ , and $A/(y,z)\cong k[x]$ is a domain, so $(y,z)$ is prime. A prime ideal is radical, so the Nullstellensatz gives $I(Z(y,z))=(y,z)$ , and prime ideals correspond to irreducible sets. The same holds for the other two axes.
Are these the components, with nothing redundant?

None of the three axes contains another, and each is irreducible. A decomposition of a noetherian space into irreducible closed sets with no redundancy is unique, so these are the irreducible components. $Y$ has dimension 1.
Show that $I(Y)=(xy,xz,yz)$ . In particular, the defining ideal is already radical.

$I(Y)=(y,z)\cap(x,z)\cap(x,y)$ . All of these are monomial ideals, so their intersection is spanned by the monomials lying in all three. A monomial lies in $(y,z)$ if it contains $y$ or $z$ , and so on. A monomial in all three therefore contains at least two different variables, so it is divisible by $xy$ , $xz$ or $yz$ .

In $A(Y)$ we have $\bar x\,\bar y=0$ with $\bar x,\bar y\neq0$ . The zero divisors are the algebraic trace of the fact that $Y$ is reducible.

Check your understanding

Which subsets of $\mathbb{A}^1$ are closed in the Zariski topology?

The finite subsets, and $\mathbb{A}^1$ itself.
Closed intervals $[a,b]$ and their finite unions.An interval is infinite, and a nonzero polynomial has only finitely many roots. Also, $k$ is algebraically closed, so it carries no ordering.
Every subset.Then every infinite proper subset would be the zero set of some polynomials. A nonzero polynomial has finitely many roots.

k[x]

is a principal ideal domain, so every closed set is

Z(f)

for one polynomial

f

. That is a finite set if

f\neq0

and all of

\mathbb{A}^1

if

f=0

. This is the cofinite topology.

Over $k=\C$ , what is $Z(x^2+y^2-1,\ x-2)$ ?

The empty set.Over $\R$ , yes. Over $\C$ , substitute $x=2$ and solve for $y$ .
Two points.
One point, where the line is tangent to the circle.The line $x=a$ is tangent only for $a=\pm1$ .

With

x=2

,

y^2=-3

, so the points are

(2,\pm i\sqrt3)

. The ideal is proper, so by the Nullstellensatz

Z

cannot be empty over an algebraically closed field. In the explorer, slide

a

past 1.

Let $\mathfrak a=(x^2,\ y)\subseteq k[x,y]$ . What is $I(Z(\mathfrak a))$ ?

$(x^2,y)$ The function $x$ vanishes at every point of $Z(\mathfrak a)$ , but $x\notin\mathfrak a$ .
$(x,y)$
$(1)$ $Z(\mathfrak a)$ is not empty: the origin is in it.

Z(\mathfrak a)

is the origin, whose ideal is

(x,y)=\sqrt{\mathfrak a}

. The ideal

(x^2,y)

records that the parabola

y=x^2

and the line

y=0

are tangent there: it equals

(y-x^2,\ y)

.

Practice

Irreducible, or not?Factor first

Decide whether each of $Z(y^2-x^2)$ , $Z(y^2-x^3)$ and $Z(x^2+y^2)$ in $\mathbb{A}^2$ is irreducible ( $k$ algebraically closed of characteristic 0).

Hint

In the UFD $k[x,y]$ , $Z(f)$ is irreducible exactly when $f$ is a power of a single irreducible polynomial.

Solution

$y^2-x^2=(y-x)(y+x)$ : two lines, reducible. $y^2-x^3$ is irreducible: as a polynomial in $y$ it is monic of degree 2, so a factorisation would be $(y-p(x))(y+p(x))$ with $p^2=x^3$ , which is impossible. So the cusp is irreducible. Finally $x^2+y^2=(x+iy)(x-iy)$ , two lines through the origin. (Over $\R$ the polynomial is irreducible but its real zero set is a single point. The dictionary only works over an algebraically closed field.)

Dense open setsThe Zariski topology is coarse

Show that every nonempty open subset $U$ of an irreducible space $Y$ is dense. Deduce that a polynomial on $\mathbb{A}^2$ that vanishes on the complement of the line $x=0$ is zero.

Solution

$Y=\overline U\cup(Y\smallsetminus U)$ is a union of two closed sets, and $Y\smallsetminus U\neq Y$ . Irreducibility forces $\overline U=Y$ . For the second part: $\{f=0\}$ is closed and contains the dense open set $\{x\neq0\}$ of the irreducible space $\mathbb{A}^2$ , so it is all of $\mathbb{A}^2$ . Then $f\in I(\mathbb{A}^2)=0$ (an infinite field has no nonzero polynomial vanishing everywhere).

More practice: Hartshorne, Exercises I.1.1 and I.1.3.

InteractiveHartshorne, Ch. I §2: Projective Varieties

Projective space and points at infinity

The affine plane is missing points: parallel lines never meet, and a hyperbola runs off in two directions. Projective space adds the missing points. Once they are there, ellipses, parabolas and hyperbolas turn out to be the same curve seen from different charts.

The idea

Projective space.

\mathbb{P}^n

is the set of lines through the origin in

k^{n+1}

. A point is written

[a_0:\dots:a_n]

, and

[a]=[\lambda a]

for

\lambda\in k^*

. A homogeneous polynomial

F

of degree

d

has

F(\lambda a)=\lambda^dF(a)

, so it has no values on

\mathbb{P}^n

, but its zero set is well defined.

Projective variety. The closed sets of

\mathbb{P}^n

are the zero sets

Z(T)

of sets

T

of homogeneous polynomials. A projective variety is an irreducible closed subset. Homogeneous radical ideals other than

(1)

and the irrelevant ideal

(x_0,\dots,x_n)

correspond to nonempty closed sets.

Affine charts.

U_i=\{x_i\neq0\}

is homeomorphic to

\mathbb{A}^n

by dividing by

x_i

. With coordinates

[X:Y:Z]

on

\mathbb{P}^2

, the chart

Z\neq0

is the plane with

x=X/Z,\ y=Y/Z

, and its complement

\{Z=0\}

is the line at infinity.

Projective closure of a plane curve. If

f\in k[x,y]

has degree

d

, the closure of

Z(f)

in

\mathbb{P}^2

is

Z(F)

, where

F(X,Y,Z)=Z^d f(X/Z,\,Y/Z)

is the homogenisation. Its points at infinity are the

[x:y:0]

with

f_d(x,y)=0

, where

f_d

is the top-degree part of

f

.

For ideals, homogenise everything. The closure of

Z(\mathfrak a)

is cut out by the homogenisations of all elements of

\mathfrak a

. Homogenising a set of generators is not always enough.

The explorer draws the real projective plane as a disc: the affine plane fills the interior, and the boundary circle is the line at infinity, with opposite points of the circle identified. The second picture is the chart $Y\neq0$ , where the old line at infinity is an ordinary line.

Explore

b, coefficient of y² c, coefficient of y

Examples

The whole of

\mathbb{P}^2(\R)

. line at infinity

Z=0

the line

Y=0

points at infinity

The chart

Y\neq0

, coordinates

(X/Y,\ Z/Y)

. Here

Z=0

is the teal line, and the gold line

Y=0

is at infinity.

Worked example

The affine curve $y=x^3$ is smooth. Find its projective closure, its points at infinity, and what the closure looks like there.

Homogenise $f=y-x^3$ .

The degree is 3, so $F=Z^3\big(Y/Z-(X/Z)^3\big)=YZ^2-X^3$ .
Find the points at infinity.

Set $Z=0$ : then $X^3=0$ , so $X=0$ and the only point is $[0:1:0]$ . This matches the rule: the top-degree part of $f$ is $-x^3$ , which vanishes only in the direction $(0,1)$ .
Look at the closure near $[0:1:0]$ in the chart $Y\neq0$ , with coordinates $u=X/Y$ , $w=Z/Y$ .

Dividing $F$ by $Y^3$ gives $w^2=u^3$ . The point at infinity is $(u,w)=(0,0)$ , and there the curve is a cusp: both partial derivatives of $w^2-u^3$ vanish. A smooth affine curve can have a singular projective closure.
Compare with the parabola $y=x^2$ .

The closure is $YZ=X^2$ , again with the single point $[0:1:0]$ at infinity. In the chart $Y\neq0$ it reads $w=u^2$ , which is smooth at the origin with tangent line $w=0$ . That line is $Z=0$ : the parabola is tangent to the line at infinity, as the explorer shows.

Check your understanding

Where do the closures of the parallel lines $y=x$ and $y=x+1$ meet in $\mathbb{P}^2$ ?

Nowhere: they are parallel.Any two distinct lines in $\mathbb{P}^2$ meet in exactly one point.
At $[1:1:0]$
At $[0:0:1]$ That is the origin of the affine chart, which lies on only the first line.

Homogenise:

Y=X

and

Y=X+Z

. Subtracting gives

Z=0

, and then

X=Y

. The common point is the point at infinity in their shared direction

(1,1)

. Try "Parallel lines" in the explorer.

Which affine conic has a projective closure that is tangent to the line at infinity?

An ellipseAn ellipse has no real points at infinity. Over $\C$ it meets the line at infinity in two distinct points.
A parabola
A hyperbolaA hyperbola meets the line at infinity in two distinct points, its two asymptote directions.

The points at infinity are the zeros of the degree-2 part. For a parabola that part is a perfect square, such as

x^2

, so there is one point at infinity and it counts twice: the closure meets the line at infinity with multiplicity 2 there, which is tangency.

Why does $F=X+Y^2$ not define a subset of $\mathbb{P}^2$ ?

Whether $F$ vanishes depends on the representative: $F(1,1,0)=2$ but $F(-1,-1,0)=0$ , and $[1:1:0]=[-1:-1:0]$ .
Because $F$ is reducible. $X+Y^2$ is irreducible. The problem is that it is not homogeneous.
Because no polynomial defines a subset of $\mathbb{P}^2$ .Homogeneous polynomials do: their zero sets are well defined.

Only homogeneous polynomials scale by a single factor

\lambda^d

under

a\mapsto\lambda a

, and that is what makes "

F(a)=0

" independent of the representative.

Practice

A hyperbola is a parabolaChanging the chart

Show that the closure of $xy=1$ is $Z(XY-Z^2)$ , find its points at infinity, and find a chart in which the closure looks like a parabola.

Solution

Homogenising gives $XY=Z^2$ . At $Z=0$ , $XY=0$ , so the points at infinity are $[1:0:0]$ and $[0:1:0]$ , the directions of the two asymptotes. In the chart $X\neq0$ , with $v=Y/X$ and $w=Z/X$ , the equation reads $v=w^2$ : a parabola, which meets the line at infinity of that chart only at $[0:1:0]$ , tangentially.

The cubic at infinityA flex on the line at infinity

Find the points at infinity of the closure of $y^2=x^3-x$ . Show the closure is smooth there and that the line at infinity meets it there with multiplicity 3.

Solution

$F=Y^2Z-X^3+XZ^2$ . At $Z=0$ , $X^3=0$ , so the only point is $[0:1:0]$ . In the chart $Y=1$ with $u=X/Y$ , $w=Z/Y$ : $w-u^3+uw^2=0$ . The partial derivative in $w$ is $1$ at the origin, so the point is smooth, with tangent line $w=0$ , the line at infinity. Restricting to $w=0$ leaves $-u^3=0$ , a triple root, so the line at infinity meets the curve there with multiplicity 3: the point is a flex.

More practice: Hartshorne, Exercises I.2.9 and I.2.16.

Animated lessonHartshorne, Ch. I §3: Morphisms · §4: Rational Maps

Rational maps and blowing up

Morphisms are maps given locally by polynomials. Rational maps only need to be defined on a dense open set, and two varieties are birational when such maps go both ways. Blowing up a point is the basic birational map, and it pulls apart the branches of a singular curve.

The idea

Regular functions. A function on a variety

Y

is regular at

P

if near

P

it equals

g/h

with

g,h

polynomials and

h\neq0

. This gives the ring

\mathcal{O}(Y)

of regular functions, the local ring

\mathcal{O}_{P,Y}

and the function field

K(Y)

. For affine

Y

:

\mathcal{O}(Y)\cong A(Y)

,

\mathcal{O}_{P,Y}\cong A(Y)_{\mathfrak m_P}

, and

K(Y)

is the fraction field of

A(Y)

.

Affine varieties are rings. A morphism is a continuous map that pulls regular functions back to regular functions.

X\mapsto A(X)

is an arrow-reversing equivalence between affine varieties over

k

and finitely generated

k

-algebras that are domains.

Rational maps. A rational map

\varphi:X\dashrightarrow Y

is a morphism on some nonempty open set, with two such identified when they agree on an open set. It is birational if it has a rational inverse.

X

and

Y

are birational exactly when

K(X)\cong K(Y)

as

k

-algebras.

Blowing up the origin of $\mathbb{A}^2$ .

X=\{((x,y),[u:v]) : xv=yu\}\subset\mathbb{A}^2\times\mathbb{P}^1

, with

\pi:X\to\mathbb{A}^2

the projection. Over a point other than the origin there is one point of

X

; over the origin there is a whole

\mathbb{P}^1

, the exceptional line

E

, one point per line through 0. On the chart

u\neq0

the coordinates are

(x,t)

with

t=v/u

, and

y=tx

.

Strict transform. For a curve

C

through the origin,

\tilde C

is the closure of

\pi^{-1}(C\smallsetminus\{0\})

. Its points on

E

are the tangent directions of

C

at the origin.

Watch

Singular curve

The plane, coordinates

(x,y)

. the curve a line through 0

The blow-up, coordinates

(x,t)

with

y=tx

. exceptional line

E

A plane curve whose only singular point is the origin.
Every point other than the origin lies on exactly one line through the origin. Label the line by its slope $t=y/x$ . In the right picture that line becomes the horizontal line at height $t$ .
Send $(x,y)$ to $(x,\,y/x)$ . Away from $x=0$ this is invertible, with inverse $(x,t)\mapsto(x,tx)$ . Follow a point along the curve and its image.
The origin is replaced by the line $E=\{x=0\}$ , one point for each slope. Substituting $y=tx$ factors the equation as $x^2\cdot(\text{strict transform})=0$ .
The strict transform meets $E$ at the tangent directions of the curve at the origin.

Worked example

Show that the conic $C: x^2-2y^2=1$ is birational to $\mathbb{A}^1$ by projecting from the point $P=(1,0)$ , and that the closure $\bar C\subset\mathbb{P}^2$ is isomorphic to $\mathbb{P}^1$ .

Intersect $C$ with the line $y=t(x-1)$ through $P$ . What equation do you get, and which root do you already know?

Substituting: $x^2-1-2t^2(x-1)^2=0$ , that is $(x-1)\big[(x+1)-2t^2(x-1)\big]=0$ . The root $x=1$ is $P$ itself.
Find the second intersection point.

$x=\frac{2t^2+1}{2t^2-1},\qquad y=t(x-1)=\frac{2t}{2t^2-1}.$

Check: $x^2-2y^2=\dfrac{(2t^2+1)^2-8t^2}{(2t^2-1)^2}=\dfrac{(2t^2-1)^2}{(2t^2-1)^2}=1$ .
Why is this birational?

The map $t\mapsto(x,y)$ is defined for $2t^2\neq1$ , and on $C\smallsetminus\{P\}$ the inverse is $t=y/(x-1)$ , a rational function on $C$ . The two maps are inverse where both are defined, so $K(C)=k(t)$ .
What happens at $t=\pm1/\sqrt2$ and as $t\to\infty$ ? Use homogeneous coordinates.

Clearing denominators, $t\mapsto[2t^2+1:2t:2t^2-1]$ . Writing $t=s_0/s_1$ , this becomes $[s_0:s_1]\mapsto[2s_0^2+s_1^2:2s_0s_1:2s_0^2-s_1^2]$ , and the three entries vanish together only when $s_0=s_1=0$ . So it is a morphism on all of $\mathbb{P}^1$ . At $t=\pm1/\sqrt2$ the image is $[\pm\sqrt2:1:0]$ : the line through $P$ is parallel to an asymptote and the second point is at infinity. At $t=\infty$ the image is $[2:0:2]=P$ : the vertical line $x=1$ is tangent at $P$ . So $\mathbb{P}^1\to\bar C$ is a bijective morphism, with inverse given by projection from $P$ (extended by the tangent line at $P$ ). It is an isomorphism.

Check your understanding

For affine varieties $X$ and $Y$ , a morphism $X\to Y$ is the same thing as:

a $k$ -algebra homomorphism $A(Y)\to A(X)$
a $k$ -algebra homomorphism $A(X)\to A(Y)$ The arrow reverses: a map $X\to Y$ pulls functions on $Y$ back to functions on $X$ .
a map that is continuous in the Zariski topologyToo weak: any bijection $\mathbb{A}^1\to\mathbb{A}^1$ is a homeomorphism for the cofinite topology, and most are not morphisms.

\varphi

gives

\varphi^*:A(Y)\to A(X)

,

f\mapsto f\circ\varphi

. Conversely, a homomorphism is determined by where it sends the coordinate functions

y_j

, and those images are the components of a polynomial map.

Choose "Node" in the animation. In how many points does the strict transform meet the exceptional line $E$ ?

OneThat is what happens for the cusp, which has a single tangent direction.
Two
NoneThe curve passes through the origin, so the closure of its preimage must reach $E$ .

The node

y^2=x^2(1-x^2)

has two tangent lines

y=\pm x

at the origin, and the strict transform

t^2+x^2=1

meets

E=\{x=0\}

at

t=\pm1

, one point for each. The two branches no longer meet.

Are $\mathbb{A}^1$ and $\mathbb{A}^1\smallsetminus\{0\}$ birational? Are they isomorphic?

Birational, but not isomorphic.
Isomorphic, by $x\mapsto 1/x$ . $1/x$ is not defined at $0\in\mathbb{A}^1$ .
Neither: one is affine and the other is not. $\mathbb{A}^1\smallsetminus\{0\}\cong Z(xy-1)$ is affine. And birationality only sees a dense open set.

Both have function field

k(x)

. But

\mathcal{O}(\mathbb{A}^1\smallsetminus\{0\})=k[x,x^{-1}]

has the nonconstant unit

x

, while the only units of

k[x]

are constants, so the rings, and therefore the varieties, are not isomorphic.

Practice

Blowing up linesDirections become points

Show that the strict transform of the line $y=mx$ is the line $t=m$ , and conclude that two lines through the origin with different slopes have disjoint strict transforms.

Solution

Substituting $y=tx$ gives $tx-mx=x(t-m)$ . The factor $x$ is $E$ , and the rest is the line $t=m$ , which meets $E$ at the single point $t=m$ . Two different slopes give the disjoint lines $t=m$ and $t=m'$ . (Vertical lines live in the other chart $x=sy$ .)

Repeated blowing upA family of cusps

How many blow-ups of the origin are needed to make the strict transform of $y^2=x^{2k+1}$ smooth, for $k\ge1$ ?

Hint

Substitute $y=tx$ and see which curve you get.

Solution

$t^2x^2=x^{2k+1}$ gives the strict transform $t^2=x^{2k-1}$ , a curve of the same shape with $k$ lowered by one. After $k$ blow-ups we reach $t^2=x$ , which is smooth. So $k$ blow-ups are needed. (The ordinary cusp $k=1$ needs one.)

More practice: Hartshorne, Exercises I.4.4, I.4.10 and I.5.6.

InteractiveHartshorne, Ch. I §5: Nonsingular Varieties · §6: Nonsingular Curves

Singular points of curves

A point of a variety is nonsingular when the variety looks like affine space there to first order. For a plane curve the test is two partial derivatives. In the cubic below a node, a cusp and an isolated point appear exactly when roots collide.

The idea

Jacobian criterion. Let

Y\subseteq\mathbb{A}^n

be an affine variety of dimension

r

with

I(Y)=(f_1,\dots,f_t)

.

Y

is nonsingular at

P

if the matrix

\big(\partial f_i/\partial x_j(P)\big)

has rank

n-r

. For a plane curve

Z(f)

, with

f

irreducible,

P

is singular exactly when

f(P)=\pd{f}{x}(P)=\pd{f}{y}(P)=0

.

It is intrinsic.

Y

is nonsingular at

P

if and only if

\mathcal{O}_{P,Y}

is a regular local ring:

\dim_k \mathfrak m/\mathfrak m^2=\dim\mathcal{O}_{P,Y}

. The singular points form a proper closed subset of

Y

.

Multiplicity and tangent cone. Move

P

to the origin and write

f=f_r+f_{r+1}+\dots

with

f_i

homogeneous of degree

i

and

f_r\neq0

. Then

r

is the multiplicity of

P

, and the linear factors of

f_r

are the tangent directions. A node is a double point with two distinct tangents; a cusp here means a double point with a single repeated tangent.

Nonsingular curves. At a nonsingular point

P

of a curve,

\mathcal{O}_P

is a discrete valuation ring, so every nonzero rational function has an order of vanishing

\operatorname{ord}_P(f)\in\Z

. Nonsingular projective curves, with dominant morphisms, are equivalent to function fields of transcendence degree 1 over

k

, with arrows reversed.

The explorer uses $y^2=g(x)=(x-e_1)(x-e_2)(x-e_3)$ . Here $\pd{f}{y}=2y$ and $\pd{f}{x}=-g'(x)$ , so a singular point needs $y=0$ and $g(x)=g'(x)=0$ : a repeated root of $g$ .

Explore

Drag the three roots

e_1,e_2,e_3

along the

x

-axis. singular point tangent cone

Jump to

Worked example

Find all singular points of the affine curve $C:\ (x^2+y^2)^2=2(x^2-y^2)$ over $\C$ , and the tangent cone at each. Then look at the points at infinity.

Write $f=(x^2+y^2)^2-2(x^2-y^2)$ and compute both partial derivatives.

$\pd{f}{x}=4x(x^2+y^2-1),\qquad \pd{f}{y}=4y(x^2+y^2+1).$
Solve $f=\partial_xf=\partial_yf=0$ . Treat $y=0$ and $x^2+y^2=-1$ separately.

If $y=0$ : $\partial_xf=4x(x^2-1)$ vanishes at $x=0,\pm1$ , and $f(\pm1,0)=1-2=-1\neq0$ , while $f(0,0)=0$ . If $x^2+y^2=-1$ : $\partial_xf=-8x$ , so $x=0$ and $y=\pm i$ , but $f(0,\pm i)=1-2=-1\neq0$ . The only affine singular point is the origin.
What is the tangent cone at the origin?

The lowest-degree part of $f$ is $-2(x^2-y^2)=-2(x-y)(x+y)$ . So the origin is a node with tangents $y=\pm x$ .
Homogenise and find the points at infinity. Are they singular?

$F=(X^2+Y^2)^2-2Z^2(X^2-Y^2)$ . At $Z=0$ , $(X^2+Y^2)^2=0$ , giving $[1:\pm i:0]$ . Near $[1:i:0]$ , set $X=1$ , $Y=i+w$ : then $1+Y^2=2iw+w^2$ , so $F=(2iw+w^2)^2-2Z^2(2-2iw-w^2)=-4w^2-4Z^2+\dots$ . The lowest form $-4(w^2+Z^2)$ has two distinct factors: another node, and the same happens at $[1:-i:0]$ . So $\bar C$ is a quartic with three nodes. (A plane quartic has arithmetic genus 3, and each node lowers the genus by one, so $\bar C$ is a rational curve.)

Check your understanding

Is the affine curve $y^2=x^3+x^2+x$ singular anywhere over $\C$ ?

Yes, at the origin.The origin is on the curve, but $\pd{f}{x}=-(3x^2+2x+1)$ equals $-1$ there.
No, it is nonsingular.
Yes, at two complex points.A singular point of $y^2=g(x)$ needs a repeated root of $g$ . Are the roots of $x(x^2+x+1)$ distinct?

g=x(x^2+x+1)

has the three distinct roots

0

and

\tfrac{-1\pm i\sqrt3}2

, so

g

and

g'

have no common root and the curve is nonsingular.

On the parabola $y=x^2$ , what is $\operatorname{ord}_P(y)$ at $P=(0,0)$ ?

1 $y$ is a coordinate on the plane, but along this curve it vanishes faster than a local parameter.
2
0Order 0 would mean $y(P)\neq0$ , but $P$ has $y=0$ .

The point is nonsingular and

x

is a local parameter (a generator of the maximal ideal of the DVR

\mathcal{O}_P

). On the curve

y=x^2

, so

\operatorname{ord}_P(y)=2

: the tangent line

y=0

meets the parabola with multiplicity 2.

Over $\R$ , $y^2=x^2(x-1)$ has an isolated real point at the origin. Over $\C$ , the origin is:

a nonsingular pointBoth partial derivatives vanish there.
a node, with tangents $y=\pm ix$
a cuspThe tangent cone $y^2+x^2$ has two distinct factors over $\C$ .

f=y^2+x^2-x^3

, whose lowest form is

y^2+x^2=(y-ix)(y+ix)

. The two branches are complex conjugate, so only their common real point is visible. In the explorer, use "Isolated point".

Practice

A one-parameter familyNode, cusp, isolated point

For each $\lambda\in\R$ , describe the singular point of $y^2=x^3+\lambda x^2$ and its tangent cone, over $\R$ and over $\C$ .

Solution

The only singular point is the origin, of multiplicity 2, with tangent cone $y^2-\lambda x^2$ . For $\lambda>0$ : a node with real tangents $y=\pm\sqrt\lambda\,x$ . For $\lambda=0$ : a cusp, tangent cone $y^2=0$ . For $\lambda<0$ : over $\C$ a node with tangents $y=\pm i\sqrt{-\lambda}\,x$ ; over $\R$ the origin is an isolated point.

Smooth, then singular at infinityCheck the closure too

Show that $y^2=x^4+1$ is a nonsingular affine curve, but that its projective closure is singular at a point at infinity. Find the tangent cone there.

Solution

Affine: $\pd{f}{x}=-4x^3$ , $\pd{f}{y}=2y$ vanish together only at the origin, which is not on the curve. Closure: $F=Y^2Z^2-X^4-Z^4$ . At $Z=0$ , $X=0$ , so the only point at infinity is $[0:1:0]$ . In the chart $Y=1$ : $z^2-x^4-z^4=0$ . Its lowest form is $z^2$ , so the point has multiplicity 2 with a single tangent $z=0$ counted twice. (It is a tacnode: two smooth branches $z\approx\pm x^2$ tangent to each other.)

More practice: Hartshorne, Exercises I.5.1, I.5.3 and I.6.2.

InteractiveHartshorne, Ch. I §7: Intersections in Projective Space

Intersections and Bézout's theorem

A line meets a plane cubic in three points, always, once you count points at infinity, complex points and tangencies correctly. Drag the line and watch each of the three corrections appear.

The idea

Hilbert polynomial and degree. For a projective variety

Y\subseteq\mathbb{P}^n

of dimension

r

, with homogeneous coordinate ring

S(Y)

, the function

m\mapsto\dim_k S(Y)_m

agrees for large

m

with a polynomial

P_Y

of degree

r

. Its leading coefficient is

\deg Y/r!

. A hypersurface cut out by an irreducible form of degree

d

has degree

d

.

Intersection multiplicity of plane curves. If the curves

f=0

and

g=0

have no common component through

P

, then

i_P=\dim_k\mathcal{O}_P/(f,g)

. It is 1 exactly when

P

is a nonsingular point of both curves and their tangent lines differ. For a line through

P

, it is the multiplicity of the corresponding root of

f

restricted to the line.

Bézout's theorem. Plane curves of degrees

d

and

e

with no common component satisfy

\sum_P i_P = de

, the sum over all points of

\mathbb{P}^2(k)

. More generally, if

Y\subseteq\mathbb{P}^n

is a variety of dimension at least 1 and

H

a hypersurface not containing it, then

\sum_j i(Y,H;Z_j)\deg Z_j=\deg Y\cdot\deg H

, summed over the irreducible components

Z_j

of

Y\cap H

.

All three corrections are needed. Count points at infinity (work in

\mathbb{P}^2

), complex points (

k

algebraically closed), and multiplicities.

Explore

Drag the two points A and B that define the line. cubic line intersection points

The cubic

Try

Worked example

Intersect the circle $C: x^2+y^2=1$ with the parabola $D: y=x^2-1$ . Find every intersection point with its multiplicity and check Bézout.

Eliminate $x$ .

From $D$ , $x^2=y+1$ . Then $y+1+y^2=1$ , so $y(y+1)=0$ .
List the affine intersection points.

$y=0$ gives $x=\pm1$ : the points $(\pm1,0)$ . $y=-1$ gives $x^2=0$ : the point $(0,-1)$ , apparently twice.
Compute $i_P$ at $P=(0,-1)$ from the local ring.

In $\mathcal{O}_P/(f,g)$ we may use $g$ to replace $y$ by $x^2-1$ . Then $f$ becomes $x^2+(x^2-1)^2-1=x^4-x^2=x^2(x^2-1)$ . Near $P$ , $x^2-1$ is a unit, so $\mathcal{O}_P/(f,g)\cong k[x]_{(x)}/(x^2)$ , of dimension 2. The curves are tangent at $P$ .
Are there intersections at infinity? Check Bézout.

The circle's points at infinity are $[1:\pm i:0]$ and the parabola's only one is $[0:1:0]$ , so they share none. Total: $1+1+2=4=2\cdot2$ .

Check your understanding

Two distinct circles meet in at most 2 points of the plane, but Bézout predicts 4. Where are the other two?

At $[1:i:0]$ and $[1:-i:0]$ , which lie on every circle.
At complex points of the affine plane.Subtracting the two equations leaves a linear equation, so the affine intersection, real or complex, has at most 2 points.
Bézout does not apply to circles.Two distinct circles are conics with no common component, so it applies.

A circle homogenises to

X^2+Y^2+aXZ+bYZ+cZ^2=0

. At

Z=0

that leaves

X^2+Y^2=0

, whatever the circle. So every circle passes through these two "circular points", and two circles share them.

A line through the node of $y^2=x^2(x+2)$ with slope $m\neq\pm\sqrt2$ meets the curve at the node with multiplicity:

1At a point of multiplicity $r$ , every line through it meets the curve with multiplicity at least $r$ .
2
3That happens only for the two tangent lines $y=\pm\sqrt2\,x$ .

Substituting

y=mx

gives

x^2(x+2-m^2)=0

: a double root at the node and a third point at

x=m^2-2

. Choose "Through the node" in the explorer.

A curve $Y\subset\mathbb{P}^3$ has Hilbert polynomial $P_Y(m)=3m+1$ . What is its degree?

1That is the constant term, which relates to the genus, not the degree.
3
4 $P_Y(1)=4$ counts independent linear forms on $Y$ , not the degree.

For a curve (

r=1

) the leading coefficient

\deg Y/1!

is 3. The constant term is

1-p_a

, so

p_a=0

. The twisted cubic has exactly this Hilbert polynomial.

Practice

A cubic and a hyperbolaTwo of the six points are at infinity

Find all intersections of $y=x^3$ and $xy=1$ in $\mathbb{P}^2$ , with multiplicities, and check that they add up to 6.

Solution

Affine: $x\cdot x^3=1$ , so $x^4=1$ : four distinct points $(\zeta,\zeta^3)$ with $\zeta^4=1$ , each of multiplicity 1 (two real, two complex). At infinity the cubic $YZ^2=X^3$ has only $[0:1:0]$ and the hyperbola $XY=Z^2$ has $[1:0:0]$ and $[0:1:0]$ . In the chart $Y=1$ : $z^2=x^3$ and $x=z^2$ . Substituting gives $z^2-z^6=z^2(1-z^4)$ , so the multiplicity at $[0:1:0]$ is 2. Total $4+2=6=3\cdot2$ .

Two skew linesA Hilbert polynomial by counting monomials

Let $Y\subset\mathbb{P}^3$ be the union of the lines $Z(x_0,x_1)$ and $Z(x_2,x_3)$ . Compute $P_Y$ , $\deg Y$ and $1-P_Y(0)$ .

Hint

$I(Y)=(x_0,x_1)\cap(x_2,x_3)=(x_0x_2,\,x_0x_3,\,x_1x_2,\,x_1x_3)$ . Which monomials survive in $S/I(Y)$ ?

Solution

The monomials not in $I(Y)$ are those involving only $x_0,x_1$ or only $x_2,x_3$ . In degree $m\ge1$ there are $(m+1)+(m+1)$ of them, so $P_Y(m)=2m+2$ . Thus $\deg Y=2$ (two lines), and $1-P_Y(0)=-1$ . A negative arithmetic genus is a sign of a disconnected curve.

More practice: Hartshorne, Exercises I.5.4 and I.7.2.

InteractiveHartshorne, Ch. II §1: Sheaves · §2: Schemes

Sheaves and the spectrum of a ring

Schemes replace the points of a variety by the prime ideals of a ring, and its regular functions by a sheaf. The first example to understand is $\operatorname{Spec}\Z$ : its points are the primes, and an integer is a "function" whose value at $p$ is its residue mod $p$ .

The idea

Sheaves. A presheaf

\mathcal F

on

X

assigns a group

\mathcal F(U)

to each open

U

, with restriction maps compatible with inclusions. It is a sheaf if, for every open cover

U=\bigcup U_i

, a section that restricts to 0 on every

U_i

is 0, and sections

s_i\in\mathcal F(U_i)

that agree on all overlaps glue to a section on

U

. The stalk

\mathcal F_P

is the direct limit of

\mathcal F(U)

over open sets

U\ni P

.

The spectrum.

\operatorname{Spec}A

is the set of prime ideals of

A

. Its closed sets are

V(\mathfrak a)=\{\mathfrak p\supseteq\mathfrak a\}

, and the sets

D(f)=\{\mathfrak p: f\notin\mathfrak p\}

form a basis of open sets. The structure sheaf

\mathcal{O}

has stalks

\mathcal{O}_{\mathfrak p}=A_{\mathfrak p}

and sections

\mathcal{O}(D(f))\cong A_f

; in particular

\Gamma(\operatorname{Spec}A,\mathcal{O})=A

.

Schemes. A scheme is a locally ringed space

(X,\mathcal{O}_X)

in which every point has an open neighbourhood isomorphic to some

\operatorname{Spec}A

.

An element $f\in A$ has a value at each point: its image $f(\mathfrak p)$ in the residue field $\kappa(\mathfrak p)=A_{\mathfrak p}/\mathfrak pA_{\mathfrak p}$ . It vanishes at $\mathfrak p$ exactly when $f\in\mathfrak p$ . For $A=\Z$ : $\kappa((p))=\mathbb{F}_p$ and $\kappa((0))=\mathbb{Q}$ .

Two new features. Points need not be closed: the closure of

(0)

in

\operatorname{Spec}\Z

is everything, so

(0)

is a generic point. And a function is not determined by its values: in

\Z/4

the element 2 is nonzero, yet its value at the only point

(2)

is 0.

Explore

numerator a denominator b

\operatorname{Spec}\Z

, with the generic point

(0)

on the left and the primes up to 47. Click a prime to read off the value there. nonzero value zero (dots show the order) not defined (a pole)

Worked example

Check the sheaf axioms on $\operatorname{Spec}\Z$ for the cover by $D(2)$ and $D(3)$ : sections on the two pieces that agree on the overlap glue to a unique global section.

Why do $D(2)$ and $D(3)$ cover $\operatorname{Spec}\Z$ ? What are the rings of sections?

A point outside both would be a prime containing 2 and 3, hence containing $1=3-2$ , which is impossible. The sections are $\mathcal{O}(D(2))=\Z[1/2]$ , $\mathcal{O}(D(3))=\Z[1/3]$ , and on the overlap $D(2)\cap D(3)=D(6)$ they are $\Z[1/6]$ .
Take $s_1=m/2^a\in\Z[1/2]$ and $s_2=n/3^b\in\Z[1/3]$ that agree in $\Z[1/6]$ . What equation does that give?

$m/2^a=n/3^b$ in $\mathbb{Q}$ , so $3^b m=2^a n$ .
Show that $s_1=s_2$ is an integer.

$2^a$ divides $3^bm$ and is coprime to $3^b$ , so $2^a\mid m$ . Hence $s_1\in\Z$ , and $s_2=s_1$ . The glued section is this integer, as it must be, since $\Gamma(\operatorname{Spec}\Z,\mathcal{O})=\Z$ .
Check the other axiom: if an integer restricts to 0 on $D(2)$ and on $D(3)$ , is it 0?

Yes: $\Z\to\Z[1/2]$ is injective. Note that both axioms are about restriction, not about values at points; the last quiz below shows why that distinction matters.

Check your understanding

What is the closure of the point $(0)$ in $\operatorname{Spec}\Z$ ?

$\{(0)\}$ A closed set $V(\mathfrak a)$ contains $(0)$ only if $\mathfrak a\subseteq(0)$ .
All of $\operatorname{Spec}\Z$
The set of nonzero primesThat set doesn't even contain $(0)$ . And it is not closed: a nonzero integer vanishes at only finitely many primes.

The closure of

\{\mathfrak p\}

is

V(\mathfrak p)

, and

V((0))

is everything. The generic point lies in every nonempty open set, which is why the picture draws it as a cloud over the whole line.

Which points does $\operatorname{Spec}\C[x]$ have?

Exactly one point for each $a\in\C$ .You are missing a prime ideal that is not maximal.
The closed points $(x-a)$ , $a\in\C$ , and the generic point $(0)$ .
One point for each nonconstant polynomial.Only prime ideals count, and $(f)$ is prime exactly when $f$ is irreducible. Over $\C$ those are linear.

\C[x]

is a PID, so its nonzero primes are

(f)

with

f

irreducible, which over

\C

means

f=x-a

up to a unit. Add the zero ideal.

On $X=\operatorname{Spec}k[x]/(x^2)$ , the section $x$ is nonzero, yet its value at the only point of $X$ is 0. Is that a contradiction?

No. On a non-reduced scheme a section is not determined by its values.
Yes, so in fact $x=0$ in $k[x]/(x^2)$ . $x\neq0$ there; only $x^2=0$ .
No: $X$ has a second point where $x\neq0$ . $k[x]/(x^2)$ has exactly one prime ideal, $(x)$ .

The value of

x

at

(x)

is its image in

\kappa=k

, which is 0. But

x

is a nonzero nilpotent. Values only see the reduced scheme; the nilpotents carry extra, infinitesimal information.

Practice

The real line as a schemePoints and residue fields of Spec ℝ[x]

Describe the points of $\operatorname{Spec}\R[x]$ and their residue fields.

Solution

$(0)$ , with residue field $\R(x)$ ; $(x-a)$ for $a\in\R$ , with residue field $\R$ ; and $(x^2+bx+c)$ with $b^2<4c$ , with residue field $\R[x]/(x^2+bx+c)\cong\C$ . Each quadratic corresponds to a conjugate pair of complex roots, so the closed points are $\C$ modulo complex conjugation: the closed upper half plane.

The spectrum of a DVRTwo points, three open sets

Let $R=k[x]_{(x)}$ . List the points of $\operatorname{Spec}R$ , the open sets, and the rings of sections over each open set.

Solution

$R$ is a DVR, with primes $(0)$ and $\mathfrak m=(x)$ . The closed point is $\mathfrak m$ and the generic point is $(0)$ . The open sets are $\varnothing$ , $\{(0)\}=D(x)$ , and the whole space. The sections are $0$ , $R_x=k(x)$ and $R$ . So the "punctured" neighbourhood already has the whole function field as its sections.

More practice: Hartshorne, Exercises II.1.1, II.2.1 and II.2.5.

InteractiveHartshorne, Ch. II §3: First Properties of Schemes · §4: Separated and Proper Morphisms

Fibres, separated and proper morphisms

A morphism of schemes has a fibre over every point, including the generic point, and the fibre is computed as a fibre product. Over $\operatorname{Spec}\Z$ , the fibres of $\operatorname{Spec}\Z[i]$ are finite schemes that are two points, one point, or one non-reduced point, depending on $p$ .

The idea

First properties. A scheme is reduced if no local ring has nilpotents, and integral if every

\mathcal{O}_X(U)

,

U\neq\varnothing

, is a domain (equivalently, reduced and irreducible). It is noetherian if it has a finite cover by spectra of noetherian rings.

f:X\to Y

is of finite type if

Y

is covered by

\operatorname{Spec}B_i

whose preimages are covered by finitely many

\operatorname{Spec}A_{ij}

with each

A_{ij}

a finitely generated

B_i

-algebra.

Fibre products and fibres. Fibre products exist, and

\operatorname{Spec}A\times_{\operatorname{Spec}R}\operatorname{Spec}B=\operatorname{Spec}(A\otimes_RB)

. The fibre of

f:X\to Y

over

y

is

X_y=X\times_Y\operatorname{Spec}\kappa(y)

, a scheme over

\kappa(y)

whose underlying space is

f^{-1}(y)

.

Separated and proper.

f

is separated if the diagonal

X\to X\times_YX

is a closed immersion, and proper if it is separated, of finite type and universally closed. Affine morphisms are separated; projective morphisms of noetherian schemes are proper.

Valuative criteria. Let

f

be of finite type with

X

noetherian. For every valuation ring

R

with fraction field

K

, and every commuting square with

\operatorname{Spec}K\to X

and

\operatorname{Spec}R\to Y

:

f

is separated if and only if there is at most one lift

\operatorname{Spec}R\to X

, and proper if and only if there is exactly one.

For a quadratic ring $\Z[x]/(x^2-D)$ , the fibre over $(p)$ is $\operatorname{Spec}\big(\Z[x]/(x^2-D)\otimes\mathbb{F}_p\big)=\operatorname{Spec}\mathbb{F}_p[x]/(x^2-D)$ , so everything depends on how $x^2-D$ factors mod $p$ .

Explore

The ring

The prime p

Bottom row:

\operatorname{Spec}\Z

. Top row: the points of the ring lying over each prime. Click a prime to select it. split, two points inert, one point ramified, one non-reduced point

Worked example

Let $X=\operatorname{Spec}k[x,y,t]/(y^2-x^3-t)$ with the morphism $f:X\to\mathbb{A}^1=\operatorname{Spec}k[t]$ , where $k$ is algebraically closed of characteristic 0. Describe the fibres, including the generic fibre, and decide whether $X$ itself is singular.

Compute the fibre over the closed point $t=c$ .

$k[x,y,t]/(y^2-x^3-t)\otimes_{k[t]}k[t]/(t-c)=k[x,y]/(y^2-x^3-c)$ . The fibre is the affine plane curve $y^2=x^3+c$ .
Which closed fibres are singular?

The partial derivatives $-3x^2$ and $2y$ vanish together only at $x=y=0$ , which lies on the curve only when $c=0$ . So every fibre is nonsingular except the cusp $y^2=x^3$ over $t=0$ .
What is the generic fibre?

Tensor with $\kappa(\eta)=k(t)$ : $X_\eta=\operatorname{Spec}k(t)[x,y]/(y^2-x^3-t)$ , a curve over the field $k(t)$ . It is nonsingular: a singular point would need $x=y=0$ , and then $t=0$ in $k(t)$ , which is false.
Is $X$ itself singular?

No. $\partial/\partial t$ of $y^2-x^3-t$ is $-1$ , never zero. In fact $t=y^2-x^3$ , so $X\cong\mathbb{A}^2$ with coordinates $x,y$ . A nonsingular total space can have a singular fibre.

Check your understanding

What is the fibre of $\operatorname{Spec}\Z[i]\to\operatorname{Spec}\Z$ over $(5)$ ?

One point, with residue field $\mathbb{F}_{25}$ .Does $x^2+1$ have a root mod 5? Try $x=2$ .
Two points, each with residue field $\mathbb{F}_5$ .
One non-reduced point.That happens only over 2, where $x^2+1=(x+1)^2$ mod 2.

x^2+1\equiv(x-2)(x-3)\pmod 5

, so

\mathbb{F}_5[x]/(x^2+1)\cong\mathbb{F}_5\times\mathbb{F}_5

by the Chinese remainder theorem.

How many points does $\operatorname{Spec}\C\times_{\operatorname{Spec}\R}\operatorname{Spec}\C$ have?

One, since each factor has one point.The underlying set of a fibre product is not the fibre product of the sets. Compute $\C\otimes_\R\C$ .
Two
One non-reduced point $x^2+1$ has the distinct roots $\pm i$ in $\C$ .

\C\otimes_\R\C\cong\C\otimes_\R\R[x]/(x^2+1)\cong\C[x]/(x^2+1)\cong\C\times\C

, whose spectrum has two points.

Which of these morphisms to $\operatorname{Spec}k$ is proper?

$\mathbb{A}^1\to\operatorname{Spec}k$ It is not universally closed: the projection $\mathbb{A}^1\times\mathbb{A}^1\to\mathbb{A}^1$ sends the closed set $xy=1$ onto $\mathbb{A}^1\smallsetminus\{0\}$ , which is not closed.
$\mathbb{P}^1\to\operatorname{Spec}k$
The line with two origins $\to\operatorname{Spec}k$ It is not separated: a map from the generic point of a DVR extends to its closed point in two ways, one through each origin.

Projective morphisms are proper. The other two each fail one half of the valuative criterion: existence for

\mathbb{A}^1

, uniqueness for the doubled line.

Practice

An arithmetic surfaceFibres of Spec ℤ[x]

Describe the fibres of $\operatorname{Spec}\Z[x]\to\operatorname{Spec}\Z$ over $(p)$ and over $(0)$ . What is $\dim\operatorname{Spec}\Z[x]$ ?

Solution

Over $(p)$ : $\operatorname{Spec}\mathbb{F}_p[x]$ , the affine line over $\mathbb{F}_p$ . Over $(0)$ : $\operatorname{Spec}\mathbb{Q}[x]$ . So $\operatorname{Spec}\Z[x]$ is a family of lines over a one-dimensional base, and it has dimension 2. A maximal chain is $(0)\subset(p)\subset(p,x)$ .

Fibres that jumpA blow-up chart

Let $f:\mathbb{A}^2\to\mathbb{A}^2$ , $(x,t)\mapsto(x,tx)$ . Describe the fibre over $(a,b)$ when $a\neq0$ , when $a=b=0$ , and when $a=0\neq b$ . Is the image closed?

Solution

For $a\neq0$ the fibre is the single point $(a,b/a)$ . Over $(0,0)$ it is the whole line $x=0$ . Over $(0,b)$ with $b\neq0$ it is empty. The image is $\{x\neq0\}\cup\{(0,0)\}$ , which is neither open nor closed (it is constructible). The dimension of the fibre jumps from 0 to 1 at the origin.

More practice: Hartshorne, Exercises II.3.10 and II.4.12.

Worked exampleHartshorne, Ch. II §5: Sheaves of Modules · §6: Divisors · §7: Projective Morphisms · §8: Differentials

Line bundles, divisors and differentials

On $\mathbb{P}^n$ there is a line bundle $\mathcal{O}(d)$ for each integer $d$ , and every line bundle is one of them. Divisors are the bookkeeping for line bundles, and differentials give every nonsingular variety a canonical one. This module is one long computation on two curves.

The idea

Quasi-coherent sheaves. On

\operatorname{Spec}A

,

M\mapsto\tilde M

is an equivalence between

A

-modules and quasi-coherent

\mathcal{O}

-modules. A sheaf of modules on a scheme is quasi-coherent if it is locally of this form, and coherent if locally

M

can be taken finitely generated (on a noetherian scheme).

Twisting sheaves. On

\mathbb{P}^n=\operatorname{Proj}k[x_0,\dots,x_n]

,

\mathcal{O}(d)=S(d)^\sim

. Its global sections are the forms of degree

d

:

\Gamma(\mathbb{P}^n,\mathcal{O}(d))=S_d

, of dimension

\binom{n+d}{n}

for

d\ge0

and 0 for

d<0

.

Weil divisors. On a noetherian integral separated scheme that is regular in codimension one, a divisor is a finite formal sum

\sum n_iY_i

of prime divisors. A rational function has

\operatorname{div}(f)=\sum v_Y(f)\,Y

, and

\operatorname{Cl}X

is divisors modulo these. Degree gives

\operatorname{Cl}\mathbb{P}^n\cong\Z

.

Cartier divisors and Pic.

\operatorname{Pic}X

is the group of invertible sheaves up to isomorphism. If

X

is noetherian, integral, separated and locally factorial (for example nonsingular), then

\operatorname{Cl}X\cong\operatorname{Pic}X

. So

\operatorname{Pic}\mathbb{P}^n\cong\Z

, generated by

\mathcal{O}(1)

.

Differentials. For

X

nonsingular of dimension

n

over

k

,

\Omega_{X/k}

is locally free of rank

n

, and the canonical sheaf is

\omega_X=\wedge^n\Omega_{X/k}

. Then

\omega_{\mathbb{P}^n}\cong\mathcal{O}(-n-1)

. Adjunction: a nonsingular hypersurface

Y

of degree

d

in

\mathbb{P}^n

has

\omega_Y\cong\mathcal{O}_Y(d-n-1)

. For a curve, the genus is

g=\dim\Gamma(X,\omega_X)

.

Worked example

On $\mathbb{P}^1$ , with affine coordinate $x$ and the point $\infty$ where $u=1/x$ vanishes, and on the plane cubic $C:\ y^2=x^3+1$ with its point $O=[0:1:0]$ at infinity: compute some divisors of functions and differentials, and read off $\operatorname{Cl}\mathbb{P}^1$ , $\omega_{\mathbb{P}^1}$ and the genus of $C$ . Work in characteristic 0.

Compute $\operatorname{div}(f)$ for $f=x^2/(x-1)$ on $\mathbb{P}^1$ . Don't forget $\infty$ .

On $\mathbb{A}^1$ : a double zero at 0 and a simple pole at 1. At $\infty$ , with $u=1/x$ : $f=\dfrac{u^{-2}}{u^{-1}-1}=\dfrac{1}{u(1-u)}$ , so $\operatorname{ord}_\infty f=-1$ . Thus $\operatorname{div}f=2[0]-[1]-[\infty]$ , of degree 0.
Why does every principal divisor on $\mathbb{P}^1$ have degree 0, and why does that give $\operatorname{Cl}\mathbb{P}^1\cong\Z$ ?

For $f=p/q$ , the finite zeros and poles contribute $\deg p-\deg q$ , and $\operatorname{ord}_\infty f=\deg q-\deg p$ , so the total is 0. Conversely, a degree-0 divisor $\sum n_i[a_i]+m[\infty]$ (with $m=-\sum n_i$ ) is $\operatorname{div}\prod(x-a_i)^{n_i}$ . So $D\sim(\deg D)[\infty]$ , and $\deg:\operatorname{Cl}\mathbb{P}^1\to\Z$ is an isomorphism.
Compute the divisor of the differential $dx$ on $\mathbb{P}^1$ .

At a finite point $a$ , $x-a$ is a local parameter and $dx=d(x-a)$ has no zero or pole. At $\infty$ , $x=u^{-1}$ , so $dx=-u^{-2}\,du$ , a pole of order 2. So $K=\operatorname{div}(dx)=-2[\infty]$ , of degree $-2$ , and $\omega_{\mathbb{P}^1}\cong\mathcal{O}(-2)$ , as the formula $\mathcal{O}(-n-1)$ says. $\Gamma(\omega)=0$ , so the genus is 0.
On $C$ , compute $\operatorname{div}(x)$ . The orders of $x$ and $y$ at $O$ are $-2$ and $-3$ .

$x=0$ at $(0,1)$ and $(0,-1)$ . There $y\neq0$ , so $\partial f/\partial y=2y\neq0$ and $x$ is a local parameter: simple zeros. The only pole is at $O$ , of order 2. So $\operatorname{div}(x)=(0,1)+(0,-1)-2O$ . (The orders at $O$ : $x$ is a degree-2 map to $\mathbb{P}^1$ and $O$ is its only point over $\infty$ , and $2\operatorname{ord}y=3\operatorname{ord}x$ from the equation.)
Show that $\omega=dx/y$ has no zeros or poles on $C$ .

Where $y\neq0$ , $x$ minus a constant is a local parameter, so $dx$ is a nonvanishing local generator and $y$ is a unit. Where $y=0$ (the three points with $x^3=-1$ ), differentiate the equation: $2y\,dy=3x^2\,dx$ , so $dx/y=2\,dy/(3x^2)$ , and $3x^2\neq0$ there, while $y$ is a local parameter. At $O$ , with local parameter $s$ : $\operatorname{ord}(dx)=\operatorname{ord}(x)-1=-3$ and $\operatorname{ord}(y)=-3$ , so $\operatorname{ord}(dx/y)=0$ . Hence $K_C=0$ , $\omega_C\cong\mathcal{O}_C$ , and $g=\dim\Gamma(\omega_C)=\dim\Gamma(\mathcal{O}_C)=1$ .
Check both answers with adjunction.

A plane curve of degree $d$ has $\omega\cong\mathcal{O}(d-3)$ . For a conic, $\mathcal{O}_C(-1)$ , of degree $-2$ , matching $\mathbb{P}^1$ . For the cubic, $\mathcal{O}_C(0)=\mathcal{O}_C$ , matching step 5. In general $\deg K=d(d-3)$ .

Check your understanding

What is $\dim\Gamma(\mathbb{P}^2,\mathcal{O}(3))$ ?

10
9Count the monomials $x^ay^bz^c$ with $a+b+c=3$ carefully.
4That is $\dim\Gamma(\mathbb{P}^1,\mathcal{O}(3))$ , the binary cubics.

Sections of

\mathcal{O}(3)

are cubic forms in three variables, and there are

\binom{5}{2}=10

monomials. This is why plane cubics form a

\mathbb{P}^9

.

On a nonsingular plane conic $C\cong\mathbb{P}^1$ , what is the degree of $\mathcal{O}_C(1)$ ?

1A line meets a conic in 2 points (Bézout).
2
$-2$ That is the degree of $\omega_C\cong\mathcal{O}_C(-1)$ .

\mathcal{O}_C(1)

is the divisor class of a line section, of degree 2. Under

C\cong\mathbb{P}^1

it is

\mathcal{O}_{\mathbb{P}^1}(2)

: the conic is the image of

\mathbb{P}^1

by quadrics.

Why is $\operatorname{Cl}\mathbb{A}^n=0$ ?

$k[x_1,\dots,x_n]$ is a UFD, so every height-one prime is principal, and every prime divisor is $\operatorname{div}(f)$ .
$\mathbb{A}^n$ has no prime divisors.It has plenty: every irreducible hypersurface.
Every divisor on $\mathbb{A}^n$ is effective. $-Z(x_1)$ is not effective. The point is that it is principal: $\operatorname{div}(1/x_1)$ .

A noetherian domain is a UFD exactly when its height-one primes are principal. If

Y=Z(f)

with

f

irreducible, then

Y=\operatorname{div}(f)

.

Practice

The genus of a plane curveFrom the canonical degree

Use $\omega_C\cong\mathcal{O}_C(d-3)$ and $\deg K=2g-2$ to find the genus of a nonsingular plane curve of degree $d$ .

Solution

$\deg\mathcal{O}_C(1)=d$ (Bézout with a line), so $\deg K=d(d-3)$ . Then $2g-2=d^2-3d$ gives $g=\frac{(d-1)(d-2)}2$ : 0, 0, 1, 3, 6 for $d=1,\dots,5$ .

Sections of a divisor on ℙ¹Riemann–Roch's first case

For $D=2[0]+[\infty]$ on $\mathbb{P}^1$ , find $L(D)=\{f:\operatorname{div}f+D\ge0\}\cup\{0\}$ and its dimension.

Solution

$f$ may have a pole of order at most 2 at 0, at most 1 at $\infty$ , and none elsewhere. So $f=p(x)/x^2$ with $p$ a polynomial, and $\operatorname{ord}_\infty f=2-\deg p\ge-1$ forces $\deg p\le3$ . So $L(D)=\{(a+bx+cx^2+dx^3)/x^2\}$ , of dimension $4=\deg D+1$ , matching $\mathcal{O}(3)$ .

More practice: Hartshorne, Exercises II.6.1 and II.8.4.

InteractiveHartshorne, Ch. III §4: Čech Cohomology · §5: The Cohomology of Projective Space · §7: The Serre Duality Theorem

Cohomology of projective space

The cohomology of $\mathcal{O}(d)$ on $\mathbb{P}^n$ can be computed by hand with the Čech complex of the standard affine cover, and the answer is a count of monomials. Slide $d$ and watch $H^0$ and $H^n$ trade places.

The idea

Sheaf cohomology.

H^i(X,-)

are the right derived functors of

\Gamma(X,-)

on sheaves of abelian groups. A short exact sequence of sheaves gives a long exact sequence in cohomology. On a noetherian space of dimension

n

,

H^i=0

for

i>n

.

Affine schemes have no higher cohomology. If

X

is a noetherian affine scheme,

H^i(X,\mathcal F)=0

for all

i>0

and all quasi-coherent

\mathcal F

. Consequently, on a noetherian separated scheme, Čech cohomology for any open affine cover computes

H^i

.

Line bundles on $\mathbb{P}^n$ . Use the cover

U_i=\{x_i\neq0\}

. Then

H^0(\mathcal{O}(d))=S_d

;

H^i(\mathcal{O}(d))=0

for

0<i<n

; and

H^n(\mathcal{O}(d))

has a basis of Laurent monomials

x_0^{a_0}\cdots x_n^{a_n}

with every

a_i\le-1

and

\sum a_i=d

. In particular

h^n(\mathcal{O}(d))=h^0(\mathcal{O}(-d-n-1))

.

Serre duality. For

X

projective and nonsingular of dimension

n

over

k

, and

\mathcal F

locally free:

H^i(X,\mathcal F)\cong H^{n-i}(X,\mathcal F^\vee\otimes\omega_X)^\vee

. On

\mathbb{P}^n

, with

\omega=\mathcal{O}(-n-1)

, this is the symmetry above.

Why monomials: on $U_{i_0}\cap\dots\cap U_{i_p}$ the sections of $\mathcal{O}(d)$ are spanned by Laurent monomials of degree $d$ whose negative exponents occur only among $x_{i_0},\dots,x_{i_p}$ . A monomial with some exponent $\ge0$ is already a section on a smaller intersection and dies in $H^n$ ; a monomial with all exponents $\ge0$ is a global section.

Explore

Each dot is a Laurent monomial of degree

d

, placed by its exponents of

x_0

(across) and

x_1

(up). basis of

H^0

basis of

H^n

Projective space

Twist d

Worked example

Let $C\subset\mathbb{P}^2$ be a curve defined by an irreducible form of degree $d$ . Compute $h^0(\mathcal{O}_C)$ and $h^1(\mathcal{O}_C)$ .

Write down a short exact sequence relating $\mathcal{O}_C$ to sheaves on $\mathbb{P}^2$ .

The ideal sheaf of $C$ is generated by its equation $f$ , of degree $d$ , so it is isomorphic to $\mathcal{O}(-d)$ :

$0\to\mathcal{O}_{\mathbb{P}^2}(-d)\xrightarrow{\ \cdot f\ }\mathcal{O}_{\mathbb{P}^2}\to\mathcal{O}_C\to0.$
Take the long exact sequence. What is $h^0(\mathcal{O}_C)$ ?

$0\to H^0(\mathcal{O}(-d))\to H^0(\mathcal{O})\to H^0(\mathcal{O}_C)\to H^1(\mathcal{O}(-d))$ . The outer terms are 0 (no forms of negative degree; no intermediate cohomology) and $H^0(\mathcal{O})=k$ . So $h^0(\mathcal{O}_C)=1$ .
Now $h^1(\mathcal{O}_C)$ .

$0=H^1(\mathcal{O})\to H^1(\mathcal{O}_C)\to H^2(\mathcal{O}(-d))\to H^2(\mathcal{O})=0$ , so $h^1(\mathcal{O}_C)=h^2(\mathcal{O}(-d))$ .
Count the monomials.

We need $a,b,c\le-1$ with $a+b+c=-d$ . Put $a=-1-a'$ and so on: $a'+b'+c'=d-3$ with $a',b',c'\ge0$ . There are $\binom{d-1}{2}=\frac{(d-1)(d-2)}2$ solutions. Set $d=3$ in the explorer on $\mathbb{P}^2$ with twist $-3$ : exactly one monomial, $x_0^{-1}x_1^{-1}x_2^{-1}$ , so a plane cubic has $h^1(\mathcal{O}_C)=1$ .

Check your understanding

What is $h^1(\mathbb{P}^2,\mathcal{O}(d))$ ?

0, for every $d$
$h^0(\mathcal{O}(-d-3))$ On a surface, Serre duality pairs $H^1$ with $H^1$ : $h^1(\mathcal{O}(d))=h^1(\mathcal{O}(-d-3))$ . That relation doesn't decide the value.
Nonzero when $d<0$ Only $H^0$ and $H^n$ of a line bundle on $\mathbb{P}^n$ can be nonzero.

The intermediate cohomology of every

\mathcal{O}(d)

on

\mathbb{P}^n

vanishes. In the monomial picture, any degree-

d

Laurent monomial either has all exponents

\ge0

, all

\le-1

, or is killed in the Čech complex.

What is $h^2(\mathbb{P}^2,\mathcal{O}(-5))$ ?

6
3Count solutions of $a+b+c=-5$ with every exponent $\le-1$ : substitute $a=-1-a'$ , etc.
0 $H^2(\mathcal{O}(d))$ vanishes only for $d\ge-2$ .

a'+b'+c'=2

has

\binom42=6

solutions, and Serre duality agrees:

h^0(\mathcal{O}(5-3))=h^0(\mathcal{O}(2))=6

. Check it in the explorer.

Why does the vanishing of higher cohomology on affine schemes matter for $\mathbb{P}^n$ ?

It lets the Čech complex of the cover $\{U_i\}$ compute $H^i(\mathbb{P}^n,\mathcal F)$ .
It shows $H^1(\mathbb{P}^n,\mathcal F)=0$ for every sheaf. $\mathbb{P}^n$ is not affine: $H^1(\mathbb{P}^1,\mathcal{O}(-2))$ is one-dimensional.
It shows every sheaf on $\mathbb{P}^n$ is quasi-coherent.That is a different question, and false.

If every finite intersection of the cover is affine (which holds on a separated scheme) and the sheaf has no higher cohomology there, the Čech complex is acyclic enough to compute derived functor cohomology.

Practice

H¹ of ℙ¹ from scratchThe Čech complex by hand

Using the cover $U_0,U_1$ of $\mathbb{P}^1$ , compute $H^1(\mathbb{P}^1,\mathcal{O}(-2))$ directly from the Čech complex.

Solution

Degree $-2$ Laurent monomials $x_0^ax_1^b$ , $a+b=-2$ . Sections on $U_0$ allow $a<0$ but need $b\ge0$ ; on $U_1$ , $a\ge0$ ; on $U_0\cap U_1$ anything. $H^1$ is the cokernel of $\Gamma(U_0)\oplus\Gamma(U_1)\to\Gamma(U_{01})$ , spanned by the monomials with both $a,b\le-1$ . With $a+b=-2$ there is only $x_0^{-1}x_1^{-1}$ . So $h^1=1$ , matching $h^0(\mathcal{O}(0))=1$ by duality.

Euler characteristic on ℙ³A polynomial in d

Check that $\chi(\mathbb{P}^3,\mathcal{O}(d))=\binom{d+3}{3}=\frac{(d+1)(d+2)(d+3)}6$ also for $d=-4$ and $d=-5$ , by counting monomials.

Solution

For $d\le-1$ only $H^3$ survives, and $\chi=-h^3$ . For $d=-4$ : all four exponents $-1$ , one monomial, $\chi=-1$ , and the formula gives $(-3)(-2)(-1)/6=-1$ . For $d=-5$ : one exponent is $-2$ , four monomials, $\chi=-4$ , and the formula gives $(-4)(-3)(-2)/6=-4$ .

More practice: Hartshorne, Exercises III.5.1, III.5.3 and III.5.5.

InteractiveHartshorne, Ch. IV §1: Riemann–Roch Theorem · §3: Embeddings in Projective Space · §5: The Canonical Embedding

The Riemann–Roch theorem

How many functions on a curve have poles only at one point $P$ , of order at most $n$ ? Riemann–Roch answers up to a correction term, and the correction disappears once $n$ is large. The pole orders that cannot occur are the Weierstrass gaps, and there are always exactly $g$ of them.

The idea

Here a curve is a complete nonsingular curve over an algebraically closed field $k$ .

Linear systems. For a divisor

D

,

L(D)=\{f\in K^*:\operatorname{div}f+D\ge0\}\cup\{0\}\cong H^0(X,\mathcal{O}(D))

, and

l(D)=\dim L(D)

. The genus is

g=l(K)=h^1(\mathcal{O}_X)

.

Riemann–Roch.

\ l(D)-l(K-D)=\deg D+1-g.

Consequences.

\deg K=2g-2

. If

\deg D>2g-2

then

l(K-D)=0

and

l(D)=\deg D+1-g

. If

\deg D\ge2g

,

|D|

has no base points; if

\deg D\ge2g+1

,

D

is very ample. For

g\ge2

the canonical map is an embedding exactly when

X

is not hyperelliptic.

Weierstrass gaps.

n\ge1

is a gap at

P

if

l(nP)=l((n-1)P)

, that is, no function has a pole of order exactly

n

at

P

and no other poles. Since

l

goes up by at most 1 per step,

l(0)=1

and

l((2g-1)P)=g

, there are exactly

g

gaps, all in

[1,2g-1]

.

Explore

Curve and point P

n, so D = nP

l(nP)

n+1-g

the correction

l(K-nP)

gaps

Worked example

Let $X$ be the genus-2 curve $y^2=h(x)$ , with $h$ of degree 5 with distinct roots, and let $P$ be its single point at infinity. Find $L(nP)$ explicitly and check Riemann–Roch. (Characteristic 0.)

What are the pole orders of $x$ and $y$ at $P$ ?

$x:X\to\mathbb{P}^1$ has degree 2, and $P$ is the only point over $\infty$ , so $\operatorname{ord}_P x=-2$ . From $y^2=h(x)$ : $2\operatorname{ord}_P y=5\operatorname{ord}_P x=-10$ , so $\operatorname{ord}_P y=-5$ . Both are regular on the affine part.
Find $l(4P)$ and check it against Riemann–Roch.

Functions regular away from $P$ are polynomials in $x,y$ , and using $y^2=h(x)$ every such function is $a(x)+b(x)y$ . Poles: $a(x)$ has even order $2\deg a$ , and $b(x)y$ has odd order $2\deg b+5$ , so no cancellation is possible. For $n=4$ : $1,x,x^2$ , so $l(4P)=3$ . Riemann–Roch: $\deg(K-4P)=2-4<0$ , so $l(4P)=4+1-2=3$ .
Use $n=2$ to find the canonical class.

$L(2P)=\langle1,x\rangle$ , so $l(2P)=2$ . Riemann–Roch: $2-l(K-2P)=2+1-2$ , so $l(K-2P)=1$ . Since $\deg K=2$ , $K-2P$ has degree 0 and a section, so $K\sim2P$ . Indeed $dx/y$ has divisor exactly $2P$ : at $P$ , $\operatorname{ord}(dx)-\operatorname{ord}(y)=-3+5=2$ , and it has no zeros or poles elsewhere.
What are the gaps at $P$ ?

The pole orders that occur are $0,2,4,5,6,7,\dots$ (even numbers from $x^k$ , and $2k+5$ from $x^ky$ ). So the gaps are $1$ and $3$ : two of them, as $g=2$ requires. Choose "Genus 2, Weierstrass point" in the explorer.

Check your understanding

$X$ has genus 3 and $D$ has degree 7. What is $l(D)$ ?

5
8That is $\deg D+1$ , the answer on $\mathbb{P}^1$ . Riemann–Roch subtracts $g$ .
It depends on $D$ .Once $\deg D>2g-2$ , $\deg(K-D)<0$ , so $l(K-D)=0$ and $l(D)$ is determined.

7>2g-2=4

, so

l(D)=7+1-3=5

.

On an elliptic curve, is there a function with a single simple pole and no other poles?

No.
Yes: $x$ in a Weierstrass equation. $x$ has a double pole at $O$ .
Yes: Riemann–Roch gives $l(P)=2$ .It gives $l(P)-l(K-P)=1+1-1=1$ , and $l(K-P)=l(-P)=0$ , so $l(P)=1$ .

l(P)=1

: only the constants. A function with one simple pole would be a degree-1 map

X\to\mathbb{P}^1

, hence an isomorphism, but

g=1\neq0

.

How many Weierstrass gaps does a point on a curve of genus 4 have?

4
7The gaps lie in $[1,7]$ , but only $g$ of those numbers are gaps.
None, for a general point.Every point has exactly $g$ gaps. At a general point they are $1,\dots,g$ .

l(nP)

climbs from

l(0)=1

to

l(7P)=7+1-4=4

in 7 steps of size 0 or 1, so 3 steps go up and 4 are gaps.

Practice

Degree 0 and degree 2 on an elliptic curveWhere the correction term lives

On an elliptic curve $X$ , find $l(D)$ for every divisor of degree 0, and for every divisor of degree 2.

Solution

Degree 0: if $f\in L(D)$ is nonzero, $\operatorname{div}f+D$ is effective of degree 0, hence 0, so $D\sim0$ . So $l(D)=1$ if $D\sim0$ and 0 otherwise. Degree 2: $2>2g-2=0$ , so $l(D)=2+1-1=2$ . The correction term $l(K-D)=l(-D)$ is nonzero only in degree 0, and only for $D\sim0$ .

Reading off the correctionGenus 4

On a curve of genus 4, a divisor $D$ of degree 5 has $l(D)=3$ . Find $l(K-D)$ , and show that $K\sim D+Q$ for some point $Q$ .

Solution

$3-l(K-D)=5+1-4=2$ , so $l(K-D)=1$ . $\deg(K-D)=6-5=1$ , and a nonzero $f\in L(K-D)$ makes $\operatorname{div}f+K-D$ an effective divisor of degree 1, a single point $Q$ . So $K\sim D+Q$ .

More practice: Hartshorne, Exercises IV.1.5 and IV.1.7.

InteractiveHartshorne, Ch. IV §2: Hurwitz's Theorem

Hurwitz's formula

A finite map of curves relates their genera through how much it ramifies. When the target is $\mathbb{P}^1$ , computing a genus turns into counting branch points.

The idea

Degree and ramification. A finite morphism of curves

f:X\to Y

has degree

n=[K(X):K(Y)]

. For

P\mapsto Q

, with

t

a local parameter at

Q

, the ramification index is

e_P=v_P(f^*t)

. Over each

Q

,

\sum_{P\mapsto Q}e_P=n

.

P

is a ramification point if

e_P>1

, and then

Q

is a branch point.

Hurwitz. If

f

is separable, then

2g(X)-2=n\,(2g(Y)-2)+\deg R

. When all ramification is tame (

\operatorname{char}k=0

, or the characteristic divides no

e_P

),

\deg R=\sum_P(e_P-1)

.

Two quick consequences. A separable map

\mathbb{P}^1\to\mathbb{P}^1

of degree

n

has

\sum(e_P-1)=2n-2

. And there is no nonconstant separable map from a curve to one of larger genus, since the right side would exceed the left.

Explore

The map

n

The sheets of

X

over the base

\mathbb{P}^1

. Over an ordinary point there are

n

points; at a branch point, sheets merge into one point with ramification index

e

. branch points

Worked example

Find the genus of the nonsingular projective model of $y^3=x^4-1$ (characteristic 0), using the map $(x,y)\mapsto x$ .

What is the degree of the map?

For a general value of $x$ , the equation $y^3=x^4-1$ has three distinct solutions $y$ . So the degree is 3.
Where is it ramified over finite values of $x$ ?

Over $x=\zeta$ with $\zeta^4=1$ , the only point is $(\zeta,0)$ . There $x^4-1=(x-\zeta)u$ with $u$ a unit, so $y$ is a local parameter and $x-\zeta=y^3u^{-1}$ vanishes to order 3: $e=3$ . Over every other finite $x$ there are three points, so no ramification.
What happens over $x=\infty$ ?

Homogenise: $Y^3Z=X^4-Z^4$ . At $Z=0$ this forces $X=0$ , so the curve has a single point $[0:1:0]$ at infinity, and it is the only point over $x=\infty$ . Hence $e=3$ there too.
Apply Hurwitz, and check the answer another way.

Five ramification points with $e=3$ : $2g-2=3\cdot(-2)+5\cdot2=4$ , so $g=3$ . Check: $F=X^4-Z^4-Y^3Z$ has partials $4X^3$ , $-3Y^2Z$ , $-4Z^3-Y^3$ , which vanish together only at $X=Y=Z=0$ . So the closure is a nonsingular plane quartic, and $g=\frac{3\cdot2}2=3$ .

Check your understanding

$X$ has genus 4 and $f:X\to\mathbb{P}^1$ has degree 2. How many branch points are there (characteristic 0)?

10
8Solve $2g-2=2(-2)+B$ with $g=4$ .
9For a double cover the number is $2g+2$ , always even.

Each branch point of a double cover has one point above it with

e=2

, contributing 1. So

6=-4+B

,

B=10=2g+2

.

Is there a nonconstant morphism $\mathbb{P}^1\to E$ to an elliptic curve?

No.
Yes, of degree 2.Degree-2 maps go the other way, $E\to\mathbb{P}^1$ .
Only in positive characteristic.Hurwitz needs separability, but a purely inseparable map preserves the genus, so no characteristic allows it.

For a separable map, Hurwitz would give

-2=n\cdot0+\deg R\ge0

, which is impossible.

For a separable map $\mathbb{P}^1\to\mathbb{P}^1$ of degree $n$ , what is $\sum_P(e_P-1)$ (tame case)?

$2n-2$
$n-1$ That is the contribution of one totally ramified point. $z\mapsto z^n$ has two of them.
0That would need $-2=-2n$ , so only $n=1$ .

-2=n(-2)+\sum(e_P-1)

. Choose "z ↦ zⁿ" in the explorer: two points with

e=n

.

Practice

A cyclic coverRamification at infinity too

Find the genus of the nonsingular projective model of $y^5=x(x-1)$ , using the map to the $x$ -line (characteristic 0).

Hint

For $y^n=h(x)$ with $n$ prime (characteristic 0), the points over $x=a$ are totally ramified when $n$ does not divide the order of $h$ at $a$ , and unramified when it does. At $\infty$ that order is $-\deg h$ .

Solution

Degree 5. Over 0 and 1, $h$ has order 1, and over $\infty$ order $-2$ ; 5 divides none of these, so each of the three is totally ramified, with $e=5$ . Hurwitz: $2g-2=5(-2)+3\cdot4=2$ , so $g=2$ .

Genus 2 over an elliptic curveCounting ramification points

Suppose a curve $X$ of genus 2 has a degree-2 map onto an elliptic curve $E$ (characteristic 0). How many ramification points does it have?

Solution

$2\cdot2-2=2(2\cdot1-2)+\sum(e_P-1)$ , so $\sum(e_P-1)=2$ . For a double cover each ramification point has $e=2$ , so there are exactly 2.

More practice: Hartshorne, Exercise IV.2.2.

InteractiveHartshorne, Ch. IV §3: Embeddings in Projective Space · §4: Elliptic Curves

Elliptic curves and the group law

A curve of genus 1 with a chosen point is a group. On a plane cubic the group law is a ruler construction: draw the line through $P$ and $Q$ , take the third point where it meets the curve, and reflect.

The idea

Elliptic curve. A curve

X

of genus 1 with a point

P_0

. Since

3\ge2g+1

, the divisor

3P_0

is very ample, and

l(3P_0)=3

: it embeds

X

in

\mathbb{P}^2

as a cubic. If

\operatorname{char}k\neq2,3

, coordinates can be chosen so that the cubic is

y^2=x^3+ax+b

with

4a^3+27b^2\neq0

, and

P_0=[0:1:0]

, a flex whose tangent line is the line at infinity.

The group.

P\mapsto\mathcal{O}(P-P_0)

is a bijection from

X(k)

to

\operatorname{Pic}^0X

, and the group structure is carried over from

\operatorname{Pic}^0

:

P+Q=R

exactly when

P+Q\sim R+P_0

as divisors. On the cubic,

P+Q+R=0

exactly when

P,Q,R

are the three intersections of a line with

X

, counted with multiplicity.

The $j$ -invariant.

j=1728\cdot\dfrac{4a^3}{4a^3+27b^2}

. Two elliptic curves over

k

are isomorphic exactly when they have the same

j

, and every

j\in k

occurs.

The formulas behind the explorer: for $P=(x_1,y_1)$ , $Q=(x_2,y_2)$ with $x_1\neq x_2$ , the slope is $\lambda=(y_2-y_1)/(x_2-x_1)$ , or $\lambda=(3x_1^2+a)/(2y_1)$ for $P=Q$ . Then $x_3=\lambda^2-x_1-x_2$ and $P+Q=\big(x_3,\ \lambda(x_1-x_3)-y_1\big)$ .

Explore

Drag P and Q along the curve. line through P and Q third point R P + Q, the reflection of R

a b

Worked example

On $E: y^2=x^3+1$ over $\mathbb{Q}$ , with $O=[0:1:0]$ , find the order of the point $P=(2,3)$ .

Compute $2P$ with the tangent line.

$\lambda=\dfrac{3\cdot4+0}{2\cdot3}=2$ , so $x_3=4-2-2=0$ and $y=2(2-0)-3=1$ . So $2P=(0,1)$ .
Compute $3P=(0,1)+(2,3)$ .

$\lambda=\dfrac{3-1}{2-0}=1$ , $x_3=1-0-2=-1$ , $y=1\cdot(0-(-1))-1=0$ . So $3P=(-1,0)$ .
What is the order of $P$ ?

$(-1,0)$ has $y=0$ , so it equals its own negative $(x,-y)$ : $2\cdot3P=O$ . Neither $2P$ nor $3P$ is $O$ , so $P$ has order 6.
So $(0,1)=2P$ has order 3. What does that say about the tangent line at $(0,1)$ ?

$3\cdot(0,1)=0$ means the tangent line at $(0,1)$ meets $E$ only there, with multiplicity 3: it is a flex. Directly, the slope there is $3x^2/(2y)=0$ , and the line $y=1$ gives $x^3=0$ , a triple root. Since $a=0$ , $j=0$ .

Check your understanding

On $y^2=x^3+ax+b$ , what is $-P$ for $P=(x,y)$ ?

$(x,-y)$
$(-x,-y)$ The curve is symmetric under $y\mapsto-y$ , not under $(x,y)\mapsto(-x,-y)$ . That point is usually not even on the curve.
The third point on the tangent line at $P$ That point is $-2P$ .

The vertical line through

P

meets the curve at

P

,

(x,-y)

and

O

. Collinear points sum to 0, so

P+(x,-y)=0

. Use "Set Q = −P" in the explorer.

Over $\C$ , how many points $P$ satisfy $2P=0$ ?

4
3Don't forget the identity $O$ .
2 $2P=0$ means the tangent at $P$ passes through $O$ , i.e. is vertical: that happens where $y=0$ .

O

and the three points

(e_i,0)

, where

e_i

are the roots of

x^3+ax+b

. Together they form

\Z/2\times\Z/2

.

Why is $P\mapsto\mathcal{O}(P-P_0)$ injective?

If $P\sim Q$ with $P\neq Q$ , a function with divisor $P-Q$ would give a degree-1 map $X\to\mathbb{P}^1$ , forcing $g=0$ .
Because $\operatorname{Pic}^0X$ is torsion-free.It isn't: the previous question found points of order 2.
Because $l(P)=2$ .On a genus-1 curve $l(P)=1$ .

\mathcal{O}(P-P_0)\cong\mathcal{O}(Q-P_0)

means

P\sim Q

. A nonconstant

f

with

\operatorname{div}f=P-Q

has a single simple pole, so it is an isomorphism onto

\mathbb{P}^1

, which is impossible for

g=1

.

Practice

Two j-invariantsNot isomorphic

Compute $j$ for $y^2=x^3-x$ and $y^2=x^3+1$ . Are these curves isomorphic over $\C$ ?

Solution

$y^2=x^3-x$ : $a=-1$ , $b=0$ , so $4a^3=-4$ and $j=1728$ . $y^2=x^3+1$ : $a=0$ , so $j=0$ . Different $j$ , so not isomorphic. These are the two curves with extra automorphisms: $(x,y)\mapsto(-x,iy)$ on the first, $(x,y)\mapsto(\omega x,y)$ on the second.

Why collinear points add to zeroA line is a divisor of degree 3

If a line $L$ meets the cubic in $P,Q,R$ , show that $P+Q+R=0$ in the group law defined through $\operatorname{Pic}^0$ .

Solution

Let $L$ also denote a linear form defining the line, and $Z$ the line at infinity, which meets the cubic in $3P_0$ . The rational function $L/Z$ has divisor $P+Q+R-3P_0$ . So $(P-P_0)+(Q-P_0)+(R-P_0)\sim0$ , which is $P+Q+R=0$ .

More practice: Hartshorne, Exercises IV.4.1 and IV.4.4.

InteractiveHartshorne, Ch. V §1: Geometry on a Surface · §2: Ruled Surfaces

Intersection numbers on a surface

On a surface, two curves meet in finitely many points, and the count depends only on their divisor classes. On the quadric $\mathbb{P}^1\times\mathbb{P}^1$ every class is a pair of numbers, and the count is a one-line formula you can check by drawing.

The idea

A surface here is a nonsingular projective surface over an algebraically closed field.

Intersection pairing. There is a unique symmetric bilinear pairing

\operatorname{Pic}X\times\operatorname{Pic}X\to\Z

,

(C,D)\mapsto C\cdot D

, such that

C\cdot D=\#(C\cap D)

when

C,D

are nonsingular curves meeting transversally. For curves with no common component,

C\cdot D=\sum_P\dim_k\mathcal{O}_P/(f,g)

.

Adjunction and Riemann–Roch. For a curve

C

on

X

:

2p_a(C)-2=C\cdot(C+K)

, and

p_a

is the genus if

C

is nonsingular. For a divisor

D

:

l(D)-s(D)+l(K-D)=\tfrac12D\cdot(D-K)+1+p_a(X)

, where

s(D)=h^1(\mathcal{O}(D))

.

Hodge index theorem. If

H

is ample,

D\cdot H=0

and

D

is not numerically equivalent to 0, then

D^2<0

.

The quadric and ruled surfaces.

X=\mathbb{P}^1\times\mathbb{P}^1

is the quadric surface in

\mathbb{P}^3

.

\operatorname{Pic}X=\Z V\oplus\Z H

, where

V

is a "vertical" fibre

\{pt\}\times\mathbb{P}^1

and

H

a "horizontal" one, with

V^2=H^2=0

,

V\cdot H=1

and

K=-2V-2H

. A class

C=aV+bH

is cut out by a form of bidegree

(a,b)

. More generally a ruled surface

\pi:X\to C

has

\operatorname{Pic}X\cong\Z C_0\oplus\pi^*\operatorname{Pic}C

for a section

C_0

with

C_0^2=-e

.

Then $(aV+bH)\cdot(cV+dH)=ad+bc$ and $p_a(aV+bH)=(a-1)(b-1)$ . The explorer represents $aV+bH$ by $a$ vertical and $b$ horizontal lines; any other curve in the class gives the same numbers.

Explore

\mathbb{P}^1\times\mathbb{P}^1

.

C=aV+bH

D=cV+dH

points of

C\cap D

C: vertical lines a C: horizontal lines b D: vertical lines c D: horizontal lines d

Worked example

Let $\varphi:\mathbb{P}^1\to\mathbb{P}^1$ be a morphism of degree $n$ , and $\Gamma\subset\mathbb{P}^1\times\mathbb{P}^1$ its graph $\{(x,\varphi(x))\}$ . Find the class of $\Gamma$ , check adjunction, and count fixed points.

Write $\Gamma=aV+bH$ . Compute $\Gamma\cdot V$ and $\Gamma\cdot H$ .

A vertical line $\{x_0\}\times\mathbb{P}^1$ meets $\Gamma$ once, at $(x_0,\varphi(x_0))$ , so $\Gamma\cdot V=b=1$ . A horizontal line $\mathbb{P}^1\times\{y_0\}$ meets it at the $n$ preimages of $y_0$ , so $\Gamma\cdot H=a=n$ . Thus $\Gamma=nV+H$ .
Compute $\Gamma^2$ and check the genus by adjunction.

$\Gamma^2=2n$ . And $\Gamma\cdot(\Gamma+K)=2n+(nV+H)\cdot(-2V-2H)=2n-2n-2=-2$ , so $p_a(\Gamma)=0$ , as it must be: $\Gamma\cong\mathbb{P}^1$ .
Fixed points of $\varphi$ are the points of $\Gamma\cap\Delta$ , where $\Delta=V+H$ is the diagonal. How many are there, counted with multiplicity?

$\Gamma\cdot\Delta=(nV+H)\cdot(V+H)=n+1$ . Example: $\varphi(z)=z^2$ fixes $0$ , $1$ and $\infty$ , which is $3=2+1$ .
Check the Hodge index theorem on $D=V-H$ , with the ample class $V+H$ .

$D\cdot(V+H)=1-1=0$ and $D$ is not numerically trivial ( $D\cdot V=-1$ ). Indeed $D^2=-2<0$ .

Check your understanding

In how many points do general curves in the classes $2V+H$ and $V+3H$ on $\mathbb{P}^1\times\mathbb{P}^1$ meet?

7
5Vertical meets horizontal, never vertical. Pair the 2 verticals of one with the 3 horizontals of the other, and so on.
6You counted only one of the two cross terms $ad$ and $bc$ .

(2V+H)\cdot(V+3H)=2\cdot3+1\cdot1=7

. Set

a=2,b=1,c=1,d=3

in the explorer and count the red dots.

A nonsingular curve in the class $3V+3H$ has genus:

4
1That is a plane cubic. On the quadric use adjunction with $K=-2V-2H$ .
18That is $C^2=2\cdot3\cdot3$ , not the genus.

p_a=(a-1)(b-1)=4

. Such a curve is the intersection of the quadric with a cubic surface in

\mathbb{P}^3

, a canonical curve of genus 4.

Can an irreducible curve on $\mathbb{P}^1\times\mathbb{P}^1$ have negative self-intersection?

No: its class is $aV+bH$ with $a,b\ge0$ , so $C^2=2ab\ge0$ .
Yes, the diagonal. $\Delta=V+H$ has $\Delta^2=2$ .
Yes, after moving it. $C^2$ depends only on the class, and moving doesn't change the class.

a=C\cdot H

and

b=C\cdot V

, and an irreducible curve meets a fibre that is not one of its components in an effective way, so both are

\ge0

. Contrast the exceptional curve of a blow-up, with

E^2=-1

(next module).

Practice

Genus-1 curves on the quadricWhich classes, and which degree?

Which classes $aV+bH$ with $a,b\ge1$ contain curves of arithmetic genus 1? Embedding the quadric in $\mathbb{P}^3$ by $V+H$ , what is the degree of such a curve?

Solution

$(a-1)(b-1)=1$ forces $a=b=2$ . The degree is the intersection with a hyperplane section: $(2V+2H)\cdot(V+H)=4$ . These are the elliptic quartic curves, intersections of the quadric with a second quadric.

The ruled surface X₁ℙ² blown up at a point

Let $X$ be $\mathbb{P}^2$ blown up at a point $p$ , with $L$ the pullback of a line and $E$ the exceptional curve ( $L^2=1$ , $L\cdot E=0$ , $E^2=-1$ , $K=-3L+E$ ). Show that the lines through $p$ become the fibres of a ruling over $\mathbb{P}^1$ with $E$ as a section, and find $e$ .

Solution

Their strict transforms have class $F=L-E$ , with $F^2=1-1=0$ : disjoint, the fibres of projection from $p$ . $E\cdot F=0-(-1)=1$ , so $E$ meets each fibre once and is a section with $E^2=-1$ , so $e=1$ . Adjunction checks: $F\cdot(F+K)=0+(L-E)\cdot(-3L+E)=-3+1=-2$ and $E\cdot(E+K)=-1-1=-2$ , so both are rational curves. Also $K^2=9-1=8$ , the same as for $\mathbb{P}^1\times\mathbb{P}^1$ .

More practice: Hartshorne, Exercises V.1.3 and V.1.5, and III.5.6 for curves on the quadric.

InteractiveHartshorne, Ch. V §3: Monoidal Transformations · §4: The Cubic Surface in ℙ³

Blowing up and the 27 lines

Blow up six points of the plane in general position and map the result to $\mathbb{P}^3$ by the cubics through them: the image is a nonsingular cubic surface. Its 27 lines can be listed from the six points, and intersection numbers say which lines meet.

The idea

Blowing up a point. Let

\pi:\tilde X\to X

be the blow-up of a point

P

on a surface, with exceptional curve

E\cong\mathbb{P}^1

. Then

\operatorname{Pic}\tilde X\cong\operatorname{Pic}X\oplus\Z E

, with

\pi^*C\cdot\pi^*D=C\cdot D

,

\pi^*C\cdot E=0

,

E^2=-1

, and

K_{\tilde X}=\pi^*K_X+E

.

Strict transforms. If

C

has multiplicity

r

at

P

, then

\pi^*C=\tilde C+rE

,

\tilde C\cdot E=r

, and

p_a(\tilde C)=p_a(C)-\tfrac{r(r-1)}2

.

The cubic surface. Let

X

be

\mathbb{P}^2

blown up at six points with no three on a line and not all six on a conic. Cubics through the six points form a 4-dimensional space of sections, and they embed

X

in

\mathbb{P}^3

as a nonsingular cubic surface with hyperplane class

-K=3L-\sum E_i

. Every nonsingular cubic surface arises this way.

The 27 lines. A line on

X

is a curve

\ell

with

\ell^2=-1

and

K\cdot\ell=-1

. There are exactly 27: the six

E_i

; the fifteen

L_{ij}=L-E_i-E_j

(lines through two of the points); the six

C_j=2L-\sum_{i\neq j}E_i

(conics through five of them). With

D=aL-\sum b_iE_i

, the pairing is

D\cdot D'=aa'-\sum b_ib_i'

.

Explore

Drag the six points. Click a point, line or conic (or a button) to pick one of the 27 lines. selected the 10 lines meeting it

Pick a line

Worked example

Let $C\subset\mathbb{P}^2$ be a cubic with a node at $P$ , and blow up $P$ . Find the class of $\tilde C$ , its self-intersection, its genus and how it meets $E$ .

Write $\pi^*C$ in terms of $\tilde C$ and $E$ , and find the class of $\tilde C$ .

A node has multiplicity 2, so $\pi^*C=\tilde C+2E$ and $\tilde C=3L-2E$ , where $L$ is the pullback of a line.
Compute $\tilde C\cdot E$ . What does it mean?

$(3L-2E)\cdot E=0-2E^2=2$ . The strict transform meets $E$ in two points, one for each tangent direction at the node, which is what the animation in the rational maps module shows.
Compute $\tilde C^2$ and the genus of $\tilde C$ by adjunction, with $K=-3L+E$ .

$\tilde C^2=9-4=5$ . Since $\tilde C+K=(3L-2E)+(-3L+E)=-E$ , we get $\tilde C\cdot(\tilde C+K)=(3L-2E)\cdot(-E)=-3\,L\cdot E+2E^2=0-2=-2$ . So $p_a(\tilde C)=0$ , matching $p_a(C)-\tfrac{2\cdot1}2=1-1$ .
What changes if $P$ is a cusp instead?

All the numbers stay the same, since a cusp also has multiplicity 2. But the two intersection points of $\tilde C$ and $E$ coincide: $\tilde C$ is tangent to $E$ at the single tangent direction.

Check your understanding

On a nonsingular cubic surface, how many of the other 26 lines meet a given line?

10
26Lines on a surface can be disjoint: $E_1\cdot E_2=0$ .
16That is the number of lines disjoint from it.

For

E_1

: the five

L_{1j}

and the five conics

C_j

,

j\neq1

, each with intersection number 1. Every line looks the same (the symmetry group of the configuration acts transitively), and the explorer counts 10 for any choice.

Blow up $\mathbb{P}^2$ at a point. The strict transform of a line through the point has self-intersection:

0
1That is $L^2$ before blowing up. The class is now $L-E$ .
$-1$ That is $E^2$ , or $(L-E_i-E_j)^2$ for a line through two blown-up points.

(L-E)^2=L^2-2L\cdot E+E^2=1-0-1=0

.

If three of the six points are collinear, the strict transform of their line has class $L-E_1-E_2-E_3$ . What is its self-intersection, and what goes wrong?

$-2$ , and $K\cdot\ell=0$ : the cubics contract this curve to a point, a singular point of the image surface.
$-1$ : it is just another line.Each blown-up point on the line lowers the self-intersection by 1.
0: it moves in a family.Compute $1-3$ .

\ell^2=1-3=-2

, and

-K\cdot\ell=3-3=0

, so the hyperplane class has degree 0 on

\ell

: the map to

\mathbb{P}^3

collapses it. Drag three points into a line in the explorer to see the warning.

Practice

The degree of the cubic surfaceK² and lines

For $X=\mathbb{P}^2$ blown up at six points, show $K^2=3$ , and that each $E_i$ has degree 1 in the embedding by $-K$ .

Solution

$K=-3L+\sum E_i$ , so $K^2=9+\sum E_i^2=9-6=3$ : the surface has degree $(-K)^2=3$ . And $-K\cdot E_i=(3L-\sum E_j)\cdot E_i=-E_i^2=1$ , so $E_i$ maps to a curve of degree 1, a line.

Two skew lines and their transversalsCounting with classes

$E_1$ and $E_2$ are disjoint. How many of the 27 lines meet both?

Solution

Lines meeting $E_1$ : $L_{1j}$ ( $j\neq1$ ) and $C_j$ ( $j\neq1$ ). Lines meeting $E_2$ : $L_{2j}$ and $C_j$ ( $j\neq2$ ). Both: $L_{12}$ and $C_3,\dots,C_6$ . That is 5. Pick $E_1$ and then $E_2$ in the explorer to see them.

More practice: Hartshorne, Exercises V.3.2 and V.4.16.