Algebraic Geometry II Homework Chapter IV Curves
A course by prof. Chin-Lung Wang 2020 Spring
Exercise 0 (by Kuan-Wen).
This is an example of proof.
Remark. This is an example for how to write in this format.
1 Riemann-Roch Theorem
Exercise 1 (by Chi-Kang).
This is equivalent to show that there exists f ∈ K(X) s,t, f ∈ H0(X, nP ) for some n > 0 and f is non-constant. By Riemann-Roch we have for any natural number n ≥ 2g − 1, h1(nP ) = 0, thus
χ(nP ) = h0(nP ) = n + 1 − g
take n ≥ g + 1 we have h0(nP ) ≥ 2, hence there is a non-constant function f ∈ H0(X, nP ).
Exercise 2 (by Chi-Kang).
Use induction on r, r = 1 is just exercise 1.1. If the consequence holds for r − 1, then for r, there exists some f s,t, f has pole at P1, ..., Pr−1 and regular elsewhere. And as 1.1 there is g s,t, g has pole at Pr and regular elsewhere. hence f + g is a function has pole at P1, ..., Pr and regular elsewhere.
Exercise 3 (by Yi-Tsung Wang).
Proof. By Nagata theorem (remark 2.7.17.2), X can be embedding as an open subset of a complete curve X, then in this case X\X is just a finite set, say X\X = {p1, . . . , pr}. By Exercise 4.1.2, take f : X → P1 such that f has poles at each of the pi and regular elsewhere. Since f is not constant, f must be surjective, then f−1(A1) = X. Moreover, f is a finite morphism, hence an affine morphism, and then X = f−1(A1) is affine.
Exercise 4 (by Yi-Tsung Wang).
Proof. Let X be a separated one-dimensional scheme of finite type over k. By Exercise 3.3.1, we may assume X is reduced. By Exercise 3.3.2, we may furthermore assume X is irreducible, hence X is integral and is not proper over k. Let Y be the normalization of X, and the natural map π : Y → X. π is finite since X is of finite type over k by Exercise 2.3.8 and is surjective since X is integral (locally, it is going-up).
If Y is proper over k, by Exercise 2.4.4, X = π (Y ) is also proper over k, contradiction. Now note that Y is also integral, separated, one-dimensional scheme of finite type over k, and is regular since Y is furthermore normal, by Exercise 4.1.3, Y is affine. By Chevalley’s theorem (Exercise 3.4.2), X is also affine.
Exercise 5 (by Shuang-Yen Lee).
By Riemann-Roch Theorem, we have
dim |D| = `(D) − 1 = `(K − D) + deg(D + 1 − g = deg(D) + (`(K − D) − g) ≤ deg(D) since K − D ≤ K =⇒ `(K − D) ≤ `(K) = g. If g = 0, then
deg(K − D) ≤ deg(K) = −2 =⇒ `(K − D) − g = 0
so the equality holds. If g 6= 0, then D = 0 =⇒ `(K − D) − g = `(K) − g = 0. Suppose D 6= 0, say D =P niPi, then K − D ≤ K − P1 ≤ K, so
0 = `(K − D) − g ≤ `(K − P1) − g ≤ `(K) − g = 0 =⇒ `(K − P1) = g.
By Riemann-Roch Theorem, `(P1) = `(K − P1) + 2 − g. So `(P1) = 2, which is impossible since g > 0.
Exercise 6 (by Shi-Xin).
Let P be a point on X, and let g denote g(X). Consider the divisor D = (g + 1)P . By Riemann-Roch Theorem, we have
`(D) ≥ deg D + 1 − g > 1.
Therefore, there is a f ∈ K(X) such that (g + 1)P + div(f ) ≥ 0, i.e. f has pole at P with order ≤ g + 1 and is regular everywhere else. Thus it induces a finite morphism ˜f : X → P1 by x 7→ f (x) which is of degree ≤ g + 1 since deg ˜f · deg ∞ = deg D.
Exercise 8 (by Shi-Xin).
(a) From 0 → OX → f∗OX˜ →P
p∈XO˜p/Op → 0 where f : ˜X → X is the normalization of X, we obtain 0 → H0(X, OX) → H0(X, f∗OX˜) ∼= H0( ˜X, OX˜) → H0(X,X
p∈X
O˜p/Op)
→ H1(X, OX) → H1(X, f∗OX˜) ∼= H1( ˜X, OX˜) → H1(X,X
p∈X
O˜p/Op) = 0
Since H0(X, OX) ∼= H0( ˜X, OX˜) ∼= k, we have 0 → H0(X,X
p∈X
O˜p/Op) → H1(X, OX) → H1( ˜X, OX˜) → 0
Thus by ex.iii.5.3, pa(X) = pa( ˜X) +P
p∈Xlength( ˜Op/Op) = pa( ˜X) +P
p∈Xδp.
(b) If pa(X) = 0, then it forces pa( ˜X) = 0 and δp = 0 for any p ∈ X. So f is an isomorphism, i.e.
X ∼= ˜X ∼= P1 which is given by Riemann-Roch Theorem.
(c)
Exercise 9 (by Ping-Hsun Chuang).
Proof. (a) Let f : Xreg → X be the inclusion. We have the short exact sequence
0 O (D) f∗Oreg(D) X
P ∈X
(f∗Oreg(D))P/O (D)P 0
Note that O (D)P = OX,P since O (D) is an invertible sheaf. Also, (f∗Oreg(D))P = OX,Pe since Xreg is normal. Moreover,
δP = length
OX,Pe /OX,P
= h0 X, (f∗Oreg(D))P /O (D)P . Finally, since χ is a additive function, we have
χ (O (D)) = χ (f∗Oreg(D)) − X
P ∈X
χ (f∗Oreg(D)P /O (D)P)
= deg D + 1 − pa(Xreg) − X
P ∈X
δP = deg D + 1 − pa(X) .
(b) Since X is projective, take a very ample divisor R on X. Then, there exists n > 0 such that O (nR + D) is generated by global section. Then, since R is very ample, O (nR + D + R) is also very ample. Now, D = M − (n + 1) R, where M = D + (n + 1) R which is very ample.
(c) Using the result in (b), it suffice to show the result in case that L is very ample. Write L = f∗O (1) for some embedding f : X → PN. Since X has finitely may irregular points, there exists a hyperplane H in PN such that H ∩ X ⊆ Xreg. Now, take D = H ∩ X.
(d) Since X is locally complete intersection, we may apply the Sere duality. We then get H1(X, O (D)) ∼= Ext0X(O (D) , ω◦X)∨
∼= Ext0X(OX, O (−D) ⊗ ωX◦)∨
∼= H0(OX, O (−D) ⊗ ωX◦)∨
Then, χ (O (D)) = h0(X, O (D)) − h1(X, O (D)) = ` (D) − ` (K − D). Finally, using the result in (a), we get the required formula.
Exercise 10 (by Ping-Hsun Chuang).
Proof. Apply exercise 4.1.9 to D = K and get
` (K) − ` (0) = deg K + 1 − ga = deg K.
Also, we have
` (K) = h0(X, ωX◦) = h1(X, OX) = pa.
Note that the second equality above holds by the Serre duality since we assume X is locally complete intersection. Thus, we get deg K = pa− 1 = 0.
Now, for any D ∈ Pic0X, apply exercise 4.1.9 to D = D + P0 and get
` (D + P0) − ` (K − D − P0) = deg (D + P0) + 1 − pa = 1.
Also, we have deg (K − D − P0) = deg K − 1 = −1 and thus ` (K − D − P0) = 0. Hence, ` (D + P0) = 1, that is, there exists a unique R > 0 such that R ∼ D + P0. Therefore, for any D ∈ Pic0X, we find a unique R such that D ∼ R − P0 and thus Xreg → Pic0X is bijection.
2 Hurwitz’s Theorem
Exercise 1 (by Pei-Hsuan Chang).
Induction on n. For n = 1, it is Example IV.2.5.3 in Hartshorne. So let’s deal with the case n > 1. Let f : X → Pn be an étale covering. We may assume that X is connected. For each hyperplane H ∼= Pn−1 in Pn, f : f∗H → H is an étale covering of H. By induction hypothesis, f∗H is disjoint union of copies of H.
Now, we are going to showing that f∗H is connected, and conclude f∗H is isomorphic to H via f . To show this, we want to show that X is normal and f∗H is ample with codimension 1, then by Corollary III.
7.9, f∗H will be connected. Notice that H is ample, so f∗H is ample since f is finite. Also, an étale covering is smooth, so X is smooth over k and thus, is normal. Hence, f∗H ∼= H, and f |f∗H is an isomorphism.
Now, deg f = deg f |f∗H = 1. An étale covering with degree 1 is an isomorphism, so X = Pn. This complete the prove.
Exercise 2 (by Yu-Chi Hou).
(a) From Exercise 1.7, we know that any curve X of genus 2 is hyperelliptic whose the degree 2 morphism f := φ|KX| : X −→ P1 coming from the canonical system. Using Riemann-Hurwitz formula, one computes directly that deg(R) = 6. If P is branched point of f , then eP = 2, for any P ∈ f−1(Q).
Since char(k) 6= 2, any ramification point P ∈ X is tamely ramified,
R = X
P ∈X
(eP − 1), and deg(R) = 6.
Hence, f is ramified exactly at 6 points.
(b) Let h(x) := (x − α1) · · · (x − α6) ∈ k[x] and K := (k(x))[z])/(z2− h). Since K/k(x) is an algebraic extension of degree 2, K/k has transcendental degree 1. This together with [K : k(x)] = 2 determines a non-singular projective curve X and a morphism f : X → P1 of degree 2. On the affine open chart U0 = Spec k[x] ⊂ P1, there exists a morphism from the affine open set V ⊆ f−1(U0) → U0 correpsonding to the inclusion k[x] ,→ k[x, ¯z] := k[x, z]/(z2− h). Hence, the function field K(V ) = K(X) = K.
Since k = ¯k, any closed point P ∈ U0 ⊂ P1 correponds to the maximal ideal (x − α) ⊂ k[x]. If α /∈ {α1, . . . , α6}, then h(α) 6= 0 and thus (x − α, ¯z ±ph(α) is a maximal ideal in k[x, z]/(z2− h).
In other words, #f−1(P ) = 2 if P ∈ U0 not corresponding to α1, . . . , α6. Thus, we have shown that f−1(U0) → U0 is only branched at α1, . . . , α6 ∈ k ∼= U0 ⊂ P1.
Next, we need to check that f does not brached at ∞ ∈ P1. To see this, we first localizing k[x](x) = k[x, x−1] ,→ (k[x, z]/(z2− h))(x)= k[x, x−1, z]/(z2− h),
which correponds to f−1(U0∩ U1) → U0∩ U1. On k[x, x−1, z], we first assume that α1, . . . , α6 ∈ k∗, then
z2− h(x) = z2− (x − α1) · · · (x − α6) = z2− x6(1 −α1
x ) · · · (1 − α6 x )
=z2− α1· · · α6x6( 1 α1 − 1
x) · · · ( 1 α6 − 1
x) = x6(x−6z2 − α1· · · α6˜h(1/x)), where ˜h(1/x) := Q6
i=1
1 αi − 1x
∈ k[x−1]. Since x is a unit in k[x, x−1, z], (z2 − h(x)) = (˜z2 − α1· · · α6h(1/x)), where ˜˜ z = x−3z. Hence, we have
k[x, x−1, z]/(z2− h) = k[x, ˜x, ˜z]/(˜z2− α1· · · α6˜h(1/x)).
Let y = 1/x, k[x, ˜x, ˜z]/(˜z2 − α1· · · α6˜h(1/x)) = k[y, y−1, ˜z]/(˜z2 = α1· · · α6h(y)). Thus, on U =˜ Spec k[y], f−1(U1) → U1 is defined by the correponding morphism from k[y] ,→ k[y, ˜z]/(˜z2 = α1· · · α6h). Same argument as above shows that f is only branched at y − α˜ i ∈ Spec k[y] for i = 1, . . . , 6. Now„ if α6 = 0, α1, . . . α5 6= 0 (since α1, · · · , α6 are distinct),
h(x) = x(x − α1) · · · (x − α5) = α1· · · α5x6( 1 α1 − 1
x) · · · ( 1
α5 − 1/x).
Repeating above argument shows that f is not branched at ∞. Thus, f is only ramified over 6 points with each ramification index 2 (since f is of degree 2). Using Riemann-Hurwitz formula, 2g(X) − 2 = 2(0 − 2) + 6 = 2 ⇒ g(x) = 2. Moreover, let P ∈ X such that f (P ) = Q ∈ {α1, . . . , α6}, then f∗P =P
P ∈f−1(QeP · P = 2P . Thus, f∗OP1(Q) = f∗OP1(1) = OX(2P ).
On the other hand, using Riemann-Roch, h0(X, OX(2P )) − h0(X, OX(KX− 2P )) = deg(2P ) − g(X) + 1 = 1. Since H0(X, OX(2P )) = H0(X, f∗OP1(1)) = H0(P1, OP1(1)) ∼= k2, h0(X, OX(KX − 2P )) = 2.
However, deg(KX − 2P ) = 2g(x) − 2 − 2 = 0. Thus, KX ∼ 2P . Hence, the map f : X → P1 is the same as the one determined by |KX|.
(c) If Pi 6= ∞ ∈ P1, for i = 1, 2, 3, then let P1 = [a : 1], P2 = [b : 1], P3 = [c : 1], then the Möbius transform φ(z) = z−az−cb−cb−a maps P1 to 0, P2 to 1, and P3 to ∞. If P1 = ∞, P2 = [b : 1], P3 = [c : 1], then we take φ(z) = z−cb−c. Since Aut(P1) = P GL(2), such φ is unique.
(d) The symmetric group S3acts on distinct element β1, β2, β3 ∈ k\{0, 1} by permuting {0, 1, ∞, β1, β2, β3), then sending the first three element to 0, 1, ∞ by Möbius transform again, then call them β10, β20, β30. Then we define [β1, β2, β3] to be the equivalence class of (β1β2, β3) modulo such S3− action.
(e) Given any genus 2 curve X, |KX| gives f : X → P1 with six distinct brached points P1, . . . , P6. Then using Möbius transform, we sends P1 7→ 0, P2 7→ 1, P3 7→ ∞, Pi 7→ βi−3, for i = 4, 5, 6. We then get an equivalence class [β1, β2, β3] modulo S3−action described in (d). Now, if φ : X −→ X∼ 0 be an isomorphism, then φ∗KX0 ∼ Kx. Thus, |φ∗KX0| gives a morphism to P1 which differ to the one from
|KX| by an ψ ∈ Aut(P1) = P GL(2). Then as in (d), the tuple (β10, β20, β30) differ by (β1, β2, β3) by an S3−action. Thus, [β10, β20, β30] = [β1, β2, β3].
Also, (b) implies that starting from six points of P1, one can construct a genus two curve X whose φ|KX| is branched exactly at the given six points. Thus, we established the isomorphism class [X]
with the tuple [β1, β2, β3] modulo S3−action.
Exercise 4 (by Yi-Tsung Wang).
Proof. Let f (x, y, z) = x3y + y3z + z3x. Then fx = z3, fy = x3, fz = y3 ⇒ f is non-singular since (0, 0, 0) /∈ P2. Since
fxx fxy fxz
fyx fyy fyz fzx fzy fzz
=
0 0 3z2 3x2 0 0
0 3y2 0
= 0 every point of X is an inflection point. For p (a, b, c) ∈ X, the tangent line at p is
c3(x − a) + a3(y − b) + b3(z − x) = 0
that is, c3x + a3y + b3z = 0. Then the natural map X → X∗ is defined by (a, b, c) 7→ (c3, a3, b3), which is a Frobenius morphism, hence is isomorphic and purely inseparable.
Exercise 5 (by Yu-Ting Huang).
(a) Let G act on X, then f−1(f (P )) is an orbit of the group action, then |f−1f (P )| = nr and each element in f−1f (P ) are of index r as P . By Hurwitz’s theorem, 2g(X) − 2 = n(2g(Y ) − 2) +P
p(ep − 1).
Then 2g−2n = n1 P
P(eP − 1) = 1nPs i=1
n
ri(ri− 1) =Ps
i=1(1 −r1
i).
(b) First, note that 2g(Y ) − 2 +Ps
i=1(1 −r1
i) = 2g(X)−2n > 0, since g(X) ≥ 2. If g(Y ) = 0, −2 +Ps i=1(1 +
1
ri) = 2g(X)−2n ≥ n2 ≥ 0. Thus, Ps
i=1(1 − r1
i) ≥ n2 + 2. Consider the minimal possibility of ri such that Ps
i=1(1 −r1
i) ≥ n2+ 2. We find that ri = 2, 3, 7. Then, −2 +12+32+67 = 421 = 2(g−1)n . i.e. n = 84(g − 1).
In the case g(Y ) = 0, n ≤ 84(g − 1).
As for g(Y ) ≥ 1, 2g(Y )−2 > 0, soPs
i=1(1−r1
i) > 0. To find maximal n, we set s = 1, r1 = 2, g(Y ) = 1.
Then 2 − 2 + (1 −12) = 2g−2n . i.e. n = 4(g − 1) ≤ 84(g − 1). Now, we can conclude that n ≤ 84(g − 1).
Exercise 6 (by Tzu-Yang Chou).
(a) Let D be effective. We first consider the short exact sequence 0 −→ OX(−D) −→ OX −→ OD −→ 0 and apply f∗. Since f is finite, so R1f∗OX(−D) vanishes and hence we have 0 −→ f∗OX(−D) −→
f∗OX −→ f∗OD −→ 0 is exact. Taking determinant, we obtain det(f∗OX) ' det(f∗OX(−D)) ⊗ det(f∗OD). Now we only need that (det(f∗OD))−1 ' OX(f∗(−D)) since then for a general divisor, we can write it as a difference of two effective ones and above formula proves the assertion. But this statement follows from Ex(II.6.11)(c).
(b) (a) tells us that OX(f∗D) ' det(f∗OX(D)) ⊗ (det(f∗OX))−1 and hence it only depends on the linear equivalence class of D. f∗◦ f∗ = n follows from their definitions, where n = deg f .
(c) By Ex(III.7.2) and Ex(III.6.10), we have the following sequence of isomorphisms: det(f∗ΩX) ' det(f∗H omX(OX, ΩX)) ' det(f∗H omX(OX, f!ΩY)) ' det(H omY(f∗OX, ΩY)) ' det((f∗OX)−1⊗ ΩY) ' (det(f∗OX))−1⊗ Ω⊗nY
(d) KX = f∗KY + R ⇒ f∗KX = nKY + B ⇒ OX(−B) ' Ω⊗nY ⊗ (OX(f∗KX))−1, and this is isomorphic to (det(f∗OX))2 by (a) and (c).
Exercise 7 (by Po-Sheng Wu).
(a) Since f is finite flat, f∗OX is locally free of rank 2. Plus, the injection OY → f∗OX is also injective on residue field, so the kernel L is locally free of rank 1. By taking det for the short exact sequence we have L ∼= detf∗OX, and then by 2.6(d) we have L2 ∼= OY since f is etale.
(b) On the affine subset U = Spec(A) ⊂ Y such that L is free, the constructed algebra is actually isomorphic to A[t]/(t2− u) via (a, bv) 7→ a + bt, where v is a generator of L(U ), and u = φ(v ⊗ v) is a unit of A. Since A[t]/(t2− u) is unramified over A, Spec(O ⊕ L) is etale over Y .
(c) Conversely, if X 7→ Y is etale of degree 2, then locally f∗OX(U ) is an unramified algebra of rank 2 over A, which is always able to be written in the form A[t]/(t2− u), so the exact sequence in (a) is splitted by f∗OX → OY where the map A[t]/(t2 − u) → A is given by taking the constant term. Now we see that (a) and (b) are converse to each other.
3 Embeddings in Projective Space
Exercise 1 (by Shi-Xin Wang).
Since deg D ≥ 5 = 2g(X) + 1, by Corollary 3.2, D is very ample. So we only need to show that if D is very ample, then deg D ≥ 5. We first show that dim |D| ≥ 3. Indeed, if dim |D| = 1, it defines a closed immersion to P1, which is impossible. Moreover, if dim |D| = 2, it defines a closed immersion from X to P2 as a plane curve, and hence by Riemann-Roch formula,
deg D = g(X) − 1 + dim |D| − dim |K − D| = 4
Therefore, g(X) = 12(deg D − 1)(deg D − 2) = 3 6= 2 is a contradiction. On the other hand, by ex.iv.1.5, deg D > dim |D| ≥ 3. Then we may assume deg D = 4. Since deg D ≥ 2g(X) − 2 = 2, dim |K − D| = −1.
However, there is a contraction
deg D = g(X) − 1 + dim |D| − dim |K − D| ≥ 5 Thus we must have deg D ≥ 5.
Exercise 2 (by Yi-Tsung Wang).
(a) Let K be a canonical divisor. Since ωX = OX(d − n − 1) = OX(1), we see that K = |K|∗L = X.L for some line L.
(b) Since X is a plane curve of degree 4, we have g (X) = 3. Since ωX = OX(1) is very ample, so is K. ` (K − D) = ` (K) − 2 = 1. By Riemann-Roch, ` (D) = deg D + 1 − g + ` (K − D) = 1, hence dim |D| = 0.
(c) Suppose not, let f : X → P1 be a finite morphism of degree 2, then D := f∗(∞) is an effective divisor of degree 2. By part (b), ` (D) = 1, and since f ∈ Γ (X,L (D)), f sends all x ∈ X to ∞ ∈ P1, contradiction. Hence X is not hyperelliptic.
Exercise 3 (by Tzu-Yang Chou).
By Ex(II.8.4), OX(K) ' OX(m) for some integer m. Moreover, deg K = 2g − 20 so m > 0 and hence K is very ample. When g = 2, K has degree 2 < 5 so cannot be very ample by Ex(IV.3.1); thus X must not be a complete intersection.
Exercise 4 (by Yu-Chi Hou).
(a) For d ≥ 1, let νd: P1 → Pdbe d−uple embedding of P1 in Pdand let X be its image. Recall the d−uple embedding is given by νd([t0, t1]) = [td0 : t0d−1t1 : · · · : t0td−11 : td1]. From Exercise I.2.12, we know that X is integral, S(X) = k[x0, . . . , xd]/I(X) is integral, and I(X) = ker(θ), where θ : k[x0, x1, . . . , xd] → k[t0, t1] is given by xi 7→ td−i0 ti1. In other words, we can write S(X) = k[td0, td−10 t1, . . . , td1]. Given r ∈ Frac(S(X)) = k(t0, t1) which in integral over S(X). Write r(t0, t1) = f (tg(t0,t1)
0,t1), where f, g ∈ k[t0, t1] and gcd(f, g) = 1, and there exists a0, a1, . . . , an−1 ∈ S(X) such that
rn+ an−1rn−1+ · · · + a1r + a0 = 0.
Repeating the proof that UFD are integrally closed (clean out the denominator g and use the relative primeness of g and f ), we know that g ∈ k∗ and hence r = f (t0, t1) ∈ k[t0, t1]. Hence, above equation reads
fn+ an−1fn−1+ · · · + a1f + a0 = 0. (1) By comparing degree, we may assume that a0, . . . , ad−1 are homogeneous of degree k0, . . . , kn−1 in degree d monomial of t0, t1 and g is homogeneous of degree m in t0, t1. Thus, equating the degree of (1) gives mn = m(n − 1) + dkn−1+ · · · = m + dk1 = dk0. Hence, m = dkn−1 ⇒ d | m. Thus,
f =P
iaitdn−1−i0 ti1. However, since each monomial tk0tj1 in a0, . . . , ad−1, these exponents congruent to zero modulo d. As a result, i ≡ 0 mod d. In other words, f ∈ S(X).
Alternatively, we see that V := Spec(k[td0, td−10 t1, . . . , td1]) is the affine toric variety associated to the cone σ := R≥0he1, e1+ de2i ⊂ R2. Since σ is strongly convex polyhedral cone, the affine monoid Sσ = Z2∩ σ is saturated, and hence the variety V = Spec(k[Sσ]) is normal. Also, observe that if V is an affine cone over a projective variety X, V is normal if and only if X is projectively normal by definition.
Next, we show that the homogenous ideal I(X) is generated by homogeneous polynomial of degree 2.
More precisely, we show that that
I(X) = hgij := xixj+1− xi+1xj : 0 ≤ i < j ≤ d − 1i ⊂ k[x0, . . . , xd].
Obviously, gij ∈ ker(θ) and hence I(X) ⊃ hgij : 0 ≤ i < j ≤ d − 1i. For the converse, given any homogeneous polynomial f ∈ I(X) of degree n, choose the lexicographic order x0 > x1 > · · · > xd as monomial ordering and let r be the remainder after division by gij’s. That is, r = f −P
0≤i<j≤d−1aijgij, where aij ∈ k[x0, . . . , xd]. By equating degree on both sides, we know that r is also a homogeneous polynomial of degree n. We now have two simple observations:
(1) r contains no monomial of the form −xli, for i = 1, . . . , d − 1. If there were such monomial, then such term can be subtracted by some multiple of gi−1,i := xi−1xi+1− x2i.
(2) Also, r contains no monomial involving variables xi, xj with j − i ≥ 2. If there were, then again such term can be subtracted by some multiple of gi,j−1:= xixj− xi+1xj−1.
Following these two observations, r can be decomposed into
r = h0(x0, x1) + h1(x1, x2) + · · · + hd−1(xd−1, xd),
where each hi is homogeneous of degree n, for all i = 0, . . . , d − 1 and contains no term like xni, for i = 1, . . . , d − 1.
Finally, for r = f −P
ijaijgij ∈ I(X), that is to say, r(td0, td−10 , . . . , td1) = 0. For each i = 1, . . . , d − 2, hi(xi, xi+1) =
n−1
X
k=1
c(i)k xn−ki xki+1 and
h0(x0, x1) = c(0)0 xn0 +
n−1
X
k=1
c(0)k xn−k0 xk1; hd−1(xd−1, xd) =
n−1
X
k=1
c(d−1)k xn−kd−1xkd+ c(d−1)d xd1. Thus, for i = 0, . . . , d, plugging xi by td−i0 ti1, we see that:
0 = c(0)0 tnd0 +
n−1
X
k=1
c(0)k tnd−k0 tk1 +
n−1
X
k=1
c(1)k tn(d−1)−k0 tr+k1 + · · · +
n−1
X
k=1
c(d−1)k tn−k0 tn(d−1)+k1 + c(d−1)d tnd1 .
Therefore, c(i)k = 0 for all i, k. That is, r = 0.
(b) Let X be a curve of degree d in Pn with d ≤ n and X * H, for any hyplerplane H in Pn. Take any hyperplane H, let D = X.H be the very ample divisor on X. Thus, deg(D) = deg(X.H) = deg(X) = d and dim |D| = n (otherwise, there exists a proper subspace V ⊂ h0(X, OX(D)) such that X ⊂ P(V∗) ( Pn). Now, since X * H, there exists P /∈ Bs|D|, then dim |D −P | = dim |D|−1 = n−1 and deg(D − P ) = d − 1.
If n > d, then pick P1, . . . , Pd ∈ Bs|D|, inductive on aboce arguement gives dim |D−/ Pd
i=1Pi| = n−d >
0 yet deg(D −Pd
i=1Pi) = 0. Therefore, D −Pd
i=1Pi ∼ 0. If so, then h0(X, OX(D −Pn
i=1Pi) = 1, contradiction. Hence, n = d. By Exercise IV.1.5, deg(d) = dim |D| if and only if D ∼ 0 or g(X) = 0.
By assumption, deg(D) > 0, we then must have g(X) = 0 and OX(H) = OP1(dH). Therefore, X ∼= νd(P1) up to Aut(Pn).
(c) If X is of degree 2 in Pn. If X is not contained in any hyperplane, then n = 2 by (b). If there exists a hyperplane H ∼= Pn−1 such that X ⊆ H, then replacing n by n − 1 and repeating the previous argument, we still get n = 2. Hence, X is a plane conic.
(d) Let X be a curve of degree 3. The same argument in (c) shows that X ⊆ P3. We now have two cases.
If X is not contained in any plane P2, then X ∼= ν3(P1) by (b). It is indeed the twisted cubic curve up to a projective transform. If X falls into some plane, then it is a plane cubic.
Exercise 6 (by Tzu-Yang Chou).
(a) Let n be the smallest integer such that X ⊆ Pn. First, Ex(IV.3.4)(b) implies that the case n > 3 is contained in (1). Also, for the case n = 2, we have g = (4−1)(4−2)2 = 3. For n = 3, we have g < 3 by Ex(IV.3.5)(b), so it remains to show that the genus cannot be 2 in this case. But X embed into P3 as a degree 4 curve, so there’s a degree 4 very ample divisor D, which contradicts to Ex(IV.3.1).
(b) Now we assume that X ⊆ P3 with g = 1. We consider the cohomology sequence of 0 −→
IX(2) −→ OP3(2) −→ OX(2) −→ 0, which is a four-term one. We see that h0(P3,IX(2)) = 10 − 8 + h1(P3,IX(2)) ≥ 2. Then the assertion follows from Bezout’s theorem.
Exercise 7 (by Yi-Heng Tsai).
Since char k 6= 2, the curve has only one node at (x, y) = (0, 0). Suppose there is a non-singular curve C which projects to it, then deg(C) = 4 and g(C) = 2 (contradicts to Ex3.6).
Exercise 9 (by Pei-Hsuan Chang).
Let H be a plane in P3. We have: H intersect X least then d distinct point ⇔ H contain a tangent line of X. Also, there are 3 intersection point of H and X are collinear ⇔ H contain a multisecant of X.
Notice that T := {H ∈ (P3)∗ | H contain a tangent line of X} is locally a subset of X × P1; thus, it has at most dimension 2. Consider S := { mulitsecants of X} ⊂ (X × X \ 4). It is a proper closed subset of X × X, so S has at most dimension 1. Hence, {H ∈ (P3 | H contains a multisecant of X} has at most dimension 2. So, T ∪ is a proper closed subset of (P3)∗. Thus, there is an open set U ⊂ (P3)∗ as desired.
4 Elliptic Curves
Exercise 1 (by Chi-Kang).
By R-R, we have h0(nP ) − h0(K − np) = n. Note that K = 0, so h0(K − nP ) is zero if n > 0, and is 1 if n = 0. So h0(nP ) = n for n > 0, and h0(0p) = 1.
Now embedded X by |3P | into P2, we say X in k[z0, z1, z2] is defined by z13 = z0(z0− z2)(z0 = λz2).
Now we choose t0 = 1 be a generator of H0(P ), x0 ∈ H0(2P ) s,t, {t0, x0} is a basis of H0(2P ), and similarly choose y0 ∈ H0(3P ) s,t, {t0, x0, y0} is a basis of H0(3P ). Then R is generated by t0, x0, y0 i,e, R = k[t0, x0, y0]/(relations). As the proof of proposition 4.6, after a change of coordinate we have y + 02 = x0(x − t0)(x − λt0). Note that in fact t0 = 1 ∈ H0(P ), so t20 = t0, thus we have the relation y20 = x0(x0− t20)(x0− λt20). Hence the map
k[t, x, y]/(y2− x(x − t2)(xλt2)) → R
is well-defined and surjective. Now the above 2 rings are intergal domain. Note that for any surjective homomorphism f : A → B between integral domain, if f is not an isomorphism we must have dim A > dim B.
But for our map both LHS and RHS has Krull dimension 2, hence it must an isomorphism.
Exercise 2 (by Yu-Chi Hou).
Let X be a genus 1 curve and D is a divisor on X with deg D ≥ 3. Since deg D ≥ 3, D is very ample (cf. Cor. IV.3.2). Hence, the complete linear system |D| gives an embedding φ|D| : X ,→ Pn, where
n = dim |D| = deg D + 1 using Riemann-Roch.
Lemma 1. X is projectively normal if and only if for any m ≥ 0, the natural map H0(Pn, OPn(m)) → H0(X, OX(m)) is a surjection.
The lemma is really a special case of Ex. II.5.14.
To check the condition of the lemma, we proceeds inductively on m. For m = 1, this follows directly from φ∗|D|OPn(1) = OX(D). Assume the induction hypothesis holds for m − 1, then we consider the following diagram
H0(Pn, OPn(m)) ⊗ H0(Pn, OPn(1)) H0(Pn, OPn(m + 1))
H0(X, OX(mD)) ⊗ H0(X, OX(D)) H0(X, OX((m + 1)D)),
where the horizontal maps are given by multiplication map and the vertical arrow is the natural map coming from X ,→ Pn. By induction hypothesis, the left arrow is surjective. If we can prove the surjectivity of the bottom horizaontal arrow, then the surjectivity of H0(Pn, OPn(m + 1)) → H0(X, OX((m + 1)D)) will follows.
Starting from here, we use the assumption that X is an elliptic curve. First of all, we can pick P ∈ X such that dP ∼ D, where d = deg D and from Riemann–Roch,
h0(X, OX(nP )) =
(1 , n = 0, 1 n , n ≥ 2.
Hence, for k ≥ 1, we have a sequence of strict inclusion
H0(X, OX(kP )) ( H0(X, OX((k + 1)P )).
Namely, there exists unique f ∈ K(X) which is regular outside P and ordP(f ) = k + 1, for each k ≥ 1.
Thus, for any f ∈ H0(X, OX((n + m)P )) there exists g ∈ H0(X, OX(nP )), h ∈ H0(X, OX(mP )) such that gh = f , for any n, m ≥ 3.
As a result, we see that the multiplication map
H0(X, OX(mdP )) ⊗ H0(X, OX(dP )) → H0(X, OX((m + 1)dP )) is surjective since d ≥ 3.
Exercise 3 (by Pei-Hsuan Chang).
Let f = y2 − x(x − 1)(x − λ). Then regular functions on X except P0 is k[x, y]/ < f > + : R.
Thus, K(X) = Frac(R) = {a(x) + b(x)y | a(x), b(x) ∈ k(x)}. Now, for each ϕ ∈ Aut(X), we can assume ϕ(x, y) = (x0, y0) = (u1(x) + v1(x), u2(x) + v2(x)y). Notice that ∀P = (x, y) ∈ X, 0 = ϕ(0) = ϕ(P + (−P )) = ϕ(P ) + ϕ(−P ) in the group law. So
P0 = ϕ(x, y) + ϕ(x, −y) = (u1(x) + v1(x), u2(x) + v2(x)y) + (u1(x) − v1(x), u2(x) − v2(x)y),
then u1(x) + v1(x) = u1(x) − v1(x) and u2(x) + v2(x)y = −(u2(x) − v2(x)y). Hence, v1(x) = u2(x) = 0, so ϕ(x, y) = (u1(x), v2(x)y).
Now, we homogenizes ϕ to get
˜
ϕ(x, y, z) = (u1x z
, v2x z
y
z, 1) = ( ˜u1(x, z), ˜v2(x, z)y, zn),
where ˜u1, ˜v2 are homogeneous rational functions of degree n and n − 1 respectively. Since ˜ϕ(P0) = P0,
˜
ϕ(0, 1, 0) = ( ˜u1(0, 0),˜v2(0, 0) · 1, 0) = (0, t, 0) for some t 6= 0. Thus, ˜v2(0, 0) 6= 0 ⇒ ˜v2(x, z) is constant, say
˜
v2(x, z) = c. Hence n = 1 ⇒ ˜u1 is linear. Now, de-homogenize ˜ϕ and get ϕ(x, y) = (x0, y0) = (ax + b, cy) for some constant a, b, c ∈ k on the affine piece.
Exercise 4 (by Tzu-Yang Tsai).
The equation equivalent to (y + a21x + a23)2 = x3+ (a2 +a412) + (a4+ a12a3)x + a6+ a423 , so by a linear transformation, we get Y3 = x3+ Ax2+ Bx + C, where A, B, C ∈ k0.
Let the roots of x3 + Ax2 + Bx + C = 0 be α, β, γ, we map {α 7→ 0
β 7→ 1 by a linear transformation, then γ 7→ γ−αβ−α = λ. Thus
j(λ) = 28(1 − λ + λ2)3 λ2(1 − λ)2
= 28(α2+ β2+ γ2 − αβ − βγ − γα)3 (α − β)2(β − γ)2(γ − α)2
, where both numerator and denominator are symmetric polynomial, which can be represented by elementary symmetric polynomial A, B, C. As a result, j is a rational function of {ai}, furthermore, j ∈ k0.
For j 6= 0, 1728, take A = 0, C = tB,
j = 28 B3
4B3+ 27C2 ⇒ B = −27jt2 4(j − 1728)
so simply take t = 1, notice that B ∈ k, we get an elliptic curve in k with j as j-invariant.
For j = 0, y2+ y = x3 is the curve; for j = 1728, y2 = x3+ x is the curve.
Exercise 5 (by Shuang-Yen Lee).
(a) By Hurwitz formula, f has no ramification points. Let P0 + Q = f∗P0, then P0 6= Q. Since
`(P0+ Q) = `(2P0) = `(2Q) = 2 (by R-R), there exist h1 ∈ L(P0 + Q), h2 ∈ L(2P0) and h3 ∈ L(2Q) which are not constant. Since `(P0) = `(Q) = 1, h21 6= L(2P0)∪L(2Q). So L(2P0+2Q) = h1, h21, h2, h3ik. Note that
(π ◦ f )∗(∞) = f∗π∗(∞) = f∗(2P0) = 2P0+ 2Q,
π ◦ f ∈ k×h21, say π ◦ f = a2h21 = (ah1)2 for some a ∈ k×. Let π0 = ah1, g = [x 7→ x2], then π ◦ f = g ◦ π0 and deg g = 2, so we get deg π0 = 2.
(b) By (a).
(c) The branch points of g are 0, ∞. ∞ is a branch point of π since π∗(∞) = 2P0. 0 is a branch point of π since f∗π∗(0) = π0∗g∗(0) = 2π0∗(0) and note that f has no ramification points. Suppose that other two branch points of π are 1, λ. Then
π0∗((1) + (−1)) = 2f∗(2Q1), π0∗((λ1/2) + (−λ1/2)) = f∗(2Q2) for some Q1, Q2 ∈ X, so 1, −1, λ1/2, −λ1/2 are branch points of π0.
Now we have two ways to count j. By the map π, we have j = 28 (λ2− λ + 1)3
λ2(1 − λ)2 . By the map π0, since the cross ratio
λ0 := (1, −1; λ1/2, −λ1/2) = 1 − λ1/2 1 + λ1/2
2 , we have
j = 28 (λ02− λ0+ 1)3
λ02(1 − λ0)2 = 28(λ2+ 14λ + 1)3 16λ(1 − λ)4 . So 16(λ2− λ + 1)3(1 − λ)2 = (λ2+ 14λ + 1)3λ.
(d) By solving the equation above, we have λ = −1, 3 ± 2√
2, 321(1 ± 3√
7i), 12(1 ± 3√ 7) and j = 26· 33, 26· 53, −33· 53, −33· 53,
respectively.
Exercise 9 (by Chi-Kang).
(a) The identity map is an isogenus, and the composition of 2 finite morphism is finite, so we only need to show if f : X → X0 is a finite morphism of degree n, then there exists X0 → X be another finite morphism. By exercise IV.4.7 we have a dual morphism ˆf : X0 → X s,t, ˆf ◦ f = nX is a finite morphism with degree n2, hence ˆf is also finite morphism with degree n, and thus isogenus is an equivalent realtion.
(b) Suppose f : X → X0, g : X → X00 are 2 finite morphism with the same (group theoritic) kernel, then X0 ∼= X00 as abelian group. So there is a natural group isomorphism g ◦ f−1: X0 ∼= X/(ker f ) ∼= X00, and this is a morphism between curves since both f, g is. Thus g ◦ f−1 is a bijective morphism between curves, hence it is an isomorphism since X0, X00 are smooth.
Now since ˆf ◦ f = nX so ker f ⊂ ker nX. And by exercise 4.7 we have f ◦ ˆf = nX0, so both f, ˆf has degree n, thus degnX = n2, so X has n2 element of order n, hence X has at most countably many subgroups G which is a subgroup of some ker nX. Hence X has at most countablley many isogenus classes.
Exercise 10 (by Shi-Xin Wang).
To construct the map φ : P ic(X × X) → R := End(X, P0), we let M ∈ P ic(X × X) and p1, p2 be two projections from X × X to X. We may guess M should be sent to M ⊗ (p∗1(M |X×{P0}) ⊗ p∗2(M |{P0}×X))−1, denoted by NM. However, NM does not lie in R. Remark that we have an isomorphism ϕ : P ic0X → X.
Therefore, we may consider
φ(M ) := [P 7→ ϕ(NM|X×P)].
This is well defined since NM|X×P has the same degree with NM|X×P0, i.e. they are both in P ic0X. Clearly, p∗1P icX ⊕ p∗2P icX ⊂ kerφ. Now let M ∈ kerφ. Since NM|X×P ∼= OX×P, by seesaw theorem, NM ∼= p∗2L for some L ∈ P icX. Therefore, M = p∗1(M |X×{P0}) ⊗ p∗2(L ⊗ M |{P0}×X), and hence p∗1P icX ⊕ p∗2P ic = kerφ.
On the other hand, for any αinR, consider the line bundle M ∈ P ic(X × X) corresponding to the divisor D = (α, idX)(X) − {P0} × X
where (α, idX) : X → X × X is the morphism given by P 7→ (α(P ), P ). Then NM still corresponds to the divisor D and
ϕ(NM|X×P) ∼= ϕ(OX(α(P ) − P0)) = α(P ) Exercise 11 (by Pei-Hsuan Chang).
(a) Let L be the parallelogram, A be the area of L. Then area of f (L) is |α2|A. Now, deg f = [L : αL] = |α2|A
A = |α|2.
(b) By exercise 4.4.7(c), we have ˆf ◦ f is an endomorphism corresponding to degf = |α|2. Thus, ˆf is an endomorphism corresponding to |α|2 · α−1 = ¯α.
(c) Let L be the lattice Z⊕τ Z. Now, if τ ∈ Q(√
−d) and integral over Z, then τ2 can be written as integral linear combination of τ and 1. Thus, Z[τ ] = Z ⊕ τ Z. Also, for a, b ∈ Z, (a + bτ )τ = aτ + bτ2 ∈ L.
Hence, ∀a + bτ ∈ Z[τ ], (a + bτ )L ⊂ L, which means Z[τ ] ⊂ R.
For any f ∈ R, say f corresponding to α ∈ C. Since αL ⊂ L and 1 ∈ L ⇒ Z ⊕ τ Z ⇒ R ⊂ Z[τ ]. To sum up, R = Z[τ ].
Exercise 12 (by Po-Sheng Wu).
(a)(b) Suppose the complex multiplication was given by α, then |α|2 = 1 for (a), 2 for (b) respec- tively. Since α is imaginary quadratic and integral, we can assume that α = (a + b√
−d)/2, b, d > 0, d squarefree, then a2 + db2 = 4( or 8, respectively). So (a, b, d) = (0, 2, 1), (±1, 1, 3) for (a), (a, b, d) = (0, 2, 2), (±1, 1, 7), (±2, 2, 1) for (b), and we get τ = i, ω for (a), τ =√
−2, (1+√
−7)/2, i for (b), respectively.
Moreover, we have j(√
−2) = 8000, j((1 +√
−7)/2) = −3375, j(i) = 1728 comparing with 4.5(d), using the fact that if Re(τ ) = 0 then j(τ ) > 0.
Exercise 13 (by Yi-Heng Tsai).
Hasse invariant = 0 i.e. hp(λ) = 0. ⇒ j = 28(λ6−3λ5(λ+6λ2−2λ+1)λ4+6λ3+6λ2 2−3λ+1) = 28(λ(2λ24−2λ+1)λ−4λ3+2λ22) = 29 = 5.
Exercise 14 (by Tzu-Yang Tsai).
By 4.21, Hasse invariant of X is 0 if and only if the coefficient of (xyz)p−1 in fp−1 is 0. Now f (x, y, z) = x3+ y3− z3, thus it’s clear that p ∈ B if and only if 3 | p − 2, thus by Dirichlet’s theorem the density of B in prime is 12.
Exercise 17 (by Ping-Hsun Chuang).
Proof. X is the curve y2+ y = x3− x in P2 with P0 = [0 : 1 : 0].
(a) Write Q = [a : b : 1] ∈ X. If a = 0, then we have y2+ y = 0 and thus Q = [0 : 0 : 1] or [0 : −1 : 1].
Case 1: Q = [0 : 0 : 1] = P . The tangent line at P [0 : 0 : 1] of X is x = −y by the implicit function theorem. Solve
(x = −y
y2+ y = x3− x and get (x, y) = (0, 0) and (1, −1). Note that the solution (0, 0) has multiplicity 2. Then, we have 2P + R ∼ 0, where R = [1 : −1 : 1]. Now, the hyperplane
x − z = 0 passing through P0 and R. Solve
(x − z = 0
y2z + yz2 = x3− xz2 and get [x, y, z] = [0 : 1 : 0], [1 : −1 : 1] and [1 : 0 : 1]. In consequence, we have R + R0 ∼ 0, where R0 = [0 : 1 : 0] and thus 2P ∼ −R ∼ R0 = [1 : 0 : 1].
Case 2: Q = [0 : 0 : 1] = P . The hyperplane x = 0 passing through P [0 : 0 : 1], Q [0 : −1 : 1], and P0[0 : 1 : 0]. Then, we have P + Q + P0 ∼ 0 and thus P + Q ∼ 0.
Case 3: a 6= 0. The hyperplane bx − ay = 0 passing through Q [a : b : 1] and P [0 : 0 : 1]. Solve (bx − ay = 0
y2+ y = x3− x and get (x, y) = (0, 0), (a, b), and
b2
a2 − a,ba33 − b
. Then, P + Q + R ∼ 0, where R =
b2
a2 − a,ab33 − b
. Now, the hyperplane x −
b2 a2 − a
z = 0 passing through P0 and R.
Solve
(x −
b2 a2 − a
z = 0
y2z + yz2 = x3− xz2 and get [x : y : z] = P0, R, R0 = h
b2
a2 − a, −1 + b −ab33 : 1i
. Hence, R + R0 ∼ 0, that is, P + Q ∼ −R ∼ R0 =h
b2
a2 − a, −1 + b −ab33 : 1i . Finally, we use the above formula to find nP for n = 1, · · · , 10:
P 2P 3P 4P 5P 6P 7P 8P 9P 10P
(0, 0) (1, 0) (−1, −1) (2, −3, ) 14,−58
(6, 14) −59 ,278 21
25,−69125 −20
49 ,−435343 161
16,−206564 (b) If p 6= 2, then the curve become y +122
= x3 − x +14. The discriminant of x3− x + 14 is 3716. Now, modulo p reduction gives non-zero discriminant if p 6= 37. This makes the curve non-singular.
If p = 37, the curve is (y + 19)2 = (x + 10) (x + 32)2 which is singular.
If p = 2, the partial derivative is given by ∂f∂x = x2+ 1 and ∂f∂y = 1 6= 0. Thus, the curve is non-singular when p = 2.
5 The Canonical Embedding
Exercise 1 (by Yu-Chi Hou).
Assume that X is complete intersection in Pn, then there exists hypersurfaces H1, . . . , Hn−1 in Pn with degree d1, . . . , dn−1 respectively such that X = H1∩ H2∩ · · · Hn−1. Using adjunction formula repeatly, one has ωX ∼= OX(Pn−1
i=1 di− (n + 1)). Let d :=Pn−1
i=1 di− (n + 1). Since g(X) ≥ 2, deg(KX) > 0. Thus, d > 0.
We then onsider d−uple embedding νd : Pn ,→ PN with N = n+dn − 1. Therfore, ωX ∼= (νd|X)∗OPN(1).
Thus, KX is very ample. However, if X is hyperelliptic, then KX cannot be very ample, and thus X cannot be complete intersection. In particular, we know that genus 2 curves are hyperelliptic (Ex.IV.1.7) and thus X cannot be complete intersection. This also proves Ex. IV.3.3.
Exercise 2 (by Yu-Chi and Pei-Hsuan Chang).
We first prove a lemma.
Lemma 2 (by Yu-Chi). Let x be a curve of genus g ≥ 2, τ ∈ Aut(X) and τ 6= 1X, then τ fixes at most (2g + 2)−points.
