5 Nonsingular Varieties

Exercise 1 (by Jung-Tao).

(a) f = x⁴+ y⁴− x², f_x = 4x²− 2x, f_y = 4y³

f_y = 0 =⇒ y = 0 =⇒ x⁴− x² = 4x³− 2x = 0 =⇒ x = 0, and the only singular point is (0, 0), its graph is a Tacnode.

(b) f = x⁶+ y⁶− xy, f_x= 6x⁵− y, f_y = 6y⁵− x

f_x = f_y = 0 =⇒ 6x⁵ = y, 6y⁵ = x =⇒ 6x⁶ = xy = 6y⁶, x = 0 iff y = 0, so we may assume x⁶ = y⁶, and f = −4x⁶ = 0 =⇒ x = 0, and the only singular point is (0, 0), its graph is a Node.

(c) f = x⁴− x³+ y⁴+ y², f_x = 4x³− 3x², f_y = 4y³+ 2y,x = 0 iff y = 0,

if y 6= 0 =⇒ 2y² = −1 =⇒ x⁴− x³ = ¹₄, which is impossible. So the only singular point is (0, 0), and its graph is a Cusp.

(d) f = x⁴+y⁴−x²y−xy², f_x = 4x³−2xy−y², f_y = 4y³−2xy−x², x = 0 iff y = 0, else fx = f_y = 0 =⇒ 4x³ = y(2x + y), 4y³ = x(2y + x) =⇒ 4xy(x + y) = 4x⁴+ 4y⁴ = xy(3x + 3y) =⇒ x + y = 0, and xy(x + y) = x⁴+ y⁴ = 2x⁴ = 0 =⇒ x = 0and the only singular point is (0, 0), its graph is a Triple point.

Exercise 2 (by Jung-Tao).

(a) f = xy²− z², fx = y², fy = 2xy, fz = −2z

f_x = f_y = f_z = 0 =⇒ y = z = 0, and its graph is a pinch point.

(b) f = x²+ y²− z², f_x = f_y = f_z = 0 =⇒ x = y = z = 0, and its graph is a conical double point.

(c) f = xy + x³+ y³, f_x = y + 3x², f_y = x + 3y², x = 0 iff y = 0,

else x = −3y², y = −3x² =⇒ −xy = 3x³ = 3y³, f = −x³ = 0 =⇒ x = 0. So the singular points are of the form (0, 0, z), and its graph is a double line.

Exercise 3 (by Yi-Heng).

(a) µP(Y ) = 1 ⇔ f1 = ax + by with a, b not all zero ⇔ Df|(0,0) = (a, b) is of rank 1.

(b) 2,2,2,3 (the multiplicity is the smallest degree in each equation) Exercise 4 (by Yi-Heng).

(a) To prove that (Y · Z) is finite, it suffices to show that OP/(f, g) is noethe-rian and has dimention = 0. Note that OP ' A_mP is noetherian and dim(OP/(f, g)) = dim(A(Y )_mP/g) = dim(Y ) − ht(g) = 0.

Next, let I = (x, y), m = µP(Y ) and n = µP(Z). Then (Y · Z)P ≥ l(O_P/(I^m+n, f, g)) ≥ l(k[x, y]/(I^m+n, f, g)) = dim_k(k[x, y]/(I^m+n, f, g)). Con-sider the exact sequence

k[x, y]/Iⁿ× k[x, y]/I^m k[x, y]/I^n+m k[x, y]/(I^n+m, f, g) 0

(A, B) Af + Bg

Therefore, dimk(k[x, y]/(I^m+n, f, g)) = dim_k(k[x, y]/I^m+n)−dim_k(k[x, y]/Iⁿ)−

dim_k(k[x, y]/I^m) = mn.

(b) Note that^OP/(f, ax + b)= (^k[x]/f (x,^−a_b x))_mP if ab 6= 0. Thus (L · Y )P = µ_P(Y ) for ab 6= 0 and fn(x,^−a_b x)) 6= 0(n = µP(Y )).

(c) We may assume L = V (x) by changing coordinate. Then P = (0, P1, P₂) ∈ Y ∩ L has P1 6= 0 or P2 6= 0.

If P1 6= 0, then (L · Y )P = (V (x) · V (f (x, 1, z)))_(0,P2) = µ_(x=0,z=P2)(f (0, 1, z)) by the discussion in (b). Similarly, if P1 = 0, then (L·Y )P = µ_(x=0,y=P1)(f (0, y, 1)) In conclusion, (L · Y )P =P

P16=0(L · Y )_P +P

P1=0(L · Y )_P = (the largest deg of z in f(0, y, z)) + (the smallest deg of y in f(0, y, z)) = d.

Exercise 5 (by Pei-Hsuan).

case 1 If p - d or p = 0, then choose f(x0, x₁, x₂) = x^d₀ + x^d₁ + x^d₂. Notice that x₀, x₁, x₂ are not all zeros, so Df = dx^d−10 , dx^d−1₁ , dx^d−1₂

has rank 1.

case 2 If p | d, then choose f(x0, x₁, x₂) = x^d−1₀ x₁ + x^d−1₁ x₂ + x^d₂. Notice that

∂f

∂x2 = dx^d−1₂ + x^d−1₁ = x^d−1. Thus, _∂x^∂f₂ = 0 ⇔ x1 = 0. So Df = (0, 0, 0) ⇔ x₁ = 0 and x0 = 0.

But f(0, 0, x2) = 0 ⇔ x₂ = 0. So, Df|p = 0 ⇔ p = (0, 0, 0) which is not in P².

Exercise 6 (by Shuang-Yen).

(a) Y = V (x³ − (y² + x⁴ + y⁴)) is a cusp. Let t = y/x, then the equation x³ = y²+ x⁴+ y⁴ can be written as (x − (t²+ x²+ x²t⁴))x² = 0, so one of the affine chart of the blowing-up image is isomorphic to V (x − (t²+ x²+ x²t⁴)). It suffices to show the nonsingularity at (x, t) = (0, 0), which is clear since f_x(0, 0) = 1 6= 0, where f(x, t) = x − (t²+ x²+ x²t⁴). Another affine chart of the blowing-up image is defined by ys³ = 1 + y²s⁴+ y², which doesn’t contain the preimage of O. Hence the blowing-up of Y is nonsingular.

(b) WLOG let P = (0, 0). Write f = f2 + · · · and fd = c_d,0x^d+ c_d−1,1x^d−1y +

· · · + c0,dy^d, then P is a node implies that f1 = 0 has distinct solutions in P¹, say λ1, λ₂. WLOG let λi ∈ D(x). Let t = y/x, then the equation f = 0 can be written as

x² X

e1,e2

c_e1,e2x^e1+e2−2t^e2

!

= 0,

so one of the affine chart of the blowing-up image is isomorphic to V ( ˜f ), where

f (x, t) =˜ X

e1,e2

c_e1,e2x^e1+e2−2t^e2.

The preimage of P in the affine chart is V ( ˜f ) ∩ V (x), which is (0, λi), so the preimage of P has two points. Since

f˜t(0, λi) = 2c2,0λi+ c1,1 6= 0,

(0, λ_i) is nonsingular. So blowing-up resolves the singularity,

(c) The variety Y defined the equation x² = x⁴+ y⁴ has a tacnode at P = (0, 0).

Let t = y/x, s = x/y, then the equation can be written as x²(1 − x²− x²t⁴) = 0, y²(s²− y²s⁴− y²) = 0, so the preimage of P appears in the affine chart which is defined by the equation g := s²− y²s⁴− y² = 0, then it’s a node since s²− y² = 0 has two solutions [1 : ±1].

(d) The multiplicity of (0, 0) on Y is 3. Let t = y/x, s = x/y, then the equation y³ = x⁵ can be written as x³(t³− x²) = 0, y³(1 − y²s⁵) = 0, so the preimage of O appears in the affine chart which is defined by the equation g := t³− x², and hence a cusp. One further blowing-up resolves the singularity by the way similar to (a).

Exercise 7 (by Wei-Ping).

(a) Clearly f is non-singular at point not equal to P . Since deg f > 1, ^∂f_∂x(P ) =

∂f

∂y(P ) = ^∂f_∂z(P ) = 0. Thus P is the only non-singular point.

(b) Consider system of equations xv = yw, xw = zu, yw = zv, f(x, y, z) = 0. now cover ¯X with open affine sets to conclude it is non-singular.

(c) φ⁻¹(p)is union of three sets: {(0, 0, 0, 1, v, w) | f(1, v, w) = 0}, {(0, 0, 0, u, 1, w) |

is independent of the choice of coordinates.

U_k := Pⁿ − Z(x_k) −→^∼

Given variety X and point P ∈ X, recall that the Zariski tangent space on P denoted by TP(X) is the k-vector space (mP/m²_P)^∧ := Hom_k(m_P/m²_P, k).

(a) We know that since mP/m²_P is finite dimensional (by Noetherianess), we have dim_kT_p(X) = dim_k(m_P/m²_P)^∧ = dim_km_P/m²_P

On the other hand, we have the inequality

dim_km_P/m²_P ≥ dim_kO_P,X where equality holds iff P is a non-singular point.

(b) Suppose given

X ^φ Y

We get a diagram

O_P,X O_{φ(P ),Y}

mP mφ(P )

m²_P m²_{φ(P )}

φ^∗_P

In order to construct a map TP(X) → T_{φ(P )}(Y ), it suffices (by duality) to construct a map mP/m²_P ← m_{φ(P )}/m²_{φ(P )}. The composition of maps

m_P/m²_P m_P ^φ m_{φ(P )} m²_{φ(P )}

∗ P

is 0 by above, so we may define mP/m²_P ← m_{φ(P )}/m²_{φ(P )} to be the map making the diagram commutative

m_P m_{φ(P )}

m_P/m²_P m_{φ(P )}/m²_{φ(P )}

φ^∗_P (TP(φ))^∧

(c) Write X, Y as the following varieties

X = V (x − y²) ⊆ A², Y = A¹ with morphism

X ^φ Y , (x, y) 7→ x Denote P as the point (0, 0) ∈ X, then φ induces a map

(k[x, y]/(y²− x))_(x,y) = O_P,X ^φ O_{φ(P ),Y} = (k[t])_t

∗ P

given by x ←[ t. To show TP(φ) is 0, it suffices to show (TP(φ))^∧ is 0, or

m_P/m²_P m_{φ(P )}/m²_{φ(P )} m_{φ(P )}

= m_P/m²_P m_P m_{φ(P )}

(T_P(φ)^∧)

φ^∗_P

is 0 (the equality above is by our definition in (b)). Choose f ∈ mφ(P ), that is, a function vanishing on φ(P ) = 0, then f = tg for some g, so

φ^∗_P(f ) = φ^∗_P(tg) = φ^∗_P(t)φ^∗_P(g) = xφ^∗_P(g) = y²φ^∗_P(g) ∈ m²_P Therefore, we have (TP(φ))^∧ = 0.

Exercise 11 (by Tai-Ning).

The statement is true only when char k 6= 2. Let’s assume char k 6= 2, Y = {[x, y, z, w] ∈ P³ : x²− xz − yw = 0, yz − xw − zw = 0}.

Z = {[x, y, z] ∈ P² : y²z − x³+ xz² = 0}

Now we show that ϕ : Y − [0, 0, 0, 1] → Z − [1, 0, −1] defined by ϕ : [x, y, z, w] 7→

[x, y, z], and we could find ϕ⁻¹ by

ϕ⁻¹([x, y, z]) =

([x, y, z,^x2−xz_y ],if y 6= 0.

[x, y, z,_x+z^yz ] ,if x + z 6= 0.

Now we check well-defined.

• For any [x, y, z, w] ∈ Y − [0, 0, 0, 1], it’s impossible that y = 0 and x + z = 0 at the same time, since otherwise, x² = xz + yw = −x², so 2x² = 0, so x, y, z are all zero, contradiction. Therefore, if y 6= 0, we have w = ^x2−xz_y , substitute and get yz = (x + z)^x2−xz_y , so y²z − x³+ xz² = 0. If otherwise x + z 6= 0, we have w = _x+z^yz and get x²− xz − y_x+z^yz , also have y²z − x³+ xz² = 0. And [x, y, z] 6= [1, 0, −1]. Thus, ϕ is a morphism.

• For any [x, y, z] ∈ Z − [1, 0, −1], it’s impossible that y = 0 and x + z = 0 at the same time, for the first case, we check x² − xz − y^x2−xz_y = 0 and yz − (x + z)^x2−xz_y = 0, which is true because [x, y, z] ∈ Z. Another case is similar. Therefore ϕ⁻¹ is a well-defined morphism.

Since Z is defined by an irreducible polynomial, so Z is irreducible. And the derivative matrix is

[−3x²+ z², 2yz, y² + 2xz]

The only possible to be all zero is x = y = z = 0. So Z is nonsingular. By the isomorphism of ϕ, Y − [0, 0, 0, 1] is also irreducible and nonsingular. So it’s sufficient to check Y is nonsingular at [0, 0, 0, 1]. The derivative matrix of Y is

 2x −w −x −y

−w z y − w −x − z



=

 0 −1 0 0

−1 0 −1 0

 . which has rank 2, so is nonsingular.

Exercise 12 (by Shuang-Yen).

(a) A quadratic form over a field k with characteristic 6= 2 has an orthogonal basis, after linear transformation, may assume f = x²0+ · · · + x²_r.

(b) For r = 0, 1, f is clearly reducible. For r ≥ 2, induction on r. Suppose not, say f = gh and g, h /∈ k^×, denote degi the degree of xi, then degig + deg_ih = 2 for each i. If degig 6= 1 for all i, then degig = 2, deg_jh = 2 for some i 6= j, then there will be a term x²ix²_j in f with coefficient in k[x₀, . . . , ˆx_i, . . . , ˆx_j, . . . , x_r], which is a contratiction. WLOG let deg0g = deg₀h = 1, write g = ax0+ b, h = cx₀+ d, then ac = 1 implies a, c ∈ k^×, after scaling, may let a = c = 1, then b + d = 0 and bd = x²1 + · · · + x²_r. If r > 2, by induction hypothesis, one of b, d is a unit, then b + d = 0 implies the another one is a unit, but bd is not a unit. If r = 2, then

−b² = bd = x²₁+ x²₂ = (x₁+ ix₂)(x₁ − ix₂), but k[x1, x₂] is a UFD. So f is irreducible if and only if r ≥ 2.

(c) Z = Sing Q is defined by the equations, f and ∂f/∂xi, which is Z = V (x²₀+ · · · + x²_r, 2x₀, . . . , 2x_r, 0, . . . , 0) = V (x₀, . . . , x_r), a linear variety of dimension n − r − 1.

(d) Let Q⁰ = V (x²₀ + · · · + x²_r, x_r+1, . . . , x_n) ⊂ V (x_r+1, . . . , x_n) ∼= P^r. For any A = (a₀, . . . , a_r, 0, . . . , 0) ∈ Q⁰ and B = (0, . . . , 0, br+1, . . . , b_n) ∈ Z, every point C on the line AB can be written as (sa0, . . . , sar, tbr+1, . . . , tbn), then

(sa₀)² + · · · + (sa_r)² = s²(a²₀+ · · · + a²_r) = 0 =⇒ C ∈ Q,

so the cone of Q⁰ and Z is contianed in Q. For any point C = (c0, . . . , c_n) ∈ Q, if at least one of c0, . . . , c_r is not 0, then C lies on the line jointing (c₀, . . . , c_r, 0, . . . , 0) ∈ Q⁰and (0, . . . , 0, b⁰r+1, . . . , b⁰_n) ∈ Z, where (tb⁰r+1, . . . , tb⁰_n) =

(b_r+1, . . . , b_n) for some t such that (b⁰r+1, . . . , b⁰_n) 6= (0, . . . , 0). If all of c₀, . . . , c_r is 0, then C lies on the line jointing (1, i, 0, . . . , 0) ∈ Q⁰ and (0, . . . , 0, cr+1, . . . , cn) ∈ Z.

Exercise 13 (by Zi-Li).

We can assume the variety X is affine, let R be the integral closure of A(X) in its quotient field. By 3.9A, R = P^k_i=1A(X)f_i, we may assume fi = g_i/F.Then, X − Z(F )is nonempty and every points of X − Z(F ) is normal because 1/F ∈ OP

for every P ∈ X−Z(F ). Besides, if P is a normal point, write fi = a_i/B, B(P ) 6= 0, then P ∈ X − Z(B) and every points of X − Z(B) are normal points. Hence, non-normal points of a variety form a proper closed set.

Exercise 14 (by Shi-Xin).

(a) Suppose P ∈ Y = Z(f) and Q ∈ Z = Z(g) are analytically isomorphic. By suitable change of variables, we can assume P = Q = (0, 0) and write

f = f_s+ h.o.t., g = g_t+ h.o.t.

where fs, g_t∈ k[x, y] are the homogeneous polynomials of minimal degree s, t in f, g respectively. WLOG, assume y is not their common tangent direction.

Denote L = Z(y). Since k[[x, y]]/(f) ∼= k[[x, y]]/(g), we have

k[[x]]/(x^s) ∼= k[[x]]/(f (x, 0)) ∼= k[[x, y]]/(f, L) ∼= k[[x, y]]/(g, L) ∼= k[[x]]/(x^t) Then it forces s = t, and hence µP(Y ) = s = t = µ_Q(Z).

(b) Consider the same process in the textbook. The idea is that for any k > r, find gk−t, h_k−s such that

f_k= g_sh_k−s+ g_k−th_t+

k−t−1

X

i=s+1

g_ih_k−i

when all the other terms are known. Then it suffices to show that for any homogeneous f of degree k > r, there are homogeneous polynomials gk−t, hk−s of degree k − t, k − s respectively such that

f = g_sh_k−s+ g_k−th_t.

Write gs = y^sg(z), h˜ _t= y^t˜h(z), f = y^kf (z)˜ where z = x/y and ˜g, ˜h, ˜f ∈ k[z]. Let m := deg ˜g, n := deg ˜h and consider the linear map

φ : P_n× P_m → P_n+m defined by φ(A, B) = ˜gA + ˜hB where Pn denote the polynomial in z of degree less than n.

Since 0 = deg(gcd(˜g, ˜h)) = m + n − rank(φ), φ is surjective, and hence there are A, B such that ˜gA + ˜hB = ˜f. Just let gk−t = y^k−tB, h_k−s = y^k−sA.

(c) (i) (2-fold) According to b, we may write f = f1f₂ with two distinct linear factors f1, f₂. Then there is an automorphism of k[[x, y]] sending f1, f₂ to x, y respectively. It induces an isomorphism

k[[x, y]]/f ∼= k[[x, y]]/(xy).

Thus every ordinary 2-fold point is analytically isomorphic to (0, 0) of Z(xy).

(ii) (3-fold) Write f = f1f2f3 with three distinct linear factors f1, f2, f3. Then every ordinary 3-fold point is analytically isomorphic to (0, 0) of Z(xy(x + y)) by considering an automorphism of k[[x, y]] sending f₁, f₂, f₃ to ax, by, x + y for some a, b ∈ k.

(iii) (4-fold) Write f = f1f₂f₃f₄with three distinct linear factors f1, f₂, f₃, f₄. For the same reason, every ordinary 4-fold point is analytically isomor-phic to (0, 0) of Z(xy(x + y)(x + αy)) for some α ∈ k by considering an automorphism of k[[x, y]] sending f1, f₂, f₃, f₄ to ax, by, x + y, c(x + αy) for some a, b, c ∈ k. Denote fα = xy(x + y)(x + αy), and consider (0, 0) in Z(fα) and (0, 0) in Z(fβ). If they are analytically

isomor-phic, then there is a map φ : k[[x, y]]/(fα) → k[[x, y]]/(f_α) such that φ(x, y) = (x, y) and φ((fα)) = (f_β). It forces that α = β. Thus there is a one-parameter family of mutually nonisomorphic ordinary 4-fold points.

(d) Given any f = f2+ h.o.t. with f2 6= 0. If f2 has two distinct linear factors, according to (c)(i), it is isomorphic to an ordinary 2-fold point, which is isomorphic to the singularity (0, 0) of Z(y² = x²). Now suppose f2 has only one linear factor, by taking a suitable automorphism, we may assume f = y²+ yg₁(x, y) + h₁(x)where deg g1 > 2 w.r.t y. By sending y + g1/2to y, we have f = y²+ yg₂(x, y) + h₂(x) where deg g2 > 3 w.r.t y. Continuing the process, we can assume f = y²+ yg(x, y) + h(x) where deg g is sufficiently large. Now we refer to some results in a useful textbook¹. Since the Milnor number of f, denoted by µ, is finite, f is right (µ + 1)-determined, which means we only need to care the part of f of degree no more than (µ + 1).

Hence by above process, f is right equivalent to y²+ h(x) for some h ∈ k[x].

Thus f defines a singularity which is isomorphic to (0, 0) of Z(y² = x^r).

1Greuel, G.M., Lossen, C., Shustin, E., 2007. Introduction to Singularities and Deformations, Springer, Berlin, ch2

在文檔中 This is an example of proof. (頁 25-34)