Exercice 2 Soit Y ⊂ P

k=C
H⊂Pn
k[X0, . . . , Xn]
1H
AB PnA B
L L0Pn
dim(LL0) = dim(L) + dim(L0)−dim(L∩L0)
dim(∅) = −1p∈Pn(n−1) Pn−1
p πp:Pn\ {p} → Pn−1πp(x) = px ∩Pn−1
k+ 1
L k Pn−k−1
πL:Pn\L→Pn−k−1πL(x) = xL ∩Pn−k−1L L0
LL0=PnL∩L0=∅M, M0
Cn+1 \ {0} → PnM⊕M0=Cn+1
πL:Pn\L→Pn−k−1
Y⊂Pnr
π:Y→Pn−1p6∈ Y π
f:Y→Pr⊂Pn
deg(Y) := deg(f)f
dim(Y)≤n−2p π :Y→Pn−1
p
y∈Y p Y y
deg(Y)
Y⊂Pn
πL(Y)⊂Pr+1 L(n−r−2)
πM:Y→PrM(n−r−1)
Y N N (n−r)

N+ 1 = dim H0(Pn,O(d)) = n+d
dM0, . . . , MNd
n+ 1 x0, . . . , xnνd:Pn→PNd
a= [a0, . . . , an] [M0(a), . . . , Mn(a)]
νd(Pn)PN
dPnνdPN
H⊂Pnf(x0, . . . , xn)=0
dPn
k\H
C=ν3(P1)ν3:P1→P3
CP3
P3C
I=ICk[x0, . . . , x3]I
A k A
A0
M A M =M0⊃M1⊃ · · · ⊃ Mr= 0
Mi/Mi+1 k r
Mlg(M)
0→M0→M→M00 →0A
lg(M) = lg(M0) + lg(M00 ).
X⊂PnV⊂X1
η A =OX,V := OX,η X
1k(X)f∈A
ordV(f) := lgA(A/f) ordV(fg) = ordV(f) + ordV(g)
Frac(A) = k(X)r∈k(X)∗r=f/g f, g ∈A
ordV(r) = ordV(f)−ordV(g) ordV(r)>0r V
ordV(r)<0r V f
div(f) = X
V⊂X
codim(V)=1
ordV(f)·V
r=x/y C ⊂P2
y2z=x3