1. Géométrie affine

(E, +, .) (E, E, +)
E
→
R4[X]
E={P∈R4[X]|P(2) = 4}
R4[X]
→δ22
4E
E= 4 + ker δ2= 4 + {P∈R4[X]|P(2) = 0}
ker δ2R4[X]
ER4[X]
Kn≥1A m n
KBKm
Kn
S={X∈Kn|AX =B}
S S
Kn
→B A S
B A X0
S=X0+ ker A
S(Kn,Kn,+) ker A
n−A
a b
S={(un)∈RN| ∀n∈N, un+1 =aun+b}
→a= 1
S={(un)∈RN| ∀n∈N, un=u0+nb}={n→nb}+R
S

a6= 1
(αn=b
1−a)S
(un)S(vn=un−b
1−a)
a(vn=v0an)
S= (αn=b
1−a)n∈N+R(an)n∈N,
Sa6= 0 a= 0
(X, ~ E, +) X
(Ai)i∈IX
→i0∈I
Ai0+ (−−−→ AiAi0, i ∈I)
Ai
(Ai)i∈I<(Ai)i∈I> < (Ai)i∈I>
−−−→
AiAi0
(X, ~ E, +)
Y Z A ∈Y B ∈Z
Y∩Z6=∅
−−→
AB ∈~ EY+~ EZ
D1=A+K~u D2=B+K~v
D1∩D26=∅ ⇔ (−−→ AB, ~u, ~v).
Y Z ~ EY~ EZ
~
E Y ∩Z
→Y∩Z M ∈Y∩Z
A M Y −−→ AM ~ EY
B M Z −−→ MB ~ EZ
−−→
AB =−−→ AM +−−→ MB ∈~ EY+~ EZ−−→
AB ~ EY+~ EZ
−−→
AB =~u +~v
Y Z N Y A +~u M
Z B −~v M =N ~u =−−→ AN ~v =−−→ MB
−−→
AB =−−→ AN +−−→ MB −−→ AB =−−→ AN +−−→ NM +−−→ MB
−−→
NM = 0 N M Y Z
→D1=A+K~u D2=B+K~v
D1∩D26=∅ ⇔ −−→ AB ∈−−→(~u, ~v)⇔(−−→ AB, ~u, ~v).

→A B Y Z ~ EY~ EZ
~
E−−→ AB ~ EY+~ EZ
A B
−−→
AB EY∩EZ
Y Z
(X, ~ E, +) Y
Z
Y∩Z6=∅
dim < Y ∪Z >= dim Y+ dim Z−dim(Y∩Z).
Y∩Z=∅
dim < Y ∪Z >= dim Y+ dim Z−dim( ~ EY∩~ EZ)+1
< Y ∪Z > Y Z
→dim < Y ∪Z >
< Y ∪Z >
•M Y ∩Z
Y=M+~ EYZ=M+~ EZ< Y ∪Z >=M+~ EY+~ EZ
Y⊂M+~ EY+~ EZZ⊂M+~ EY+~ EZ,
Y∪Z⊂M+~ EY+~ EZ< Y ∪Z >⊂M+~ EY+~ EZ
~
E<Y ∪Z> < Y ∪Z >
< Y ∪Z >=M+~ E<Y ∪Z>.
Y=M+~ EY⊂M+~ E<Y ∪Z> ~ EY⊂~ E<Y ∪Z> ~ EZ⊂~ E<Y ∪Z>
~
EY+~ EZ⊂~ E<Y ∪Z>
•A∈Y B ∈Z
< Y ∪Z >
−−−−−−−→
< Y ∪Z > =~ EY+~ EZ⊕K−−→ AB.
~
EY⊂−−−−−−−→ < Y ∪Z > ~ EZ⊂−−−−−−−→ < Y ∪Z > K−−→ AB ⊂−−−−−−−→ < Y ∪Z >
~
EY+~ EZ⊕K−−→ AB ⊂−−−−−−−→ < Y ∪Z >
A+~ EY+~ EZ+K−−→ AB X Y
Z < Y ∪Z >
−−−−−−−→
< Y ∪Z > =~ EY+~ EZ+K−−→ AB.
−−→
AB ~ EY+~ EZ
−−→
AB =~u +~v

~u ∈~ EY~v ∈~ EZ
A+ (~u +~v) = A+−−→ AB =B
A+~u =B−~v
A+~u Y B −~v Z Y ∩Z=∅
→ P AD P A+~ D
ADD0D0
D A D0=A+~ D,
→
→D1D2
−→
D1
−→
D2
→
R3
(a, b)∈R2
D:x= 2z+ 1
y=az −2, D0:x=bz −3 y= 4z+ 1
→
D=
1
−2
0
+R
2
a
1
D0=−3
1+R
b
4
1
D D0
a= 4 b= 2
R3a∈R
D:x=az −1
y= 2z+ 3 , D0:x=z−2 y= 3z−1

→
D=
−1
3
0
+R
a
2
1
D0=
−2
−1
0
+R
1
3
1
A=
−1
3
0
∈ D B=
−2
−1
0
∈ D0
det(−−→ AB, ~ D,~ D0) = det 
−1a1
−4 2 3
0 1 1
=−3+4a= 0, a = 3/4.
D D0P D P 0
D0P P 0
P∩P0D
P P 0Q P
P∩Q P 0∩Q
P Q R
P P 0∆ = P∩P0D1
D2A1A2
A0
1A0 2D1D2P
P0(A1A2) (A0
1A0 2)
∆
→~ P∩~ P0
D D0P P 0
P P 0
~
P∩~ P0=~ D P ∩P0D D0
→
P∩Q P 0∩Q
~
P∩~ Q=~ P0∩~ Q
P P 0
→P∩Q P ∩R P ∩Q∩R
P∩Q P ∩R Q ∩R