Problème : Résultant de deux polynômes

1t2`+B+2 R

G2b `û~2t2b M2 bQMi Tb +[mBb ,

ě /Mb mM Z/nZ BMiĕ;`2 Un T`2KB2`V- QM `ûbQmi H2b û[miBQMb 2M 6*hP_AaLh 2i 2M miBHBbMi [mǶmM T`Q/mBi /2 7+i2m`b 2bi MmH bbB HǶmM m KQBMb /2b 7+i2m`b 2bi MmH

ě /Mb mMZ/nZMQM BMiĕ;`2 UnMQM T`2KB2`V- QM T2mi miBHBb2` H2 i?ûQ`ĕK2 /2b `2bi2b +?BMQBb UpQB` THmb

#bV Qm #B2M i2bi2` iQmi2b H2b TQbbB#BHBiûbX .Mb +2 +b- TQm` `ûbQm/`2x⁴= 1QM T2mi M2 i2bi2` [m2 H KQBiBû /2b +Hbb2b +`( x)⁴=x⁴XXX G2 MQK#`2 /ǶQTû`iBQMb 2bi HBKBiû 2i QM MǶQm#HB2 Tb /2 `û/mB`2 KQ/mHQ R8 TQm` ûpBi2` H2b KmHiBTHB+iBQMb /2 i`QT ;`M/b MQK#`2bX S` 2t2KTH2- TQm` +H+mH2`8⁴ KQ/mHQ R8- QM +QKK2M+2 T` +H+mH2` 8² = 64 = 1 KQ/mHQ R8 TmBb 8⁴ = (8²)² = ( 1)² = 1 KQ/mHQ R8XXX

RX RN 2bi T`2KB2`- /QM+Z/19Z2bi mM +Q`Tb /QM+ BMiĕ;`2X BMbB QM T2mi 7+iQ`Bb2` 2M( ˙x ˙1)·( ˙x²+

˙

x+ ˙1) = ˙02i }MH2K2Mi-x˙ = ˙1Qm #B2Mx˙²+ ˙x+ ˙1 = 0X PM +?2`+?2aet˙ b˙ i2Hb [m2a˙+ ˙b= 12i

˙

ab˙= 1X PM i`Qmp2 ˙72i11X˙

6BMH2K2Mi- H2b bQHmiBQMb bQMi R-d- 2i RRX

kX S` +QMi`2-15MǶ2bi Tb T`2KB2`X S` H2 i?ûQ`ĕK2 /2b `2bi2b +?BMQBb- bB 2bi HǶBbQKQ`T?BbK2 2Mi`2 Z/3Z2iZ/5Z- QM (1mod15) = (1mod3,1mod5)X

PM `ûbQmi HQ`bx⁴₃= 1mod3/MbZ/3Z2ix⁴₅= 1mod5/MbZ/5ZX PM i`Qmp2x32{1mod3,2mod3} 2i x5 2{1mod5,2mod5,3mod5,4mod5}X SQm` +?[m2 +QmTH2(x3, x5)- QM `2i`Qmp2 H +Hbb2 KQ@

/mHQ R8 +Q``2bTQM/Mi2 2M +QMbB/û`Mi <sup>1</sup>(x3, x5)X PM i`Qmp2 ¹(1mod3,1mod5) = 1mod15-

1(1mod3,2mod5) = 7mod15- ¹(1mod3,3mod5) = 13mod15- ¹(1mod3,4mod5) = 4mod15-

1(2mod3,1mod5) = 11mod15- ¹(2mod3,2mod5) = 2mod15- ¹(2mod3,3mod5) = 8mod15 2i ¹(2mod3,4mod5) = 14mod15X

6BMH2K2Mi- HǶ2Mb2K#H2 /2 bQHmiBQMb 2bi{˙1,˙2,˙4,˙7,˙8,11,˙ 13,˙ 14˙}X jX /27 `+BM2*m#B[m2U-MV,

`2bmHii4()

7Q` F BM `M;2UMV, O H /2`MBĐ`2 pH2m` i2biö2 2bi UM@RVXXX B7 UF jV $W M 44 $W M , O Qm #B2M B7 UF j @ V$W M 44 y,

`2bmHiiXTT2M/UFV

`2im`MU`2bmHiiV

9X *Ƕ2bi /m +Qm`b ,x2i n bQMi T`2KB2`b 2Mi`2 2mt, x2bi mM ûHûK2Mi BMp2`bB#H2 /2 Z/nZ ,x 2 (Z/nZ)^⇤X

HQ`bx^{Card(Z/nZ)⇤}= 12iCard(Z/nZ)^⇤='(n)X

8X x 2bi ;ûMû`i2m` /2 G ,H2 bQmb ;`QmT2 2M;2M/`û T` x 2bi : , iQmi ûHûK2Mi /2 G 2bi mM2 TmBbbM+2 U2MiBĕ`2 `2HiBp2V /2x,G={x^k|k2Z}X

G2 MQK#`2 /2 ;ûMû`i2m`b /2U20162bi û;H ¨'(2016)X G¨ 2M+Q`2- Tb #2bQBM /2 +H+mH2ii2 , 2016 = 2⇥1008 = 2²⇥504 = 2³⇥252 = 2⁴⇥126 = 2⁵⇥63 = 2⁵⇥3²⇥7X

_TT2HQMb 2MbmBi2 [m2'(p^k) =p^k p^{k 1}TQm`pT`2KB2` 2i [m2 bBn2imbQMi T`2KB2`b 2Mi`2 2mt- /ǶT`ĕb H2 i?ûQ`ĕK2 /2b `2bi2b +?BMQBb-'(nm) ='(n)⇥'(m)X

6BMH2K2Mi-'(2016) = (2⁵ 2⁴)⇥(3² 3)⇥(7¹ 7⁰) = 16⇥6⇥6 = 576

S`Q#HĕK2 R

S`iB2 R

RX

kX _a bm7 [mǶBH 2bi BMmiBH2 /2 bmTTQb2` [m2 H2 #mi 2bi mM 2bT+2 p2+iQ`B2H /2 /BK2MbBQM }MB2X a2mH2 H bQm`+2 /QBi HǶāi`2X

X G T`2mp2 b2 7Bi /B`2+i2K2Mi aLa `û+m``2M+2X .ǶBHH2m`b- +2mt [mB QMi +`m 7B`2 mM2 `û+m``2M+2

`2K`[m2`QMi [mǶBHb MǶQMi Tb miBHBbû H2m` ?vTQi?ĕb2 /2 `û+m``2M+2 /Mb HǶ?û`û/BiûXXX

aQBix2NkX HQ`bf<sup>k</sup>(x) = 0E2i 2M +QKTQbMi p2+fHBMûB`2-f^k+1(x) = 0E- /QM+x2Nk+1X aQBi 2MbmBi2 y2Ik+1X AH 2tBbi2 HQ`bx2E i2H [m2 f<sup>k+1</sup>(x) =yX SQbQMbz =f(x)X HQ`b T`

+QMbi`m+iBQM-f^k(f(x)) =yX .QM+y2IkX

#X PM pQBi bb2x bQmp2Mi H }M /2 H T`2mp2 ,A={p|Np+1=Np}2bi mM2 T`iB2 MQM pB/2 /2N /QM+ /K2i mM THmb T2iBi ûHûK2MiX

S` +QMi`2- BH 2bi ``2 /2 pQB` mM2 T`2mp2 +Q``2+i2 /2A6=;X

PM T2mi TQm` +2H `2;`/2` H2b /BK2MbBQMb , H bmBi2 (/BK(N_k))2bi mM2 bmBi2 +`QBbbMi2 Um b2Mb H`;2V /ǶT`ĕb H [m2biBQM T`û+û/2Mi2X *Ƕ2bi mbbB mM2 bmBi2 /Ƕ2MiB2`b 2i KDQ`û2 T`

/BK(E) +` Nk ⇢ EX *2ii2 bmBi2 2bi /QM+ ahhAPLLA_1 5 .QM+ BH 2tBbi2 p 2 N i2H [m2 /BK(N<sub>p</sub>) =/BK(N<sub>p+1</sub>)X *QKK2 /2 THmbNp⇢Np+1-Np=Np+12iAMǶ2bi /QM+ Tb pB/2XXX +X _`2K2Mi 7Bi , #Bx``2XXX

G `ûTQMb2 2bi MQM 5 S` 2t2KTH2 HǶ2M/QKQ`T?BbK2 /2 H T`iB2 j 2bi mM +QMi`2 2t2KTH2- KBb HǶ2M/QKQ`T?BbK2 /2R[X]f :P 7!P⁰ +QMpB2Mi û;H2K2Mi , TQm` H2b /2mt-N_k=Rk[X]

2bi mM2 bmBi2 bi`B+i2K2Mi +`QBbbMi2 /2 MQvmt 5 /X #Q`/û- KBb ``2K2Mi #B2M 7Bi

GǶBM+HmbBQMNp0⇢Np0+k pB2Mi /2 H +`QBbbM+2 /2 H bmBi2 /2b MQvmtX

_û+BT`Q[m2K2Mi- KQMi`QMb [m2 bBN_p0 =N_p0+1- HQ`bN<sub>p0+1</sub>=N<sub>p0+2</sub>X aQBix2N<sub>p0+2</sub>X HQ`b f^p0+1(f(x)) = 0 2i f(x) 2 Np0+1 = Np0X .QM+ f^p0(x) = 0 2i f^p0+1(x) = 0X 6BMH2K2Mi x2Np0+1X

S` `û+m``2M+2- QM KQMi`2 HQ`b [m2Np0=Np0+k TQm` iQmikX

jX X GǶûMQM+û 2bi bQmp2Mi KH +QKT`Bb , BH 7mi KQMi`2` B+B [m2 (Ik)2bi biiBQMMB`2 2i [m2 H2 `M;

+Q``2bTQM/Mi 2bi H2 KāK2 [m2 TQm` H bmBi2(Nk)X

PM bBi [m2 H bmBi2(Ik)2bi +`QBbbMi2X AH 2M 2bi /2 KāK2 TQm` H bmBi2 /2b /BK2MbBQMb(/BK(Ik))X

*2ii2 bmBi2 2bi KDQ`û2 T` /BK(E)2i ¨ pH2m`b 2MiBĕ`2b- /QM+ biiBQMM2 ¨ T`iB` /ǶmM `M; MQiû q0X

S` i?ûQ`ĕK2 /m `M; TTHB[mû ¨ f<sup>k</sup> U[mB 2bi mM 2M/QKQ`T?BbK2 T` +QKTQbû2 /Ƕ2M/QKQ`@

T?BbK2bV /BK(Nk) +/BK(Ik) =/BK(E)X

.QM+ /BK(I_k) =/BK(E) /BK(N_k)2i H bmBi2(/BK(I_k))biiBQMM2 2t+i2K2Mi m KāK2 `M;

[m2 H bmBi2 /BK(Nk)X

*QKK2Ik⇢Ik+1- H bmBi2(Ik)biiBQMM2 mbbB 2t+i2K2Mi ¨ T`iB` /2 +2 `M; 5

#X S2m #Q`/û2- KBb bb2x #B2M 7Bi 2M ;ûMû`HX

S` mM `;mK2Mi /2 /BK2MbBQM Ui?ûQ`ĕK2 /m `M; 2M+Q`2V- BH bm{i /2 KQMi`2` [m2 H bQKK2 2bi /B`2+i2- /QM+ [m2 HǶBMi2`b2+iBQM 2bi `û/mBi2 ¨{0}X

aQBix2Np0\Ip0X HQ`bf^p0(x) = 0 2i x=f^p0(y)X .QM+ f^p0(f^p0(y)) = f^2p0(y) = 0X .QM+

y2N2p0=Np0X 6BMH2K2Mi-x=f^p0(y) = 0X

9X Ip02bi /2 /BK2MbBQM }MB2X G #BD2+iBpBiû /ǶmM 2M/QKQ`T?BbK2 2bi /QM+ û[mBpH2Mi2 ¨ b bm`D2+iBpBiû Qm #B2M bQM BMD2+iBpBiûX A+B- H bm`D2+iBpBiû 2bi THmb bBKTH2 +` T` ?vTQi?ĕb2-Ip0+1 =Ip0- +2 [mB bB;MB}2 2t+i2K2Mif(Ip0)) =Ip0XXX

8X JQMi`QMb [m2 H `2bi`B+iBQM /2f ¨Np0 2bi mM 2M/QKQ`T?BbK2 , bQBix2Np0 HQ`b f^p0(x) = 0X .QM+f^x2Np0+1=Np0X

G `2bi`B+iBQM /2f ¨ N_p0 2bi MBHTQi2Mi2 /ǶQ`/`2 BM7û`B2m` Qm û;H ¨ p₀ +` TQm` iQmi x 2 N_p0- f^p0(x) = 0X

.2 THmb- HǶQ`/`2 2bi bmTû`B2m` ¨p0 +` T` biimi /2p0- BH 2tBbi2x2E i2H [m2 f^{p0 1}(x)6= 0 2i f^p0(x) = 0X *2x2bi mM ûHûK2Mi /2N_p0 i2H [m2f^{p0 1}(x)6= 0/QM+ H `2bi`B+iBQM /2f ¨N_p0 MǶ2bi Tb MBHTQi2Mi2 /ǶQ`/`2 BM7û`B2m` ¨p0 1XXX

k

eX .ûKQMi`2` [m2 H bmBi2(dim(Nk+1) dim(Nk))_k2N2bi /û+`QBbbMi2X

*Ƕ2bi H [m2biBQM /B{+BH2 /2 HǶ2t2`+B+2X AH bǶ;Bi /2 KQMi`2` [m2 HǶû+`i 2Mi`2 /2mt /BK2MbBQMb bm++2bbBp2b /2 MQvmt Biû`ûb p 2M /BKBMmMiX _2K`[mQMb T` H2 i?ûQ`ĕK2 /m `M; [m2 +2i û+`i 2bi H2 KāK2 [m2 +2HmB 2Mi`2 /2mt /BK2MbBQMb /ǶBK;2b Biû`û2b bm++2bbBp2b ,dim(Nk+1) dim(Nk) = dim(Ik) dim(Ik+1)X _2;`/QMb HQ`b +QKK2Mi +2 +QKTQ`i2 +2i û+`iek= dim(Nk+1) dim(Nk)X AMimBiBp2K2Mi- /2Ik+1=f(Ik)⇢Ik- QM T2`/ek /BK2MbBQMbX *Ƕ2bi ¨ /B`2 [mǶBH 2tBbi2 mM 2bT+2 p2+iQ`B2Hk i2H [m2Ik Fk=Ik+1p2+dim(Fk) =ekX Zm2 b2 Tbb2@i@BH /QM+ 2Mi`2Ik+12iIk+2\ PM Ik+1=f(Ik)2iIk+2=f(Ik+1)2i 2M T`iMi /2Ik Fk=Ik+1- QM Q#iB2Mi 2M +QKTQbMi T`

f,Ik+1+f(Fk) =Ik+2X ii2MiBQM , H bQKK2 MǶ2bi THmb 7Q`+ûK2Mi /B`2+i2 /Mb H /2`MBĕ`2 û;HBiûX .QM+ T` H 7Q`KmH2 /2 :`bbKM ,e<sub>k+1</sub>=dim(I<sub>k+2</sub>) dim(I<sub>k+1</sub>)dim(f(F<sub>k</sub>))dim(F<sub>k</sub>) =e<sub>k</sub> +`f 2bi mM2 TTHB+iBQM HBMûB`2 U/QM+ [mB #Bbb2 H2b /BK2MbBQMbVX

G bmBi2(ek)2bi /QM+ #B2M /û+`QBbbMi2X

dX aBg 2bi mM 2M/QKQ`T?BbK2 MBHTQi2Mi- H bmBi2(Nk)/2b MQvmt Biû`ûb /2g 2bi mM2 bmBi2 ah_A*@

h1J1Lh +`QBbbMi2 Dmb[mǶ¨ mM `M;p0 ¨ T`iB` /m[m2H 2HH2 2bi biiBQMMB`2 ¨Np0 =E +`g2bi MBHTQi2MiX

.QM+ H bmBi2 /2b /BK2MbBQMb(dim(Nk))2bi mbbB mM2 bmBi2 /Ƕ2MiB2`b bi`B+i2K2Mi +`QBbbMi2 /2 y ¨ dim(E)TmBb biiBQMMB`2 ¨ T`iB` /m `M;p0X .QM+p0dimE 2ig^p0=g^dimE= 0_L(E)X

GǶQ`/`2 /2 MBHTQi2M+2 /2g2bi /QM+ BM7û`B2m` Qm û;H ¨ /BK(E)X

S`iB2 j , mM mi`2 2t2KTH2

aQBiE=R[X]2iP 2EX PM /û}MBi (P) =P(X+ 1) P(X)X

RX 2bi HBMûB`2 UiQmi H2 KQM/2 HǶ KQMi`ûX *Ƕ2bi mM 2M/QKQ`T?BbK2 +` (P) 2bi mM2 bQKK2 /2 +QKTQbû2 /2 TQHvMƬK2b- /QM+ mM TQHvMƬK2X

AMmiBH2 /2 T`H2` /2 /2;`û /Mb +2ii2 [m2biBQM 5 kX S`iB[m2K2Mi T2`bQMM2 MǶ `;mK2Miû _A:Pl_1la1J1Lh 5

ker ={P2K[X]|P(X+ 1) P(X) = 0}X lM i2H TQHvMƬK2 2bi /QM+ R@Tû`BQ/B[m2 2i TQm` mM i2H TQHvMƬK2-Q=P P(0) 2bi mM TQHvMƬK2 /K2iiMi mM2 BM}MBiû /2 `+BM2b Um KQBMb iQmb H2b 2MiB2`bV- /QM+ MmH 5U+2i `;mK2Mi `2pB2Mi +QMbiKK2Mi /Mb H2b 2t2`+B+2b 555V

6BMH2K2Mi-P=P(0)2bi mM TQHvMƬK2 +QMbiMiX

_û+BT`Q[m2K2Mi- iQmi TQHvMƬK2 +QMbiMi +QMpB2Mi /QM+ker( ) =R0[X]X

SHmb /B{+BH2 ,ker( ^k) =Rk 1[X]X SQm`Pk=X<sup>k</sup>- QM KQMi`2 T` `û+m``2M+2 [m2 deg( (X<sup>k</sup>)) = k 1X .QM+ T` HBMû`Biû /2 - <sup>2</sup>(P) = 0 , deg(P)  1 2i T` `û+m``2M+2- ^k(P) = 0 , deg(P)k 1X 6BMH2K2Mi- F2`( ^k) =Rk 1[X]X

GǶ2M/QKQ`T?BbK2 MǶ2bi Tb MBHTQi2Mi +` TQm` iQmin2N- F2`( ^k)6=R[X]XXX

jX m +b Qɍ QM MǶm`Bi Tb 2m HǶB/û2 /Mb H [m2biBQM T`û+û/2Mi2- bBPk=X^k- HQ`bdeg( (P)) = k 1X hQmDQm`b T` HBMû`Biû- (Rn[X])⇢Rn 1[X]X

X S` i?ûQ`ĕK2 /m `M;-dim(Im( )) = (n+ 1) dim(ker( )) =n kX P` ^k(Rn[X])⇢Rn k[X]X S` BM+HmbBQM /BK2MbBQM- ^k(Rn[X]) =Rn k[X]X

HQ`bIm( ) = R[X]+` TQm` iQmi TQHvMƬK2 P /2 /2;`ûn- BH 2tBbi2 iQmDQm`b m KQBMb mM TQHvMƬK2Q/2 /2;`ûn+ 1i2H [m2 (Q) =PX

#X .ǶT`ĕb +2 [mB T`û+ĕ/2- ⁿ⁺¹= 0X .QM+ HǶQ`/`2 /2 MBHTQi2M+2 /2 2bi BM7û`B2m` Qm û;H ¨n+ 1X .2 THmb- ⁿ(Xⁿ)6= 0X .QM+ HǶQ`/`2 /2 MBHTQi2M+2 2bi û;H ¨n+ 1X

9X X ii2MiBQM , +Q[mBHH2b /Mb H2 bmD2i , HB`2 ǴJQMi`2` [m2 TQm` iQmi 2MiB2` Mim`2H p- ^p(P) = Xp

k=0

( 1)^{p k} Çp

k å

P(X+k).ǴX *2mt [mB QMi #Q`/û H [m2biBQM QMi ;ûMû`H2K2Mi +Q``B;û S` `û+m``2M+2 bm`pX C2 M2 7Bb [m2 HǶ?û`û/Biû ,

p+1(P) = ( ^p(P)) = Xp

k=0

( 1)^{p k} Çp

k å

P(X+k)

!

=

= Xp

k=0

( 1)^{p k} Çp

k å

P(X+k+ 1) Xp

k=0

( 1)^{p k} Çp

k å

P(X+k)

= Xp

k=0

( 1)^{p k} Çp

k å

P(X+k+ 1) + Xp

k=0

( 1)^{p k+1} Çp

k å

P(X+k)

= Xp+1

k⁰=1

( 1)^{p k0+1} Ç p

k⁰ 1 å

P(X+k⁰) + Xp

k=0

( 1)^{p k+1} Çp

k å

P(X+k)

= ( 1)^p+1 Çp

0 å

P(X)+

Xp

k=1

( 1)^{p k+1}P(X+k) ñÇ p

k 1 å

+ Çp

k åô!

+( 1)⁰ Çp

p å

P(X+p+1)

= ( 1)^p+1 Çp

0 å

P(X) + Xp

k=1

( 1)^{p k+1}P(X+k) Çp+ 1

k å!

+ ( 1)⁰ Çp+ 1

p+ 1 å

P(X+p+ 1)

=

p+1X

k=0

( 1)^{p+1 k} Çp+ 1

k å

P(X+k)

_û+m``2M+2 ûi#HB2 UiQmi2 `2bb2K#HM+2 p2+ H2 #BMƬK2 /2 L2riQM M2 b2`Bi 7Q`imBi2XXXV

#X PM bBi [m2 ⁿ⁺¹= 0X .QM+ TQm` iQmiP- ⁿ⁺¹(P) = 0X

.Mb +2ii2 7Q`KmH2- 2M BbQHMi H2 i2`K2k= 0- QM Q#iB2Mi , ( 1)ⁿ⁺¹P(X) =Pn+1

k=1( 1)^{n+1 k n+1}_k P(X+

k)X

PM TQb2 k= ( 1)^{k+1 n+1}_k XXX

9

3. SoitP la matrice de passage de la base(e1, e2, e3) à la base(u, v, w). On aS=PS^′P⁻¹. DéterminonsP⁻¹.

⎧⎨

⎩

u=e1−2e3

v=e2−3e3

w=2e₁+3e₂+e₃

⇔

⎧⎨

⎩

e1=u+2e3

e2=v+3e3

w=2(u+2e₃) +3(v+3e₃) +e₃

⇔

⎧⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎩

e3= 1

14(−2u−3v+w) e1=u+ 2

14(−2u−3v+w) e₂=v+ 3

14(−2u−3v+w)

⇔

⎧⎪

⎪⎪

⎪⎪

⎨

⎪⎪

⎪⎪

⎪⎩

e3= 1

14(−2u−3v+w) e₁= 1

14(10u−6v+2w) e2= 1

14(−6u+5v+3w)

DoncP⁻¹= 1 14

⎛

⎝

10 −6 −2

−6 5 −3

2 3 1

⎞

⎠puis

S=PS^′P⁻¹= 1 14

⎛

⎝

1 0 2

0 1 3

−2 −3 1

⎞

⎠

⎛

⎝

1 0 0 0 1 0 0 0 −1

⎞

⎠

⎛

⎝

10 −6 −2

−6 5 −3

2 3 1

⎞

⎠= 1 14

⎛

⎝

1 0 −2

0 1 −3

−2 −3 −1

⎞

⎠

⎛

⎝

10 −6 −2

−6 5 −3

2 3 1

⎞

⎠

= 1 14

⎛

⎝

6 −12 −4

−12 −4 −6

−4 −6 12

⎞

⎠= 1 7

⎛

⎝

3 −6 −2

−6 −2 −3

−2 −3 6

⎞

⎠.

S=1 7

⎛

⎝

3 −6 −2

−6 −2 −3

−2 −3 6

⎞

⎠.

Problème : Résultant de deux polynômes

Partie I : Définitions et propriétés 1. Cas où uest bijective

(a)Soit(A, B)∈F². deg(P) =pet deg(A)!q−1et donc deg(PA)!p+q−1. De même, deg(QB)!p+q−1et donc deg(PA+QB)!p+q−1.uest effectivement une application deEversF.

Soient((A₁, B1),(A₂, B2))∈E² et(λ₁,λ2)∈C².

u(λ₁(A₁, B1) +λ2(A₂, B2)) =u((λ₁A1+λ2A2,λ1B1+λ2B2)) =P(λ₁A1+λ2A2) +Q(λ₁B1+λ2B2)

=λ1(PA₁+QB1) +λ2(PA₂+QB2) =λ1u((A₁, B1)) +λ2u((A₂, B2)). u∈L(E, F).

(b) Siu est bijective, l’élément1de Fadmet un antécédent par udans E. Donc ∃(U, V) ∈ C_q−1[X]×C_p−1[X]tel que UP+VQ=1. Le théorème deBézoutpermet alors d’affirmer que les polynômesPetQsont premiers entre eux.

(c)SupposonsPetQpremiers entre eux. Soit(A, B)∈E.

(A, B)∈Ker(u)⇔PA+QB=0⇔PA=−QB.

Donc, si(A, B)∈Ker(u),Qdivise−QB=PAet puisquePetQsont premiers entre eux,QdiviseAd’après le théorème deGauss. DoncA∈QC[X]∩C_q−1[X] ={0} car deg(Q) =q > q−1. Par suite, Aest nul puisBest nul carQB=0et Q̸=0.

On a montré que Ker(u)⊂{(0, 0)}et donc Ker(u) ={(0, 0)}.

Ainsi,uest une application linéaire injective. Comme de plus dim(E) =dim(F) =p+q <+∞, on sait queuest bijective.

uest bijective si et seulement siPetQsont premiers entre eux.

2. Matrice deu

(a)Soitk∈!0, q−1".

u% (X^k, 0)&

=PX^k= 'p

m=0

akX^m+k.

De même, pourk∈!0, p−1",u% (0, X^k)&

= 'q

m=0

b_kX^m+k. Donc

Mat^B,B′(u) =MP,Q. (b)D’après les questions 2.(a), 1.(b), et 1.(c),

Res(P, Q)̸=0⇔det(MP,Q)̸=0⇔ubijective⇔PetQpremiers entre eux.

3. Racine multiple (a)SoitPdeC[X]de degré supérieur ou égal à1. On sait qu’une racine multiple dePest encore racine deP^′et qu’une racine commune àPetP^′ est racine multiple deP. Donc, d’après la question 2.(b),

Padmet une racine multiple⇔PetP^′admettent une racine commune⇔Res(P, P^′) =0.

(b)SiP=X³+aX+balorsP^′=3X²+apuis

Res(P, P^′) = ' ' ' ' ' ' ' ' ' '

b 0 a 0 0 a b 0 a 0 0 a 3 0 a 1 0 0 3 0 0 1 0 0 3 ' ' ' ' ' ' ' ' ' '

=b ' ' ' ' ' ' ' '

b 0 a 0 a 3 0 a 0 0 3 0 1 0 0 3

' ' ' ' ' ' ' '

+a ' ' ' ' ' ' ' '

a b a 0 0 a 0 a 1 0 3 0 0 1 0 3

' ' ' ' ' ' ' '

=3b ' ' ' ' ' '

b a 0 0 3 0 1 0 3 ' ' ' ' ' '

+a² ' ' ' ' ' '

a 0 a 0 3 0 1 0 3 ' ' ' ' ' '

+a ' ' ' ' ' '

b a 0 a 0 a 1 0 3 ' ' ' ' ' '

=9b ' ' ' '

b a 0 3 ' ' ' '

+3a² ' ' ' '

a a 1 3 ' ' ' '

−a² ' ' ' '

a a 1 3 ' ' ' '

=27b²+4a³. Ainsi,

X³+aX+ba une racine double si et seulement si27b²+4a³=0.

Partie II : Applications 4. Equation de Bézout. (a)

Res(P, Q) = ' ' ' ' ' ' ' ' ' ' ' ' ' '

1 0 0 1 0 0 0

0 1 0 −1 1 0 0

0 0 1 0 −1 1 0

1 0 0 1 0 −1 1

1 1 0 0 1 0 −1

0 1 1 0 0 1 0

0 0 1 0 0 0 1

' ' ' ' ' ' ' ' ' ' ' ' ' '

= ' ' ' ' ' ' ' ' ' ' ' ' ' '

1 0 0 0 0 0 0

0 1 0 −1 1 0 0

0 0 1 0 −1 1 0

1 0 0 0 0 −1 1

1 1 0 −1 1 0 −1

0 1 1 0 0 1 0

0 0 1 0 0 0 1

' ' ' ' ' ' ' ' ' ' ' ' ' '

(C₄←C₄−C₁)

= ' ' ' ' ' ' ' ' ' ' ' '

1 0 −1 1 0 0

0 1 0 −1 1 0

0 0 0 0 −1 1

1 0 −1 1 0 −1

1 1 0 0 1 0

0 1 0 0 0 1

' ' ' ' ' ' ' ' ' ' ' '

= ' ' ' ' ' ' ' ' ' ' ' '

1 0 0 1 0 0

0 1 −1 −1 1 0

0 0 0 0 −1 1

1 0 0 1 0 −1

1 1 0 0 1 0

0 1 0 0 0 1

' ' ' ' ' ' ' ' ' ' ' '

(C₃←C₃+C₄)

= ' ' ' ' ' ' ' ' ' '

1 0 1 0 0

0 0 0 −1 1 1 0 1 0 −1

1 1 0 1 0

0 1 0 0 1

' ' ' ' ' ' ' ' ' '

= ' ' ' ' ' ' ' ' ' '

0 0 1 0 0

0 0 0 −1 1 0 0 1 0 −1

1 1 0 1 0

0 1 0 0 1

' ' ' ' ' ' ' ' ' '

(C₁←C1−C3)

=− ' ' ' ' ' ' ' '

0 1 0 0

0 0 −1 1 0 1 0 −1

1 0 0 1

' ' ' ' ' ' ' '

= ' ' ' ' ' '

0 −1 1 1 0 −1

0 0 1

' ' ' ' ' '

=− ' ' ' '

−1 1 0 1 ' ' ' '

=1̸=0.

Donchttp ://www.maths-france.fr 3 ⃝^c Jean-Louis Rouget, 2009. Tous droits réservés.

P etQsont premiers entre eux.

(b)Soit(A, B)∈C₂[X]×C₃[X]. PosonsA=α₀+α₁X+α₂X² etB=β₀+β₁X+β₂X²+β₃X³ de sorte que (A, B) =α₀(1, 0) +α₁(X, 0) +α₂(X², 0) +β₀(0, 1) +β₁(0, X) +β₂(0, X²) +β₃(0, X³).

PA+QB=1⇔u((A, B)) =1⇔MP,Q

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝ α0

α1

α2

β0

β1

β₂ β₃

⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠

=

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝ 1 0 0 0 0 0 0

⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠

Maintenant

MP,Q

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝ α0

α₁ α2

β0

β1

β2

β3

⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠

=

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝ 1 0 0 0 0 0 0

⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠

⇔

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝

1 0 0 1 0 0 0

0 1 0 −1 1 0 0

0 0 1 0 −1 1 0

1 0 0 1 0 −1 1

1 1 0 0 1 0 −1

0 1 1 0 0 1 0

0 0 1 0 0 0 1

⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝ α0

α₁ α2

β0

β1

β2

β3

⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠

=

⎛

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎜

⎝ 1 0 0 0 0 0 0

⎞

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎟

⎠

⇔

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎩

α0+β0=1 α1−β0+β1=0 α2−β1+β2=0 α0+β0−β2+β3=0 α0+α1+β1−β3=0 α1+α2+β2=0 α2+β3=0

⇔

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎩

β0=1−α0

β3=−α2

α0+α1+β1=1 α2−β1+β2=0

−β₂−α2=−1 α0+α1+β1+α2=0 α1+α2+β2=0

⇔⇔

⎧⎪

⎪⎪

⎪⎪

⎪⎪

⎪⎨

⎪⎪

⎪⎪

⎪⎪

⎪⎪

⎩

α2=−1 β3=1 β₂=2 β1=1 α1=−1 α0=1 β0=0

⇔A=1−X−X²etB=X+2X²+X³.

On peut prendre(A0, B0) = (1−X−X², X+2X²+X³).

(c)Soit(A, B)∈C[X].

AP+BQ=1⇔AP+BQ=A0P+B0Q⇔P(A−A0) =Q(B₀−B).

Nécessairement,PdiviseQ(B₀−B)et donc, puisqueP∧Q=1,PdiviseB0−Bd’après le théorème deGauss. Donc il existe un polynômeStel queB0−SP. De même, il existe un polynômeRtel queA=A0+QR.

Réciproquement, siA=A0+QRetB=B0−PS,

AP+BQ=1⇔A0P+B0Q+QP(R−S) =1⇔QP(R−S) =0⇔R=S.

S =)

(1−X−X²−S(X³−X+1)), X+2X²+X³+S(X⁴+X³+1)), S∈C[X]&

.

4. Equation d’une courbe (a)Pourt∈R,

−−dM→ dt (t) =

* 2t+1 2t−1

+

. En particulier,∀t∈R,

−−dM→

dt (t)̸=−→0 etΓest un arc régulier.

Tableau de variations conjointes dex ety

t −∞ −1 2

1

2 +∞

x^′(t) − 0 +

+∞ +∞

x

−1

+∞ 4 +∞

y

3 4

y^′(t) − 0 +

Etude des branches infinies.Quandttend vers±∞,x(t)ety(t)tend vers+∞puis y(t)

x(t) =t²−t+1

t²+t tend vers1et enfiny(t)−x(t) =−2t+1tend vers−∞quandttend vers+∞et vers+∞quandttend vers−∞.

On en déduit queΓ admet une direction asymptotique d’équationy=xmais n’admet pas de droite asymptote.

Tracé deΓ

1 2 3 4 5 6 7 8 9 10 11 12

1 2 3 4 5 6 7 8 9 10 11 12

(b)M(x, y)∈Γ⇔∃t∈R/

+ x=x(t)

y=y(t) ⇔∃t∈R/

+ A(t) =0

B(t) =0 ⇔AetBont une racine commune dansR. Maintenant,

AetBont une racine commune dansC⇔Res(A, B) =0

⇔ ' ' ' ' ' ' ' '

−x 0 1−y 0 1 −x −1 1−y

1 1 1 −1

0 1 0 1

' ' ' ' ' ' ' '

=0

⇔−x ' ' ' ' ' '

−x −1 1−y

1 1 −1

1 0 1

' ' ' ' ' '

+ (1−y) ' ' ' ' ' '

1 −x 1−y

1 1 −1

0 1 1

' ' ' ' ' '

=0

⇔−x((−x+1) + (y)) + (1−y)(2−(−x−1+y)) =0

⇔x²−2xy+y²−4y+3=0.

http ://www.maths-france.fr 5 ⃝^c Jean-Louis Rouget, 2009. Tous droits réservés.

En résumé,

M(x, y)∈Γ ⇔AetBont une racine commune dansR

⇒AetBont une racine commune dansC

⇔x²−2xy+y²−4y+3=0.

SiM(x, y)∈Γ alorsx²−2xy+y²−4y+3=0.

Remarque.

M(x, y)∈Γ⇔∃t∈R/

+ x=t²+t

y=t²−t+1 ⇔∃t∈R/

+ x−y=2t−1 y=t²−t+1

⇔∃t∈R/

⎧⎪

⎨

⎪⎩ t=1

2(x−y+1) y=1

4(x−y+1)²−1

2(x−y+1) +1

⇔y=1

4(x−y+1)²−1

2(x−y+1) +1

⇔x²−2xy+y²−4y+3=0.

(c) La matrice de q dans la base canonique est

* 1 −1

−1 1 +

. Cette matrice est de rang 1 et on sait que la courbe d’équationx²−2xy+y²−4y+3=0est une conique du genre parabole et donc soit une parabole, soit une droite, soit est vide. En tenant compte du tracé de la question (a), cette courbe est une parabole.

6. Nombre algébrique. Déterminons le résultant deP etQ.

Res(P, Q) = ' ' ' ' ' ' ' '

−3 0 y²−7 0 0 −3 −2y y²−7

1 0 1 −2y

0 1 0 1

' ' ' ' ' ' ' '

=−3 ' ' ' ' ' '

−3 −2y y²−7

0 1 −2y

1 0 1

' ' ' ' ' '

+ ' ' ' ' ' '

0 y²−7 0

−3 −2y y²−7

1 0 1

' ' ' ' ' '

=−3(−3+ (3y²+7)) + ((3y²−21) + (y⁴−14y²+49)) =y⁴−20y²+16.

Siy=√ 3+√

7, alorsPetQont une racine commune dansCà savoir√

3et donc Res(P, Q) =0. Ainsi, poury=√ 3+√

7, on ay⁴−20y²+16=0et donc

√3+√

7est racine du polynômeX⁴−20X²+16.

Ensuite, pourz∈C,

z⁴−20z²+16=0⇔(z²−10)²=84⇔z²=10+2√

21ouz²=10−2√ 21

⇔z²= (√ 3+√

7)²ouz²= (√ 3−√

7)²

⇔z∈{√ 3+√

7,−√ 3+√

7,√ 3−√

7,−√ 3−√

7.

Les racine du polynômeX⁴−20X²+16sont√ 3+√

7,−√ 3+√

7,√ 3−√

7et−√ 3−√

7.

Remarque.α=√ 3+√

7⇒α²=10+2√

21⇒(α²−10)²=84⇒α⁴−20α²+16=0.