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I : Theorem Great Galois ' when polynomial a theorem that - PowerPoint PPT Presentation

Dec 30, 2022 •222 likes •471 views

Applications of statement I : Theorem Great Galois ' when polynomial a theorem that down write today a expresses in something analogous to expressible roots that are has the gvntmhi formula : quadratic formula , cubic formula ,

  1. Applications of statement I : Theorem Great Galois '

  2. when polynomial a theorem that down write today a expresses in something analogous to expressible roots that are has the gvntmhi formula : quadratic formula , cubic formula , quartic this formula for Motivation Q ) are Cin - O iwmdn roots of ax 't bxtc - : - b ± ME x - - - Z a held elements , felt ops , in form of expressible roots Note : are extraction and not

  3. formula root at Cubic ax 't b x' texted is a - ✓ qtT t Vq-Tt tp q= p 't bcjaazd - %a - %a , where - r p - - , elements , field operations , of field terms in formula is Note : extinctions root and in terms of roots same takeaway Quartic formula are : root extractions field elements , field operating , -

  4. " solvable a formula of " nations by generalize these To ( Radical Extension ) Def radical called extension extension is Elf a A field fields - intermediate if exist there - E Eee , E Ee = E = Eo E E , E Ez E - - F ei E Ei - i Kiel have for all so that some we - Ei - i ( nine ; ) with ni EIN Ei and - some - ei EE ; - did ) hifi root of " is just some ( where x

  5. Ix ) splitting field of x ' -2 € the EI . we be Ut E w= -11¥ , wt ) = Q ( Vz know where E w ) ( Vz = :÷÷÷÷÷:i÷÷÷÷ : :* :* , ( RE , Fs ) Vz ) a = . .

  6. Define ( solvable by radicals ) A polynomial flxltflx ) solvnbkbgmdicab if is radical contained splitting field some its in is Kf extension ÷i÷÷÷÷÷iem¥÷÷÷÷ . : expressible roots are its field terms at in elements , field open heirs root extinctions - and .

  7. - Qfsryfs ) radicals by Kf solvable since Exe ID X3 - 2E is radical extn ④ ( vz , Fs ) 1a we've where is seen a . be given char (F) Hn ctf EI , and let . suppose radicals since by soluble " f- ( x ) - c Then - X is - , Wn ) = FIVE Kf - extension it radical Flora , un ) is and a church In , even works if basic idea ( same roots of unity ) with but need car some

  8. polynomial quadratic , cubic , quartic EI Evey charles -42,33 ) radicals ( if by solvable are express the cook have formulas That because we operates , element , field field terms at in extinction ) root and " ? " other side of the coin the about Whit not solvable by medicals , does Thet f Ie , if ' is aren't expressible in not that f has root extractions ? mean element , field ops , and of field terms

  9. composition ) cloud under ( radical extensions are extensions of , Tha F radical k and If are E of F . extension a radical Ev K is - ÷ :÷ , Een ' te . emirate .im , - " - - * ÷ , ( Vez ) ,E Ez = - FIVE , ) Ei - iii. Imai . - Ii - I ¥ :* , - - - :* , -

  10. no analog radicals , then There not soluble is Cos If f is for polynomials of degree self ) formula quadratic of . roots formula , Thu all such PI If There were a terms of Held expressible F , in be would of f extinctions . Ie root if , and root a is a • pontian , medical extension . a subfield of some FK ) then is f ke =¥÷Ik of ) , result , means this last But by our radical extension . Ie , f indica Is , is solvable also by in some is BB

  11. higher degree analogs of quadratic formula looking fr So : solvable by radicals ? is f when asking to amounts : can figure Higginson soluble if f out is How roots ? a formula for its looking fer w/o medicals by excursion into need to take answer this we an To , group Theory .

  12. ( solvable group ) Deth called soluble a sequence it there is G A is group of subgroups - ⇐ Hn - f- HE G geg ) ⇐ Hos HEHE . - and fer all His Hit , oeisn so that Hit ya , cyclic is . . By solvable Fundamental are groups abelian ⇐ All finite . > ne HN ④ Renato for - ④ Ink 4 , ne , theorem , GI kn some . - , - - ⑦ let Eka .to?LnztofeHo--tOlelE---EKn.to.--tOKnk { e) skatole Ho and so

  13. since solvable EL S , is E S3 { e) E ( ( 123 ) ) size 6 size 3 site I 412317053 have Iska ) > 1=-115,1 we Sime , 53/4123,7=12 and 4123%3=83 { e) a ( CRH ) Also and

  14. EI solvable toe : Sy is , 1137124 ) , 1147123 ) ) E Ay s Su { e) s ( ( 12713477 E le , 1127134 ) , desired of subgroups Sats bees chain this Fun exercise : properties - gon ) because ( symmetries of solvable is Exe Dn a { e) E { rotations ) E Dn bill . fits the

  15. solvable ) Thin ( As not is solvable . As not is If ( sketch ) - cycles , which means ⑨ As by 3 generated is - cycle } = { o , 3 Tiff is and KEN some As - ok : . - lab c) c- As As if - cycles conjugate in ⑤ all r , - ie 3 are , c- As F - lxyz ) c- As is there so thin Sonu and rz - " that = FTZF F . ,

  16. - cycle has H 3 ④ if Ho Ag some Atle ) and , then enough ? is this why Ayy not cyclic , then - le ) and H As H If - . need solvable , be some fee As to we Hence H t le ) with . torments Hotty subgroups in As 3- ' 5k contains " consecutive H some w/ . " " II. 1¥ . ! tins Ii : ' { e) DA 5 pm

  17. - ish theorem tem group theory Deep - subgroup of G and solvable , then G evey If is every quotient at solvable too G is . are not A . and n > 5. Then Sn CI If solvable . subgroup isomorphic to As Pf in find a One can - { Gesu : Olli ) - i " S ( namely An " ↳ Sn Sn 's since or for all i > 5) inside this ) tnke the As and .

  18. THEOREM ! BIG THE FOR Now - - Great Theorem ) Thon ( Galois ' Tt let Kf , let flat flex ) , and chart F ) - O - solvable by its splitting field radicals Then f be is . solvable iff Gal ( KH F ) is . PI Next time .

  19. A few takeaways to prove not all quirks solvable was first Abel are • for polynomials certain [ Ya that could Abel prove ° hnd.am?i::neiemic " . " ! ! " " " " , ay poly solvable , , 53 , Sy so • we Sa knew are radicals at solvable degree by E 4 is

  20. all - Thompson that proved Feit EX 1962 , In n 250 pgs ) solvable . ( proof was odd order are groups odd , then 164 ( KHE ) ) Hence it is indicants f- is by solvable .

  21. " unsolvable quintic " EI exist There that's quintic polynomial be flx ) E Ix ) any let root real precisely 3 with irreducible and . radicals . solvable by : f not is Claim says That group they 514 Gal ( KHAN Ff : so , order 5 . element r of God ( KHQ ) has an that get Gul LKFIQ ) ↳ 55 we Since , ( thought of in It ) . - cycle that r 5 is a

  22. exchanges the complex conjugation complex 2 But its roots fixed , so roots keeps but rat the 3 transposition a . Ss ) inside ( thought of Gul ( Kha ) So - cycle and 2 has 5 cycle a . a with a 5. gale subgroup of Si : ay Group Thay faut . So . Gul ( KH @ HSI of Ss 2- cycle all ' and is a not solvable by . So f solvable not is so radicals . -

  23. 5 - 4xt2 For example x :

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