Theeren ; "-x = TTI, (x) where fi (x)E (F. [ x] are the manic irreducible polynomials with dey (5, ) | n Proof Suppose flx) ElFLAJ is manic & irreducible & dog (f) In Then K' FPL x] / (food) is a field with pay' elements. Write a for image of x in k Then or ". dog => flallypygm . - x x ! " - x Support flu) monic & irreducible f (x) XF"- x Let E / IF. split x -x. Let S=6 EE; P=) IFpm => 3 6s nt. Fla) = 0 => IF, () es =IF = > deg [f ) | Theorem; IF is cyclic Proof : By induction on h. Claimi For Jill, # {d El,"; d =192d Prooffor =1 ad satisfies x- 1=0, x -1 / x P" -d - 1/ -x =) d hear roots , since xx1 splits over IFp " Claim For dip"-1, # [x ElF* and ( ) = 1)= # ED Can , " and (p)=1) Proof: Induction on d. Dare Case Jel V mild = # {BEZ band (p)=d) .MAT401 Lecture 17 March (5, 2013 Finite Fields Definition/ Afinite field is a field with finitely many elements Exi 7 : R, [ x 1 / ( Hx x ) ={0, 1, d, s + ly Let k be a finite field . Lemmal charck) #0 Proof: If charck)=0 , then 2 -K but Z is infinite & K is finite Let p=cher(k ) " .2 ck Lenna; LK: 7, coo je Kis a frite extension of Ep degree Prent; K is a vector space over Zo. Since K is finite it has a finite basis. Let , = LK : 7 ) are dog ( K / 2) Corollary IK = p" Thegren; For every prime power poy] finite field K with kap Proof : Let f ( x ) = x P' - x EZ,[x] Nate, F(x) = * is a ring hampmorphism in char(p) " fix) =" -x = F"Yx)-x Let E / Tip be a splitting field for f (x], s = for ER; fla) = 0} 1 ) Mate ? fixed .. ged ( f, 5 ' ) = 1 . . flack has distinct roots 21 0,1 ES 3) a, bes => lab ear !"=ab =>abES 4) a, beS = >[ath] " = F" (ath) = ["Cal- Fom (6) =ap" +1" = ath = >atb ES 5)at5, at0 -> Calpe peas hes =)SCE is a field and (51=p"Lemming Let K be a field, I kleph then VaEK , al"ed Proof: Note " K* is an abelian group, of size p"-1 Also, (JP" = 0) Corollary; Every field of size p" is a minimal splitting field for finite Corollary; Every 2" fields K, Ke with (K, = Ki are isomorphic Proportion? IF captains a copy of pm iff men. Proof ; Suppose into & Fomc => EmcF. =3 pm -Lly"-L => [ . .] pm-1 1pel, for reremainder of n module my contradiction Suppose win. Let E/ Z be a field extension splitting *f"-x and x -x 5. = (o GE; OF" = ); Sm= [BEE; A"-= BY S, Sm both fields Let hem s, rEIN. BES.=> BEBPRE-( 80myFACE -1) Alternatively ; B " = F " ( B ) = Fomr ( p ) ( For )" ( B) = pLinear Algebra; Let F be a field V- wetor space over F, don n =)fe, e CV s. to Every .. EV can be written uniquely and of = for , f,EF Exi f ( x ) = * ' 4x +/ EF, Lx] irreducible Exifled,all, day atall, bases for IF/ IF 2 Consider xox = * [x7 - 1) = xx (x+1) [*6+85+x"+x"4x +*+/ [ x ' +x + 1 ) = (x - 2 ) ( x " + d x to " +/ ) ( x " + 2 + all + (d ] + 2+1) =) x 14x +1 =(x to) [x to') (xtal+2) * 3 + x 2 +1 = ( x +* +1) ( x + 2 +1) ( * + 2 +2+/)Q3 (10 points) Consider the polynomial at) : 3:3 m + l E $3M]. Since m) has no roots in E3 and it has degree 3 it is irreducible. Thus the ring R :2 Zg[$]f{f($)) is a finite fieldsof order 2?. Let 0: E R denote the element :2: + [mlj . so that rm) 2 0. We proved in class that RX is cyclic. Find (with proof!) an element 1|" E Rx which is a generator. In other 1wordsr find an element 5" such that every element in RX is of the form \"r\" for an integer n