APPLICATIONS OF GALOIS THEORY 1. Finite Fields

1 CHAPTER IXAPPLICATIONS OF GALOIS THEORY1. Finite FieldsLetFbe a nite eld. It is necessarily of nonzero characteristicpand its prime eld is the eld a vector space overFp,itmusthaveq=prelements wherer=[F:Fp]. Moregenerally, ifE Fare both nite, thenEhasqdelements whered=[E:F].As we mentioned earlier, the multiplicative groupF ofFis cyclic (because it is a nite subgroup of themultiplicative group of a eld), and clearly its order isq 1. Hence each non-zero element ofFis a root ofthe polynomialXq 1 1. Since 0 is the only root of the polynomialX, it follows that theqelements ofFare roots of the polynomialXq X=X(Xq 1 1). Hence, that polynomial is separable andFconsists ofthe set of its roots. (You can also see that it must be separable by nding its derivative which is 1.) Wemay now conclude that the nite eldFis the splitting eld overFpof the separable polynomialXq Xwhereq=jFj. In particular, it is unique up to isomorphism. We have proved the rst part of the a prime.

2 For eachq=pr, there is a unique (up to isomorphism) nite eldFwithjFj= have already proved the uniqueness. Supposeq=pr, and consider the polynomialXq X2Fp[X]. As mentioned aboveDf(X)= 1sof(X) cannot have any repeated roots in any extension, itis separable. LetFbe its splitting eld [x1;:::;xq]wherex1;:::;xqare theqdistinct roots off(X). However, it is not hard to see that the setfx1;:::;xqgis in fact already a eld. For,xis a root off(X)=Xq Xif and only ifxq=x, and the fact that raising to theqth power is a ringhomomorphism in characteristicp( , (x+y)q=xq+yqand (xy)q=xqyq) tells us that the set of rootsis a subring. Ifx6=0,wehave(x 1)q=(xq) 1=x 1, so every element in this subring is invertible in thesubring. HenceFis the set of roots off(X) and hasqelements as claimed. We want to t all these nite elds in the same eld and show how they are related to one another. Tothis end, we shall use a result to be proved later. Namely, ifFis any eld, then we shall show later thatthere is an algebraic extensionFwhich is algebraically closed and which is unique up an extension is called analgebraic closureofF, or with abuse of terminologythealgebraic closure ofF.

3 Let pdenote the algebraic closure of the prime eldFp. For anyq=prthere is a unique sub eld of pisomorphic to every eld withqelements, namely the splitting led ofXq XoverFp. We shall denotethis instance of a eld withqelements nite eldsFqare coherently related. Namely, rst suppose thatFq Fq0. The the latter may beviewed as a vector space over the former of dimensiond=[F0q:Fq]. That is,F0qis isomorphic to a directsum ofdcopies ofFqa, whenceq0=jF0qj=jFqjd=qd,orpr0=prd. It follows , supposerdividesr0, ;q=pr,andq0=pr0=qd. We claim thatFq this note that the latter is the splitting eld overFpofXqd Xand the former is the splitting eld ofX1 X. However, dividing yieldsXqd XXq X=Xqd 1 1Xq 1 1=X(q 1)(k 1)+X(q 1)(k 2)+ +Xq 1+1 Typeset byAMS-TEX8990IX. APPLICATIONS OF GALOIS THEORY wherek=(qd 1)=(q 1) =qd 1+qd 2+ +q+ 1. (Note the confusing double use of the formula fora geometric sum!) It follows thatXq XdividesXq0 Xso any splitting eld of the latter contains asplitting eld of the former.

4 Hence, by the uniqueness of splitting elds in p,Fq0 Fqas pis in fact the union of the nite sub eldsFqwhereq=pr. For, it is algebraic overFpbyde nition, so any element is in a nite extension ofFp, hence in a nite sub eld of p. However, the aboveanalysis assures us that the eldsFqare the only such sub are now in a position to calculate the GALOIS groupG(Fq0=Fq)inthecaseq0=qd. First, de ne q: p! pby q(x)=xq. It is not hard to see that qis a ring homomorphism, so at the very least it isa monomorphism. It is in fact also an epimorphism and hence an automorphism of p. (See the exercises.) a given primep. The restriction of qtoFq0is an automorphismofFq0and it generates the GALOIS groupG(Fq0=Fq)which is cyclic of note that q xesFqwhich is the set of elements in psatisfyingxq= to see that it carriesFqdinto itself. Since it is a monomorphism and that eld is nite, its restrictionis an automorphism ofFqd. Moreover, we have iq(x)=xqiso q=Id()xqi=xfor allx2 Fqd:If this were to happen for somei<d,Fqwould in fact haveqielements instead which is nonsense.

5 Hence,the restriction of qhas orderd. On the other hand, [Fq0:Fq]=d, so by GALOIS 's main theoremjG(Fq0=Fq)jhas orderdand hence is generated by q. explicitly that q: p! ppreserves sums, products, and the identity element 12 that q: p! pis an epimorphism, hence an automorphism. (You already know from thediscussion in the text that it is a monomorphism.)3.(a) An automorphism of pis speci ed if we know its restriction dto everyFqwithq= ,ifd0=cd;q=pd,andq0=pd0, then the restriction of d0toFqmust be d. Suppose conversely, that weare given for each positive integerdan automorphism dofFpd, and that the family of all dsatis es theconsistency condition just enunciated. Show that there is an automorphism of pwhich restricts to donFpdfor everyd.(b) Is every such a power of ; , isG( p=Fp) cyclic with generator ? Hint: This is not Extension of the base eldLetK Fbe a nite, normal separable extension. We often want to know what happens to the Galoisgroup when we extendFto a larger eldLwhich may not even be algebraic overF.

6 For example, theextensionQ[i] Qis normal and separable with GALOIS group cyclic of order 2. Similarly, we can say thesame forR[i] R, and in fact there is a natural way to identify the two GALOIS groups. The rst problem indealing with this situation in general is that there is no reason even to assume thatKandLcan be imbeddedin the same eld. Assume then that there is a eld which contains bothKandL. (In the example, all the elds are contained inC.) We denote byKLthe smallestsubringof which contains bothKandL. (Thisde nition is not quite standard!)KLas before is the set of nite sumsPxiyiwithxi2 Kandyi2 Landit would ordinarily not be a sub eld of ; we would have to form fractions in order to get a eld. However,ifK Fis nite thenK=F[x1;:::;xn] for appropriate elements ofK,soKL=L[x1;:::;xn]isinfact nite overLand is thereby a eld. In this case we shall could have extensionsK0 FandL0 Fwhich areF-isomorphic to the previous extensions withK0andL0contained in some eld 0but withK0L0not isomorphic toKL.

7 This seems paradoxical, but it canhappen since the construction ofKLdepends to some extent on the common enclosing eld . We shallanalyze this situation in more detail when we discuss ring THEORY later in this EXTENSION OF THE BASE FIELD91 Theorem.(Natural Irrationalities). LetK Fbe a nite, normal, separable extension, and letLbe an extension ofFsuch thatKandLare contained in a common eld. ThenKL Lis a nite,normal, separable extension. Moreover, restrictingL-automorphisms ofKLtoKyields a monomorphismG(KL=L)!G(K=L),andtheimageis G(K=K\L).LK1 1 KLG(KL=L) G(K=K\L)K\LjjG(K=F) note thatKLis normal overL. For, we can writeK=F[x1;:::;xk] as generated by theroots of a separable polynomial with coe cients inF; henceKL=L[x1;:::;xk] is also such a splitting 2G(KL=L). The restriction of toKcertainly xesFand by normality it must carryKinto itself; so it is an element 02G(K=F). is certainly completely determined by its e ect onx1;:::;xkand since these are inK, it is determined by its restriction 0toK.

8 Hence 7! 0is a monomorphism. Anelementx2 Kis xed by the subgroup which is the image of this monomorphism if and only if it is xed byall 2G(KL=L), , if and only ifx2L. But sincex2 Kin any case, this holds if and only ifx2K\ GALOIS 's Main Theorem, the image subgroup must beG(K=K\L). Example =3p2andlet!=e(2 i)=3as earlier. LetL=Q( ). We use the fact that is notalgebraic to conclude thatL\K=Qin this case. (Can you prove it?) Hence the diagram looks likeL[ ;!]1 1Q[ ;!]LS3 S3 QExample Qbe a nite, normal, separable extension contained \Ris themaximal real sub eld real thenKR=Rand nothing interesting occurs. IfKis not contained inR, then we must haveKR=Csince it its a proper extension ofRandCis the only such algebraic follows thatG(K=Q) has a subgroup of order 2 generated by the restriction of complex (a) LetKbe a eld and letPandQbe groups of automorphisms ofK. Show thatKPQ=KP\KQ.(b) LetK Fbe a eld extension and letLandMbe intermediate elds.

9 Show thatG(K=LM)=G(K=L)\G(K=M).(c) Suppose thatK Fis a nite normal separable extension. Conclude that in the GALOIS correspondencebetween subgroups ofG(K=F) and intermediate sub elds, intersections of subgroups correspond to compositaof sub elds and products of subgroups correspond to intersections of sub that ifKandNare nite normal extensions ofFboth contained in the same eld , then thecompositumKNis normal APPLICATIONS OF GALOIS THEORY3. Cyclotomic extensionsArootofthepolynomialXn 1forsomen>0 is called aroot of unity. For example, inC, if we put =2 =n,thenXn 1hasthenrootseik ,k=0,1,:::,n 1. These appear in the complex plane as thevertices of a regularn-gon inscribed in the unit the base eld have (Xn 1) =nXn 1=0soXn 1 is not separable. Conversely, ifpdoes not dividen,thenXn 1andnXn 1clearly have noroots in common, soXn 1 is separable. For this reason, we shall always assume when discussingnth rootsof unity that gcd(p;n)=1ifp>0.

10 Of course, if the characteristic is 0, there is no need for any an extension ofF. The roots ofXn 1inKare distinct (assuming as above that gcd(p;n)=1),and it is clear that they form a subgroup ofK under multiplication. Hence, the set of roots forms a cyclicgroup of order n. Assume further thatXn 1 splits completely inK. Then, the order of this group isn. In that case, a generator of the group ofnth roots of unity inKis called aprimitiventh root of root is then a power iwith 0 i n 1, and such a power is also primitive ( a generator) if andonly if gcd(i;n) = 1. It follows that the number of primitive roots is (n)where is the Euler { , again since the roots are all powers of ,wehaveinK[X]Xn 1=n 1Yi=0(X i)Note that this splitting already takes place inF[ ] which is a splitting eld forXn [ ] is called acyclotomicextension ofF. Notice that the existence of a primitiventh root of unity in some extensionofFimplies thatXn 1 has distinct roots, so it is separable and necessarilynis relatively prime to thecharacteristic relatively prime to the characteristic be a primitiventh root of (F[ ]=F)is isomorphic to a subgroup ofU(Z=nZ)(the group of units ofZ=nZ) and hence is abelianof order dividing (n).}