A crash curse in Galis thery First versin 0.1 14. september 2013 klkken 14:50 In these ntes K dentes a field. Embeddings Assume that is a field and that : K! and embedding. If K L is an extensin, we say that the embedding : L! extends if K =. In the special case when K, and is the inclusin f K, suchanembedding is said t be an embedding ver K. A fact that makes life easy wrking with fields, is that ring hmmrphism between fields always are injective. And if in additin the tw fields are f the same finite dimensin ver sme grund field K and the hmmrphism is an embedding ver K (that is, it is K-linear), the hmmrphism is als surjective. We intrduce the ntatin Emb (L, ) fr the set f embeddings : L! extending.incase is the inclusin f K in, we write Emb K (L, ) fr the set f embeddings ver K. That is, we have Emb (L,!) ={ embeddings : L!! such that K = } Emb K (L, ) = { embeddings : L! such that (x) =x fr x 2 K} The Galis grup An imprtant special case is when =L, andl is a finite extensin f K. Then Emb K (L, L) is a grup. Indeed the cmpsitin f tw embeddings ver K is an embedding and this gives Emb K (L, L) amnidstructure.sincel is f finite dimensin ver K, everyembeddingverk is invertible, S Emb K (L, L) is a grup. This grup is the famus Galis grup f L ver K. WedenteitbyGal(L/K). In the mre general case when L nt necessarily is a finite extensin, Emb K (L, L) is nt always a grup, but the subset f invertible embeddings is a grup. And in this case, this is the Galis grup. Prblem 1. Let n 2 be an integer and p aprime(frsimplicity).shwthat Gal(Q( np p)/q) =1if n is dd and Gal(Q( np p)/q) ' Z/2Z if n is even. Prblem 2. Let K be any field and let a 2 K be an element. Let n 2 be an integer and let f(x) =x n a and L = E f the stem field. When is f(x) irreducible? What are the pssible Galis grups Gal(L/K)? Prblem 3. The set up is the standard ne: Tw extensins K L and K. Shw that the tw Galis grups Gal(K/L) and Gal( /K) bth act n Emb K (L, ), respectively frm the left and frm the right. 1
The stem field In this paragraph f(x) 2 K[x] will be an irreducible plynmial whse degree we dente by n. The plynmial being irreducible the qutient ring E f = K[x](f(x)) is a field, and we call it the stem field f f. ItisanextensinfK. The residue class f x, say, isbydefinitinartff(x) in E f.se f cmes equipped with a preferred rt f f(x). The degree f E f ver K equals the degree n f f, andthepwers1,x,...,x n 1 frm a K-basis fr E f. If K is a field extensin, any rt! that f(x) might have in, givesriset an embedding f E f in. Indeed,sendingx t! induces a hmmrphism K[x]! which factrs thrugh E f = K[x]/(f(x)) since it maps f(x) t f(!) =0.Henceweget amape f!, which autmatically is an embedding ver K. Cnversely, if : E f! is an embedding, the image f ( ) under is a rt f f in. Altgether,thereisa (cannical) identificatin: Emb K (E f, ) '{rts f f in }. In case K is nt cntained in, butthereisgivenanembedding : K!, thereisa similar identificatin. If f(x) = P i a ix i,weletf = P i (a i )x i, which is a plynmial with cefficients in (K). Wethenhave: Emb (E f,!) '{rts f f in }. If! 2 L is a rt f f(x), thenthecrrespndingembeddinggivesanismrphism E f ' L(!) which is an ismrphism ver K, inthesensethatitisk-linear. Hence if! and! 0 are tw rts f f in L, thetwextensinfieldsk(!) and K(! 0 ) are ismrphic ver K. The ismrphism sends a plynmial expressin p(!) t the plynmial expressin p(! 0 ). Example. AbasicexampleisK = R and f(x) =x 2 +1. Then ne has C = E f (this might be taken as a definitin!). The residue class f x is nrmally dented by i, and f curse i 2 = 1. The plynmial x 2 +1 has anther rt in E f,namely i. This shws that f might have several rts in E f,butnefthem,theresidueclassfx, is singled ut as the rt. e Prblem 4. Let p be a prime and let p(x) =x p 1 +x p 2 + +x+1 = (x p 1)/(x 1). This is called the p-th cycltmic plynmial. a) Shw that p is irreducible Hint: Apply Eisenstein s criterin t a twist f p(x). b) Shw that p(x) has p rts in the stem field E f. c) Shw by an example, that n is nt irreducible if n is nt prime. Prblem 5. The setup is as in the previus exercise. Assume p>2. Check that 2 is a rt f p(x), andlet be the crrespnding ismrphism f E f t E f sending 2
t 2.Whatisthematrixf in the basis frmed by the pwers 1,...,x p 1?Dthe same with 2 replaced by q. Existence f splitting fields We draw sme easy cnsequences. It is interesting t nte that what tday is merely an bservatin nce was a serius issue in mathematics invlving the mst prminent mathematicians at that time. In fact in prving that C is algebraically clsed, Lagrange used that any plynmial has a rt smewhere, but he was very vague abut where smewhere was, and Gauss criticizes this severely he did nt like mysterius rts in the clud. Anyhw, we have: Prpsitin 1 Given a field K and a plynmial f(t) 2 K[t]. Thereexistsafield extensin K E in which f has a rt. A slightly mre cmplicated statement is the fllwing. Recall that asplittingfield fr a plynmial f(t) 2 K[t] is field extensin E f K ver which f splits as a prduct f linear factrs, and such that E is generated ver K by the rts f f. Prpsitin 2 Every plynmial f(t) 2 K[t] has a splitting field E. Ifn =degf(t), then [E : K] <n! Prf: Inductin n deg f. LetF be a field where f has the rt, andletg(t) = f(t)/(t ). Then a splitting field fr g(t) ver K( ) F is a splitting fr f(t). Furthermre deg g = n 1, andtherefrewegetbyinductinthat[e : K] =[E : K( )][K( ) : K] apple (n 1)!n = n! Embeddings f the stem field The number f rts f f is bunded by the degree n f f,andf has n distinct rts in if and nly if f splits a prduct f n distinct linear factrs in [x]. In that case we say that f splits simply in. The fllwing prpsitin, which finally is merely an bservatin, is ne f the pillars f the fundatinal Galis thery. Cmbined with an inductin prcedure, invlving twers f fields, it leads mre r less directly t half f the fundamental therem f Galis thery, the ther half hinging n a result similar t Artin s therem n characters. Prpsitin 3 Assume : K! is an embedding. Then The number f embeddings f E f in extending is bunded by deg f. Thatis #Emb (E f, ) apple [E f : K] Equality hlds if and nly if f splits simply in. 3
Climbing twers The prperty f the stem field described in the previus prpsitin generalizes t any field extensin. The technic f prf is, as indicated abve, inductin n the degree [L : K] applied t intermediate fields K E L that is three-stry twers cmbined with the stem field case. Prpsitin 4 Given en embedding : K!. AssumethatK L is a finite field extensin. Then the number f extensins f t L is bunded by the degree [L : K], that is: #Emb (L, ) apple [L : K] Prf: Inductin n [L : K]. The prpsitin is true fr stem fields. Let x 2 L and x/2 K. The embedding can by the stem field case be extended t K(x) in at mst [K(x) : K] ways, and each f these extensins can by inductin be further extended in at mst [L : K(x)] ways. Hence the prpsitin, since [L : K(x)][K(x) :K] =[L : K]. The mral cntent f this argument is that the tw sides f the inequality bth t a certain extent behave multiplicatively with respect t twers f field. That is, the right side is multiplicative in twers, but the left side is nly submultiplicative. Hence t get cnditins n L that guaranteed equality, we need cnditins that make the number f embeddings behave well. And it turns ut, as the secnd statement abut the stem fields indicates, that the gd cnditin is that any plynmial in K[t] having a rt in L splits simply in. Wehave Prpsitin 5 Let : K! be an embedding and K L afieldextensins.then the tw fllwing cnditins are equivalent: #Emb (L, ) = [L : K] Every plynmial f(t) 2 K[t] having a rt in splits simply in Prf: Assuming the cnditin 2, we use inductin n [L : K] t prve 1. We start by prving a lemma: Lemma 1 Let E be an intermediate field between K and L. Assumethattheextensin K L ver K satisfies cnditin 2. Then the extensin E L satisfies cnditin 2 as well. Indeed, pick an element x 2 L and let f K and f E be the minimal plynmials f x ver respectively K and E. Then f E is a factr in f K.Indeed,f K 2 E[t] and f K (x) =0. Therefre f E has nly simple rts since f K nly has simple rts, and it splits in linear factrs in as f K des. This prves the lemma. By inductin it fllws that, at least if E ( L, anyembeddingfe in extends in exactly [L : E] ways t L. 4
If nw E = K(x) fr sme x 2 L, theminimalplynmialf K splits simply in, and hence by the stem field case, any embedding f K in! extends t K(x) in exactly [K(x) :K] ways. This gives all tgether [L : E][K(x) :K] =[L : K] extensins f t L, andwearethrugh. The ther way arund: If cnditin 2 des nt hld, there is by the stem field case an element x 2 L with #Emb (K(x), ) < [K(x) :K], andsinceeachfthesecan be extended t L in at mst [L : K(x)] ways, it fllws that #Emb (L, ) < [L : K(x)][K(x) :K] =[L : K] Prblem 6. Use inductin n [L : K] cmbined with the stem field case, t shw that any embedding f K int an algebraically clsed field can be extended alng any finite extensin K L. Use Zrns s lemma t shw that the result still hlds true when L is merely algebraic ver K (i.e., nt necessarily finite, but all elements algebraic ver K). Splitting fields Recall that if f(x) 2 K[x] is a plynmial, a splitting field fr f ver K is a field extensin L f K with the fllwing tw prperties. Firstly, f shuld split as a prduct f linear factrs in L, thatis,allitsrtsshuldbeinl. Secndly, the field L shuld be generated ver K by the rts f f. The splitting fields play a very special rle in the thery f equatins, and the fllwing is a central result: Prpsitin 6 Assume that f(x) 2 K[x] is a plynmial and that L is a splitting field fr f. Ifthertsff are all distinct, then fr any embedding f K in L, nehas #Emb (L, L) =[L : K]. Inparticular#Gal(L/K) =[L : K]. Prf: We use inductin n [L : K] and apply a variant f the twer-climbing technic. Let be ne f the rts f f nt in K, andlete = K( ). Then L is als a splitting field fr f viewed as ply in E[x]. NwE 6= K, s[l : E] < [L : K], andbyinductin #Emb (L, L) =[L : E] fr any embedding f E in L. The minimal plynmial f ver K is a factr in f, hence splits cmpletely in L as L is a splitting field fr f, anditsrtsaredistinct, since thse f f are. Frm prpsitin 3 n page 3 it fllws that fr any embedding f K in L it hlds true that #Emb (E,L)=[E : K]. The prpsitin then fllws by multiplicativity in twers. Example. The splitting field f x 3 2 ver Q is Q(, 3p 2) where is a primitive third rt f unity. Indeed, the three rts 3p 2, 3p 2 and 3p 2 all lie in Q(, 3p 2), and since = 3p 2/ 3p 2, L is generated by the rts. e Prblem 7. Shw that the splitting field f f(x) 2 K[x] is unique up t ismrphism ver K. 5
Prblem 8. Let p be a prime and let be a primitive p-th rt f unity. Let f(x) = x p 1 + x p 2 + + x +1.Shwthatthesplittingfieldff is equal t Q( ). Prblem 9. Let a and b be t different members f the field K neither being a square. Shw that L = K( p a, p b) is the splitting field f f(x) =x 4 (a+b)x 2 +(a b) 2.What is thew Galis grup Gal(E/K) and hw des it act n L? FieldslikeK( p a, p b) are classically called biquadratic fields. Prblem 10. Let a be a ratinal number. Determine the splitting field f x 4 a. Galis fields The cnditin 2 is naturally split in tw. The extensin L is separable if the minimal plynmial f any x 2 L ver K nly has simple rts. And n may say that L is nrmal twards if any plynmial in K[t] having a rt in K factrs in a prduct f linear factrs ver, r s t say,has all its rts in. If =L, this last cnditin is what is called nrmal. FieldextensinsK L satisfying the cnditin 2 abve with =L are called Galis fields. Recalling that Emb K (L, L) is the Galis grup Gal(L/L), wehave Prpsitin 7 Let K L be a field extensin. Then the rder f the Galis grup Gal(L/K) is equal t [L : K] if and nly if the extensin is separable and nrmal. One has Prpsitin 8 The field extensin K L is Galis if and nly if L is the splitting field ver K f sme plynmial, Prf: Any splitting field is Galis by prpsitin 6 n page 5. AssumethatL is Galis ver K. Taketheminimalplynmialf f any element in L. SinceL is nrmal and separable, the rts 1,..., s f f are all in L. LetE = K( 1,..., s ). Then L is Galis ver E by lemma 1 and is by inductin the splitting field f sme g. Then L is the splitting field f fg. T a subgrup H Gal(L/K) ne assciates the fixed field L H f L cnsisting f elements untuched by any element in Galis grup. That is L H = { x 2 L (x) =x fr all 2 Gal(L/K) }. This is clearly a field, the s being ring hmmrphisms. Therem 1 Let G be a grup f autmrphisms f the field L and let K = L G.Then [L : K] apple G A therem f Artin n autmrphisms f fields. Let L be a field and let G be a finite grup f autmrphism f L. The fixed field f G is the subfield L G = { x 2 L (x) =x fr all 2 G }. The fllwing therem is clsely related t Artin s result n characters, althugh it ges in a different directin. 6
Therem 2 The degree f L ver the fixed field L G is bunded by the rder f G. That is [L : L G ] apple G Prf: Assume that a 1,..., m are elements in L and that m> G. Wehavet shw that they are linearly dependent ver K. Nw, as m> G, thereisann-trivial slutin ( 1,..., m ) in L f the system f equatins where i s are the elements in G 1( 1 )x 1 + + 1 ( m )x m =0... n( 1 )x 1 + + n ( m )x m =0, indeed, the number f equatins being less than the number f unknwns. Since 1 2 G is amng the s, any slutin =( 1,..., m ) f the system gives the linear relatin 1 1 + + m m =0 where the i s are in L. Our task is t find ne slutin with i s all in L G. The set f slutins f the system is invariant under the actin f G. Applyingany 2 G t the equatins, and using that runs thrugh G when des, we see that if =( 1,..., m ) is any slutin, ( ) =( ( 1 ),..., ( m )) is als a slutin. Nw, we pick a slutin with the greatest number f i s different frm zer. By renumbering and scaling we may assume that 1 2 K. Assumethat k /2 L G,thatis fr ne f the elements 2 G we have ( k ) 6= k. Then ( ) =( ( 1 ),..., k ( k ),...)=(0,..., k ( k ),...) is a nn-trivial slutin with fewer nn-zer crdinates than. Cntradictin. Cmbining with xxx we get Therem 3 Let G be a finite grup acting n the field L. ThentheextensinL f L G is Galis and [L : L G ]= G The fllwing therem, which n may call the first Main therem in Galis Therem fllws frm this cmbined with xxx. Therem 4 (The first main therem) Let K L be an extensin. Then the fllwing are equivalent K is the fixed field f the Galis grup, that is K = L Gal(L/K) The rder f the Galis grup equals the degree, that is Gal(L/K) =[L : K] The extensin K L is separable and nrmal. 7
The secnd main therem f Galis thery Of the mre acclaimed therems in algebra is the fundamental therems f Galis thery. Their histry is lng and glrius ging back t Galis, with its dramatic twists and nn-mathematical ingredient: lve stries, duels and early death. It is an very imprtant and deep therem, and tuches the grundwater f mathematics. It illustrates very well the deep analgy between algebra and gemetry uncvered by frmalizatin and abstractin. The classificatin f cvering spaces f a tplgical spaces by subgrups f the fundamental grup, is strikingly similar and indeed the tw therems are just different faces f the same principle. The setting f The fundamental therem is Galis extensin K L. It states that there is a ne-ne-crrespndence between n the ne hand subgrups f the Galis grup Gal(L/K) and n the ther hand intermediate fields E lying between K and L, that is a three-stry twer K E L. Given a subgrup H G, theassciatedintermediatefieldisthe fixed field f H. That is L H = { x 2 L (x) =x fr all 2 H }. Given an intermediate field E, thecrrespndingsubgrupfthegalisgrupisthe Galis-grup Gal(L/E) which bviusly is cntained in Gal(L/k). Itcnsistsfthe elements f Gal(L/K) nt mving any member f E, infrmula-ling: Gal(L/E) ={ 2 Gal(L/K) (x) =x fr all x 2 E } Clearly, by definitin, there are inclusins E L Gal(L/E) H Gal(L/L H ) and the cntent f the Fundamental Therem is that these tw inclusins are equalities: Therem 5 Let K L be a Galis extensin (i.e., separable and nrmal) then E = L Gal(L/E) H =Gal(L/L H ) Thus the assignment H 7! L H and E 7! Gal(L/E) set up mutually inverse maps that ne may depict as fllws: { Subgrups f Gal(L/K) }!{Intermediate fields E, i.e., twers K E L } Prf: We first attack the inclusin E L Gal(L/)E.Bylemma1n page 4,theextensin E L is nrmal and separable, and hence by prpsitin 5 n page 4 we have [L : E] = Gal(L/ ) E.Onthetherhand,Artin stheremgivesus[l : L Gal(L/E) ]= Gal(L/K), and therefre [L : L Gal(L/E) ]=[L: E], andthetwfieldse and L Gal(L/E) are equal. Nw, the secnd equality H Gal(L/L H ) is just therem 4. 8
Inclusins are reversed The crrespndence in the main therem has several nice prperties, it reverses inclusins and preserves indices. If H and H 0 are tw subgrups f Gal(L/K) the it hlds true that H H 0 is and nly if L H0 L H. [H 0 : H] =[L H : L H0 ]. Cnjugate subgrups and nrmal subgrups Let 2 Gal(L/K) be an element and let H be a subgrup. The cnjugate subgrup H 1 has as fixed field the image (L H ) f the fixed field f H.Indeed,frany 2 H,if 1 (x) =x,clearly 1 (x) = 1 (x). Hence L H 1 = L H This bservatin leads the the fllwing therem, that we might cnsider as the third in the rw f main therems. It states that in the Galis crrespndence nrmal subgrups crrespnd t sub fields nrmal and separable ver K. Separabilityhldsfrevery intermediate field, s the real cntent f this statement is that nrmal subgrups and nrmal field extensins crrespnd. Therem 6 Assume that K L is a Galis extensin and that H Gal(L/K) is a subgrup. The fixed field L H is Galis ver K if and nly if H is a nrmal subgrup. In that case the Galis grup Gal(L H /K) is naturally ismrphic t the qutient Gal(L/K)/H. Prf: Since Gal(L/L H 1 )= H 1 and Gal(L/L H )=H the subgrup H and its cnjugate H 1 are equal if (and nly if) the crrespnding fixed fields are. That is, if and nly (L H )=L H.FrmthiswecncludethatH being a nrmal subgrup is equivalent t the fixed field L H being invariant under the actin f the whle Galis grup Gal(L/K). Assume that L H is invariant. The subgrups H acts by definitin trivially n L H which implies that the qutient Gal(L/K)/H acts n L H and since Gal(L/L H )=H, this actin is faithful. It fllws that Gal(L/K)/H is cntained in Gal(L H /K), butthe tw are f the same rder, and therefre equal. Indeed Gal(L H /K) apple[l H,K]= [L : K] [L : L H ] = Gal(L/K) Gal(L/L H ) = Gal(L/K)/H. This shws as well that L H is galis ver K, since Gal(L H /K) =[L H : L]. Versjn: Saturday, September 14, 2013 2:50:36 PM 9