Chapter 1 Field Extensions - University of Washington

1 Chapter 1 Field ExtensionsThroughout this chapterkdenotes a Field andKan extension Field Splitting FieldsDefinition polynomialsplits overkif it is a product of linear polynomialsink[x]. Let :k Kbe a homomorphism between two fields. There is a uniqueextension of to a ring homomorphismk[x] K[x] that we also denote by ;explicitly, (n i=0 ixi)=n i=0 ( i) it makes sense to ask if a polynomial ink[x] has a zero inK. Similarly,it makes sense to ask if a polynomial ink[x] splits inK[x].Definition k[x] be a polynomial of degree 1. An extensionK/kiscalled asplitting fieldforfoverkiffsplits overKand ifLis an intermediatefield, sayk L K, andfsplits inL[x], thenL=K.

2 The second condition in the definition could be replaced by the requirementthatK=k( 1,.., n) where 1,.., nare the zeroes main result in this section is the existence and uniqueness up to isomor-phism of splitting L K, andKis a splitting Field forf k[x], thenKis also a splitting Field forfoverL. The converse is false as one sees by takingf=x2+ 1 andk=Q L=R K= a splitting Field forfoverk. IfFis an extension ofKand isa zero offinF, then K. To see this, writef= (x 1)..(x n) with kand 1,.., n K, and observe that 0 =f( ) = ( 1)..( n),so = ifor 1.

3 Field a splitting Field forfoverk, and let be a zero (x )gfor someg K[x]. Becausefsplits inK, so doesg. HenceKis a splitting Field forgoverk( ).Theorem a Field andf k[x]. Thenfhas a splitting Field , sayK/k, and[K:k] (degf)!. onn= degf. If degf= 1, thenf= x+ with , k,sofalready splits ink, so we can takeK= thatn >1. Iffis already split we may takeK=k, so we mayassume thatfhas an ireducible factor, sayg, of degree 2. By Proposition??.??,ghas a zero in the extension fieldk( ) =k[x]/(g); the degree of thisextension is degg n. Now writef= (x )hwhereh k( )[x].

4 Sincedegh=n 1, the induction hypothesis says there is an extensionL/k( ) overwhichhsplits, and [L:k( )] (n 1)!. Certainlyfalso splits overL, and[L:k] = [L:k( )][k( ) :k] n!. If 1,.., nare the zeroes offinL, thenk( 1,.., n) is a splitting Field forfoverk. The proof of Theorem involves a choice of an irreducible factor off. It isconceivable that choosing a different factor might produce a different splittingfield. Before showing that is not the case, and hence that a splitting Field isunique up to isomorphism, we need the following :k k be an isomorphism of fields.

5 Letf k[x]beirreducible. If is zero offin some extension ofkand is an extension of (f)in some extension ofk , then there is an isomorphism :k( ) k ( )such that |k= idkand ( ) = . map extends to an isomorphismk[x] k [x] and sends (f) to( (f)), so induces an isomorphism between the quotient rings by these composition of the obvious isomorphismsk( ) k[x]/(f) k [x]/( (f)) k ( )is the desired isomorphism. Theorem a Field andf k[x]. Let :k k be an isomorphismof fields. LetK/kbe a splitting Field forf, and letK /k be an extension suchthat (f)splits inK.

6 Then1. there is a homomorphism :K K such that |k= ;2. ifK is a splitting Field for (f)overk , thenK = (1) We argue by induction on [K:k]. Writef= a product ofirreduciblespi k[x]. Then (f) = (p1).. (pn), and each (pi) is irreducibleink [x].If [K:k] = 1, thenK=k, and we can take = . NORMAL EXTENSIONS3 Suppose that [K:k]>1. Then somepi, sayp1, is not linear. Let Kbe a zero ofp1, and let K be a zero of (p1). By Lemma , there is anisomorphism :k( ) k ( ) such that |k= .NowKis a splitting Field forfoverk( ), and (f) splits overk ( ). Since[K:k( )]<[K:k] we can apply the induction hypothesis to obtain :K K such that |k( )=.

7 Hence(2) Certainly is injective, so it remains to show it is surjective. However,iff= ( x 1)..(x n) then (f) = ( x ( 1))..(x ( n)); sinceKandK are splitting fieldsK=k( 1,.., n) andK =k ( ( 1),.., ( n)), is also surjective, and hence an isomorphism. Theorem polynomial of positive degree has a unique splitting Field up Normal extensionsDefinition finite extensionK/kisnormalif every irreducible polynomialink[x] that has a zero inKactually splits overK. Theorem extensionK/kis normal if and only if it is the splitting fieldof a ( ) WriteK=k( 1.)

8 , n). The minimal polynomialpiof ihas azero inKso splits inK. Hencef=p1 pnsplits inK. ButKis generatedby the zeroes off, soKis the splitting Field offoverk.( ) SupposeK=k( 1,.., n) is the splitting Field of a degreenpolynomialg kwhere 1,.., nare the zeroes ofg. Letf k[x] be irreducible andsuppose that Kis a zero off. We must show thatfsplits offg K[x], and letLbe the splitting Field forfgoverK. Writef= (x )(x )..(x ) where , .., L. Since and are zeroes off, there is an isomorphism :k( ) k( ) such that ( ) = and |k= idk(Lemma ).

9 Now think ofg k( )[x] andg= (g) k( )[x]. NowKis a splittingfield forgoverk( ) and (g) splits inK( ) so, by Theorem , there is a map :K K( ) such that |k( )= . Hence |k= |k= idkand (g) =g. Now0 = (g( i)) =g( ( i)) so ( i) belongs toK. Hence (K) K. In particular, = ( ) K; the same argument shows thatallthe zeroes offbelong toK. The next result shows that a finite extensionK/kcan be embedded in aunique smallest normal extensionL/k. The extensionL/kis called a finite extension. Then there is a finite extensionL/Ksuch that4 Chapter 1. Field normal overk, and2.

10 IfK F LandFis normal overk, thenF=L, and3. ifL /Kis a finite extension such thatL satisfies (1) and (2), then thereis aK-isomorphism :L L .Proof.(1) WriteK=k( 1,.., n), letpi k[x] be the minimal polynomialof i, setp=p1p2 pn, and letLbe the splitting Field ofpoverK. Thenk K L,Lis the splitting Field forpoverk, and [L:k]< . By ,Lis normal overk.(2) IfK F LandFis normal overk, then eachpisplits inFbecauseit is irreducible and has a zero inF, whencepsplits inF, soF=L.(3) IfL /Kis a finite extension such thatL satisfies (1) and (2), theneachpisplits inL , so by Theorem there is a map :L L such that |K= idK.