Separability. Version september 2013 klokken 10:28. Separability MAT4250 Høst 2013

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1 Separability Versin september 2013 klkken 10:28 Multiple rts and separable plynmials Let as usual K be a field, and let a plynmial f 2 K[t] f degree m is given. It is cnvenient and we lse nthing t suppse that f is mnic. Let be a field cntaining K such that! 2 is a rt f f. We may then write f(t) =(x!) r g(x) where g(x) 2 [t] is a plynmial nt vanishing at! and r 1. The rt is asimple rt if r =1,andinthecntrarycase,i.e., if r 2, wecall! amultiplert.one has the well knwn and elementary, but usfull lemma Lemma 1 The rt! is multiple if and nly if f 0 (!) =0. Prf: The prduct rule fr the derivative gives f 0 (x) =r(x!) r 1 g(x)+(x!) r g 0 (x), which shws that f 0 (!) =0if and nly if r 2. If the plynmial f splits cmpletely ver, thereisafactrizatin f(x) =(x! 1 ) (x! m ) with all the rts! i in. Whentherts! i are distinct, we say f splits simply ver and call f(x) a separable plynmial. Ifnt,thatis,ifatleastnertccursat least twice, the plynmial is said t be inseparable. Prblem 1. Verify that if f splits simply in ne extensin f K it splits simply in any ther extensin 0 where it splits. Hint: Chse a cmmn algebraic clsure f and 0. It is f curse an easy matter t find plynmials with multiple rts, but t find inseparable, irreducible plynmial is a lt mre subtle 1. As the prpsitin belw will shw, ne has t search in psitive characteristic t find such species. Amng the many new and cntra-intuitive phenmenn that can happen in psitive characteristic, is that the derivative f a plynmial may vanishes identically withut the plynmial being cnstant. The simplest example f a plynmial with this behavir is t p a. Mre generally, if f(x) =g(x pm ),byapplyingthechainrule,weseethat f 0 (x) =p m x pm 1 g 0 (x pm )=0 since the characteristic f K is p. Nw, these examples are in fact the nly examples. We have: 1

2 Lemma 2 Let K be a field f characteristic p and assume f(x) is a plynmial in K[x] whse derivative vanishes identically. Then f(x) =g(x pn ) fr sme plynmial g(x) in K[x] whse derivative des nt vanish identically. PlyWithVanishDe Prf: Let f(t) = P 0appleiappler a ix i.sincethederivativef 0 (x) = P 0appleiappler ia it i 1 vanishes identically, ia i = 0 fr 0 apple i apple r. This implies that either a i = 0 r i = 0,the latter means that i has p as factr. Hence the nly nn-vanishing terms f f are thse whse degree is divisible by p, and we may write f(x) = P 0applejappler/p a pjt pj.or,phrased differently, f(x) =g(x p ) fr sme plynmial g. Ifg has a nn-vanishing derivative, we are happy, if nt, we repeat the prcess until happiness. Prpsitin 1 Let K be a field and f(x) 2 K[x] an irreducible plynmial. Then f(x) is inseparable if and nly if f 0 (x) vanishes identically, i.e., f 0 (x) =0as a plynmial. In particular, f(x) is separable if ne f the fllwing cnditins is fulfilled: The characteristic f K is zer. The degree f f(x) is prime t the characteristic f K. Prf: Let E be the stem field f f(x), thatise = K[x]/(f(x)). Then f has a rt in E, andf(x) is the minimal plynmial t. The degree f f 0 (x) being ne less than he degree f f(x), it fllws that if f 0 ( ) =0,thenf 0 (x) =0identically. In characteristic zer this can f curse nt nt happen since f(x) is nt cnstant, and lemma 2 shws that in case it happens in psitive characteristic, the degree f f must be divisible by the characteristic f K. Prblem 2. Assume that f(x) is an inseparable, irreducible plynmial. Shw that ne may write f(x) =g(x pn ) where g(x) is separable and irreducible. Cnclude that all rts f f have the same multiplicity p n. Prblem 3. Let K be a field f characteristic p and let a 2 K be an element. Shw that x pn a is irreducible ver K if and nly if a is nt p n -th pwer in K. Hint: Use that the Frbenius map x 7! x pn is additive t shw that if a pn = b pn in a field f characteristic p, thena = b. Henceallthertsff in a splitting field are equal. Separable field extensins Recall that giving an embedding ver K f the stem field E f f an irreducible plynmial f(x) 2 K[x] int a field, isequivalenttgivingartff in. Assume that is a splitting field fr f. Frseparableplynmials thatis,plynmials that have as many different rts in as the degree n indicates this means that there are exactly n emddings int, andfrtheinseparableplynmialsthere are less. Hence 2

3 StemFieldSeparable Prpsitin 2 Suppse that K be a field and let f(x) 2 K[x] be an irreducible plynmial f degree n. Let be a splitting field fr f. Thenf is separable if and nly there are n different embeddings ver K f the stem field E f int. A small, but useful, extensin f this result is t the case when K is nt cntained in, but just embedded. That is, there is given an embedding : K!. Then the result abve reads as f(x) is separable if and nly if there are exactly n extensins f t L. The prf is nt difficult and left as an exercise. Prpsitin 2 abve may be generalized t fields nt necessarily being stem fields (but in the end f the day they will be, due t what is called the Therem f the primitive element ). Assume that L is a field extensin f K. Anelement 2 L is said t be a separable element if the minimal plynmial is separable. The extensin is called a separable extensin if any element in L is separable. All this shuld be understd relatively t the field K. Hwever,nehasthethe lemma Lemma 3 Assume that K E L is a twer f finite field extensins. If an element f L is separable ver K, itisseparablevere. Prf: Let f(x) be the minimal plynmial f ver K. Then f curse f(x) 2 E[x], but it is nt necessarily irreducible there. We may therefre factr f(x) =g(x)h(x) ver E, where g(x) is the minimal plynmial f ver E and where h( ) 6= 0.Applying the prduct rule, we find f 0 (x) =g 0 (x)h(x)+g(x)h 0 (x). Hence if f 0 ( ) 6= 0,theng 0 ( ) 6= 0(since h( ) 6= 0and g( ) =0). SepInTwers Prblem 4. Shw by a (stupid) example that the cnverse f the lemma is nt true Hint: Take E = L. Prpsitin 3 Let L be a finite extensin f the field K. Assumethat is an algebraic clsed field and that is an embedding f K in. ThenL is a separable extensin if and nly if the number f different embeddings f L in extending equals the degree [L: K] f L ver K. Prf: Pick an element 2 L. IfL = K( ) we are dne, since then L is ismrphic t the stem field E f and is separable ver K.Ifnt,wehavethetwerK K( ) L. The number f embeddings f K( ) in equals the degree [K( ): K], andbyinductin 2, each f these can be extended t L in [L: K( )] ways since L is separable ver K by lemma 3, andwearethrughsincethedegreeismultiplicativeintwers. 1 Just fr this reasn, many authrs apply the terms separable and inseparable nly t irreducible plynmials. 2 In fact, we need t prve a slightly mre general statement t make the inductin wrk: Any embedding f K int can be extended t L in exactly [L: K] ways. 3

4 Crllary 1 In a twer K E L f field extensins, L is separable ver K if and nly if bth E is separable ver K and L is separable ver L. Prf: We cunt extensins f embedding f the fields int sme algebraically clsed field : AssumeL separable ver E and E separable ver K. Then by the prpsitin, an embedding f K int can be extended t L in [E : L] different ways, and each f these can be further extended t L in [L: E] ways. Hence the ttal number f extensins t L equals [E : K][L: E], andthisisequalt[l: K]. The ther way rund is just lemma 3 abve. MinRelInSep Perfect fields A perfect field is a field K having the prperty that any extensin L f K is separable. All fields f characteristic zer are perfect, and the gd news fr number therists is that all finite fields are perfect. One has the fllwing characterisatin f perfect fields relating t existence f p- rts in K: Prpsitin 4 Assume that K is a field f characteristic p. ThenK is perfect if and nly if every ne f its elements is a p-th pwer, i.e., every element in K has a p-th rt in K. Prf: Assume that every element has a p-th rt, and suppse that L is an inseparable extensin f K. Then there is an element 2 L, ntink, satisfyingaminimal relatin nq + a n 1 (n 1)q +...a 1 q + a 0 =0 (K) where q = p m,andthea i s are in K. SinceeveryelementinK has a p-rt, they all have a q-th rt as well, and we may find b i s in K with b q i = a i. The relatin then takes the frm nq + b q n 1 (n 1)q +...b q 1 q + b q 0 =0, but, as rising elements t q-th pwer is additive in characteristic p, thisgives ( n + b n 1 n b 1 + b 0 ) q =0, cntradicting the minimality f the relatin K. The ther way arund, if a 2 K is nt a p-th pwer, the plynmial x p a is irreducible and the extensin E f = K[x]/(x p a) f K is inseparable. Finite fields are perfect The fields f finite characteristic we shall meet in this curse, are are almst exclusively the finite fields. They are all perfect, s we very seldm will have t cpe with inseparable fields extensins hwever w ll meet inseparable algebras, but that is anther stry we cme back t later. T see that a finite field F q where q = p n is a prime pwer is separable. We check that every element is a p-th pwer. The Frbenius map x 7! x p is additive (this is true in characteristic p) andinjective x p =0bviusly implies that x =0.Henceitmust be bijective since F q is finite. 4

5 All this fuzz abut nthing? The reader deserves at least ne example f a extensin that is nt separable. The easiest is prbably the fllwing. Let K be any field f characteristic p and let x be a variable. Then K(x p ) K(x) is nt separable, since x p is nt a p-th pwer in K(x p ) (any p-th pwer in K(x p ) is f degree p 2 in K(x)). Multiple rts and the discriminant We knw since childhd (at least mathematically speaking) that a quadratic plynmial ax 2 + bx + c has a duble rt if and nly if the discriminant b 2 4ac vanishes. Indeed, the discriminant is the square f the difference f the tw rts. The discriminant is a very cnvenient tl; withut actually knwing the rts, ne can decide if they are equal r nt. In general, if the mnic plynmial f(x) f degree n has the n rts! 1,...,! n in splitting field, nedefinesthediscriminantff as the prduct Y (! i! j ) 2, f = 1applei<japplen If f(x) is nt mnic, but has leading cefficient a, welet Y f = a 2n 2 (! i! j ) 2. 1applei<japplen Obviusly j vanishes precisely when tw rts cincide. The discriminant is invariant under any permutatin f the rts, and thus it is a symmetric plynmial in the! i s. Hence by the fundamental therem n symmetric plynmials, it may be expressed as aplynmialinthecefficientsff, justlikeb 2 4ac in the quadratic case, but the actual expressin is much mre cmplicated fr ply s f higher degree. In fact the frmula fr the discriminat is what we call a universal frmula. It is the same whatever the! i s. One can, if ne wants t be 100% frmal, frmulate it in terms f independent indeterminants w 1,...,w n and prve it ver the ring Z[w 1,...,w n ]. There are many frmulas invlving such a nice creature as the discriminant. We shall give tw, ne where we relate the discriminant t the values f the derivative f f at the rts, and ne where it is related t the van der Mnde determinant. Discriminant and derivatives Since f(x) = Q j (x! j),neeasilyseesbyapplying the prduct rule, that f 0 (! i )= Q j, j6=i (! i! j ).Hence Y f 0 (! i )= Y! Y (! i! j ) = Y (! i! j ) 2 i i i<j j, j6=i where is a sign. After a mment f reflectin ne cnvinces ne self that = ( 1) n(n 1)/2 every ne f the n(n 1)/2 pairs f indices i, j cntributes twice t the prduct, nce with the factr! i! j and nce with! j! i.hence f =( 1) n(n 1)/2 Y i f 0 (! i ). (H) 5

6 In case f(x) is nt mnic, but has leading cefficient a, we find DiskPrduktAvDer f =( 1) n(n 1)/2 a n 2 Y i f 0 (! i ). This relatin between the discriminant and the derivative f f has nice frmulatin in terms f the nrm. Recall that if K L is a separable extensin, the nrm N L/K ( ) f equals the prduct Q i i( ) f the value at f the different embeddings i f L in sme sufficiently big field. Let be a splitting field fr f and let! be ne f the rts. Then if L = K(!), thevalues i(!) are just the different rts! i f f in and i(f 0 (!)) = f 0 (! i ). Cmbining this with equatin (H) abve,webtain Prpsitin 5 Let f(x) be a mnic plynmial in K[x] and let! 1,...,! n be the rts f f in sme splitting field. Then f =( 1) n(n 1)/2 N K(!)/K (f 0 (!)). The case f the trinme x n + ax + b Cmputing discriminants by hand can be achallenge,butluckily,tdaynehassftwarethatcancandthedirtyjbsfrus. Hwever, in the special case f the the trinmes x n + ax + b, the cmputatins with sme cleverness are nt hrrible at all, and is a classic. S we shall d it. This furnishes us we many examples, and it reminds us that cumputatin by hand smetimes is an ptin, and prbably mst imprtantly, yu have a chance t becme wiser by ding the calculatuns yur self and nt nly pushing the buttn. S let a and b be elements f the field K and let f(x) =x n + ax + b. Let! be a rt f f. The derivative is given as f 0 (!) =n! n 1 + a. Nw! n 1 = a b! 1,s renaming the derivative x and using x = (n 1)a nb! 1,wecanslvefr! and btain nb! = x +(n 1)a. This shws that nb f( x +(n 1)a )=0. Clearing the denminatr, we get (x +(n 1)a) n na(x +(n 1)a) n 1 +( 1) n b n 1 n n =0. Interpreting x as a variable, the left side is the minimal plynmial f f 0 (!), andthe nrm f f 0 (!), which we are lking fr, is the cnstant term with the factr ( 1) n. The binmial therem and sme cleaning up gives f =( 1) n(n 1)/2 (n n b n 1 +( 1) n 1 (n 1) n 1 a n ) Specializing t case f a cubic x 3 + ax + b, negetstheubiquiusfrmulafrthe cubic discrimant f = 27b 2 4a 3 6

7 Prblem 5. Cmpute the discriminant f the tw cubic plynmials x 3 + x +1and x 3 x 1. Prblem 6. Let g(x) =x 3 + ax 2 + b shw that g = 27b 2 4a 3 b. Hint: Relate g t the discriminat f f(x) =x 3 g(x 1 ) Prblem 7. Cmpute the discriminat f x 3 2x Prblem 8. The aim f this exercise is t shw that if x 3 + ax + b is cubic plnmial whse discriminant is negative, then it has a unique real rt given by! = 3 q ( b + /3 p 3)/2+ 3 q( b /3 p 3)/2 (J) where is a square rt f the discrimant 4a 3 27b 2. RealRt a) Let be a primitive third rt f unity. Shw that there are real numbers s and t such that s + t and s + t are the tw cmplex rts f f(x). b) Shw that the real rt equals s + t. Hint: The sum f the rts is zer. c) Shw that b = s 3 + t 3. Hint: The prduct f the rts is b. d) Shw that the prduct f the differences f the rts (i.e., )is±3 p 3(s 3 t 3 ). e) Shw the frmula (J). Hint: Slve fr s and t. Prblem 9. If f(x) =ax 3 +bx 2 +cx+d, shwthat f = b 2 c 2 4ac 3 4b 3 d 27a 2 d abcd. The Van der Mnde determinant T treat the van der Mnde determinant we d it generically, as we did with the discriminant. We use n independent variables w 1,...,w n be variables, s the frmulas we btain are valid in the plynmial ring Z[w 1,...,w n ].Subsequentlynemaysubstitutefrthew i whatever elements ne wants, as lng as they are elements f a cmmutative ring, hence the frmulas are universally valid. V n = V n (w 1,...,w n )= w 1 w w n w1 2 w wn w1 n 1 w1 n wn n 1 Setting w i = w j kills the determinant tw clumns then being equal. This means that w i w j divides V (and this is where we use the ring Z[w 1,...,w n ] and that it is a ufd ). It fllws that the prduct P n = Q i>j (w i w j ) is a factr f V. 7

8 This prduct and the van der Mnde are bth plynmials f the same degree. Indeed, every term f the van der Mnde is f degree P n 1 i=0 i = n(n +1)/2, andthe prduct has as many (linear) factrs as there are pairs with i>j,thatisn(n +1)/2. Hence the tw expressins agree up t a scalar factr, say n.settingw n =0and develping V n alng the last clumn gives V n wn=0 =( 1) n 1 w 1 w n 1 V n 1 and similarly P n wn=0 =( 1) n 1 w 1 w n 1 V n 1.Hence n = n 1,andfcurse 2 =1, s the scalars n all reduce t 1. We have established the fllwing: Prpsitin 6 Assume that f(t) is a plynmial f degree n whse rts are! 1,...,! n in sme extensin field f K. Thenwehavethefllwingrelatinbetweenhediscriminant f f and the van der Mnde determinant made ut f the rts: f = V 2 n (! 1,...,! n ) Nilptency and trace There is a clse cnnectin between nilptency f a linear peratr A n a vectr space and the tracelessness f all the pwers f A. Infact,verafieldfcharacteristiczer,A being nilptent is equivalent t the vanishing f the traces tr A i fr i 1. Ifthegrund field is f psitive characteristic, hwever, the situatin is a little mre cmplicated due t the fact that there are nn-cnstant plynmials with a vanishing derivative. Even if ur main bjects f study rings f algebraic integers in number fields all live in characteristic zer, we have a great interest in cnsidering the case f psitive characteristic. The reasn is as fllws. If A B is an extensin f say number rings fr example culd B be the integtral clsure f A in sme finite field extensin f the fractin fielsd f A and p A aprimeideal,wearecertainlyinterestedingettinga hand n the set f prime ideals q in B with q \ A = p that is, the fibre f the map spec B! spec A ver p. This fibre is in ne-t-ne-crrespndence with the primes in the ring B/pB, andthisringisanalgebraverthefinitefielda/p. The lessn is: We are als interested in finite dimensinal algebras ver finite fields! Example. An illustrative example is when B is generated by ne element ver A; t be cncrete, say A = Z and B = Z[ ] where belngs t sme finite extensin L f Q, and is integral ver Z. Then B = Z[x]/(F (x)) where F (x) is the minimal plynmial f ver Q, andthisplynmialhasintegralcefficients, being integral. If nw p 2 Z is a ratinal prime, we can perfrm the fllwing small cmputatin B/pB = Z[x]/(F (x),p)=f p [x]/(f(x)), where f(x) 2 F p [x] dentes the plynmial btained frm F by reducing all cefficients md p. Ofcurse,f(x) is nt necessarily irreducible (even if F is). Sme times it is and sme times nt. If f(x) =g 1 (x) v1 g r (x) vr is a factrizatin f f where the g i (x) s are different irreducible ply s in F p [x], then by sme well knwn chinese therem B/pB = F p [x]/(g 1 (x) v1 g r (x) vr )= Y F p [x]/(g i (x) v i ) 8

9 Each factr F p [x]/(g i (x) v i ) is an algebra ver F p,anditisafield if and nly if v i =1, that is, if and nly if it is withut nilptent elements. S we see that nilptency is cuple t multiple factrs f f, rincasethefactrsarelinear,tmultiplertsf f. Ply s with multiple rts in sme extensin f the base field, are the inseparable ply s, sthelessnis:nilptencyiscupledtinseparability. e Prblem 10. Shw that x 2 +1is irreducible md p if and nly if p 1 md 4. Linear peratrs Cming back t linear peratrs, we fix a field K and a linear peratr A n the finite dimensinal vectr space V ver K. The characteristic plynmial f A is, as we knw, defined by P A (t) = det(t I A). OversmeextensinL f K the characteristic plynmial splits int a prduct P (t) = Q i (t i) f linear factrs. The i s are the eigenvalues f A in. In additin t the characteristic plynmial we shall make use f the fllwing assciated plynmial: Q A (t) = Y (1 it) =t n P A (t 1 ) It still has cefficients in K. The degree f Q A may drp cmpared t the degree f P A.Itisequaltthenumberfnn-zereigenvaluesfA. Infact,thertsfQ are the inverses f the nn-vanishing eigenvalues f A. A cnsequence f Cayley-Hamiltn is that the peratr A is nilptent if and nly if Q A is cnstant. Our next lemma is cmpletely elementary, n the level f a first curse in calculus, but will be very usual: Lemma 4 Assume that Q(t) = Q 1appleiapplen (1 it) is a plynmial where the i s are in sme cmmutative ring A. Thenthefrmulabelwhldstrueinthepwerseriesring A[[t]]: Q 0 (t) Q(t) = 1appleiapplen,kapple1 k i t k 1 The identity in the lemma is cmpletely frmal. That is, we may assume that i s are independent variable, and the identity hlds in the ring f frmal pwer series 3 Z[ 1,..., n][[t]]. It is therefre valid whenever the tw sides have a meaning. Prf: The prduct rule gives Q(t) 0 = P i ( i) Q j6=i (1 jt). Then FundFrmel Q 0 (t) Q(t) = i i 1 t i and the lemma fllws frm the the ubiquitus frmula fr the sum f a gemetric series. 3 Admittedly, this frmulatin is nt very much first-calculus-curse-ish. 9

10 Lemma 5 If A is an peratr n a finite dimensinal vectr space V f dimensin n ver the field K, then Q 0 A (t) Q A (t) = tr A k t k 1 1applek hlds in the pwer series ring K[[t]], whereq A (t) =t n P A (t 1 ) and P A (t) dentes the characteristic plynmial f A. Prf: Q If the i s with 1 apple i apple n are the eigenvalues f A, thennehasq A (t) = i (1 it) and tr A k = P k i i. Prpsitin 7 Fr the peratr A, thetwfllwingcnditinsareequivalent Fr all k apple 1, thetracefa k vanishes, i.e., tr A k =0. Q A (t) 0 =0 The trace frm and separability We have nw cme t main therem f this sectin. Let as usual K be a field and L afiniteextensinfk. The trace gives us a bilinear and symmetric frm n L with values in K,namelythefrm(x, y) =tr L/K (xy) fr x, y 2 L. Indeedtr L/K (xy) is clearly symmetric, and the prduct being linear in each variable, tr L/K (xy) is bilinear. This frm is very useful since it detects inseparability; mre precisely the trace frm is nn-degenerate if and nly if the extensin L f K is separable: TraceFrmNndegerate Therem 1 Let K L be a finite separable field extensin. Then the bilinear frm tr L/K (xy) is nn degenerate if and nly if the extensin L f K is separable. Prf: Assume first that L is separable ver K. We use inductin n n =[L: K], andletl be a cunter example f least degree ver K. The field L being a cunter example, there is an element x 2 L nt in K with tr L/K (xy) =0fr all y. HenceL = K(x) since L was the smallest bad guy; indeed K(x) is separable ver K, and,fcurse,tr K/L xy =0fr all y 2 K(x). By putting y = x k 1 we see that tr L/K (x k )=0fr all k. Since x n = a 0 + a 1 x + + a n 1 x n 1 with the a i s in K, anda 0 6=0,thisgivestr a 0 = na 0 =0,andn must be divisible by the characteristic p f K. Let nw Q(t) =t n P (t 1 ) where P (t) dentes the characteristic plynmial f the multiplicatin map x. Then by lemma 4 we have Q 0 (t) Q(t) = tr L/K (x k )t k 1 =0, k 1 10

11 and Q 0 vanishes identically. Nw 0=Q(t) 0 = nt n 1 P (t 1 )+t n 2 P 0 (t 1 )=t n 2 P 0 (t 1 ) as p divides n. HenceP 0 vanishes identically. But the characteristic and the minimal plynmial f x cincide (since L = K(x)), and by cnsequence x is nt separable ver K. Fr the prf f the implicatin in the ther directin, assume that L is nt separable, and let x 2 L be inseparable ver K. WefirstlkattheextensinK(x) generated by x. If P (t) is the characteristic plynmial (which is equal t the minimal plynmial f x ver K)fmultiplicatinbyx in K(x),thenP 0 =0.HenceQ 0 =0, and by what we just did, tr K(x)/x (y) =0fr all y 2 K(x). It fllws that tr L/K =tr K(x)/L tr L/K(x) =0. Just ne remark abut this prf. Mst f it is there t deal with the characteristic p case. In characteristic zer, ne cncludes immediately frm the fact that tr L/K (x k )= fr all k 1, thatq 0 =0,henceQ(t) cnstant, i.e., that x is nilptent, which means that x =0since K is a field. And thus the trace frm is nn-degenerate. The dual basis and finiteness f the integral clsure Assume that v 1,...,v n is a K basis fr the sperable extensin L f K. Then by the general thery f nn-degenerate quadratic frms, there exsists a dual basis. This is a basis v 0 1,...,v 0 n such that tr L/K (v i v 0 j)= ij. The prf f the fllwing result illustrates the usefulness f the dual basis. The result is really the basis fr the whle algebraic number thery. It is fundamental that the integral clsure f a Dedekind ring in a finite extensin f the field f fractin, still is Dedekind. The main ingredient in the prf f this result, and the nly subtle ne, is that the integral clsure remains netherian. This fllws frm ur next prpsitin. Prpsitin 8 Let A be netherian dmain, integrally and clsed in its fields f fractins K. AssumethatL is a finite extensin f K such that the bilinear frm tr L/K (xy) is nn-degenerate. Then integral clsure B f A in L is a finitely generated mdule ver A. Prf: Let v 1,...,v n be a basis fr L ver K. Wemayassumethatthev i s are cntained in B (multipliedbyacmmndenminatr,theystillcnstituteabasis). Let v1,...,v 0 n 0 be the dual basis with respect t the bilinear frm tr L/K (xy). Then we claim that B is cntained in the A-submdule f L generated by the v i s. This submdule is by definitin finitely generated, and A being ntherian, it fllws that B is finitely generated. Since v1,...,v 0 n 0 is a K-basis fr L,any 2 B may be written as = a 1 v1+ +a 0 n vn 0 with a i 2 K. Ourtaskistverifythatthea i s are in A. Nw tr L/K ( v j )= a i tr L/K (v 0 iv j )=a j 11

12 and since bth and v i are in B their prduct is, and hence tr L/K ( v i ) 2 A by prpsitin 7 n 6 in the sectin Nrms and Traces.. Therem 2 Let A K be a dedekind ring and its field f fractins. Let L be a finite, separable extensin f K and dente by B the integral clsure f A in B. ThenB is a Dedekind ring. Prf: We knw that B is integrally clsed and finitely generated as an A-mdule. Hence it is netherian and f the same Krull-dimensin as A (It is generally true that if B is an A-algebra which is finitely generated as an A-mdule, then if ne f A and B is nteherian, the ther is, and they have the same Krull-dimensin ). Withut the hypthesis that L be separable ver K, thedmainb need nt be finite ver A, but it will be netherian and f Krull dimensin ne. This is called the Krull- Akizuki-therem, and we shall nt prve it. S, the therem remains true even if we skip the assumptin that L is separable ver K. Etale r separable algebras One f the main prblems in algebraic number thery is t describe hw a ratinal prime p factrizes in the ring A f integers in an algebraic number field K. Aspart f this, it is f great imprtance t decide which primes p have a multiple factr in A. Such primes are said t ramify in A. One way f attacking this prblem, is t study the F p -algebra A/pA. Lemma 6 The prime p ramifies in A if and nly if A/pA has nn-zer nilptent elements Prf: If pa = Q i p i the chinese therem gives A/pA = Y i A/p i which clearly is withut nilptents if and nly if all the nn-zer i s are equal t ne. In the case that A is geneated by ne element ver Z,thatis,fthefrmZ[x]/(f(x)) we culd test nilptency in A/pA by testing f(x) fr multiple rts md p. This happends if and nly if the discriminant f f has p has a factr. Hwever nt all rings f integers are generated by ne element! Hence we smething mre general than just the discriminant f a plynmial. Let A be an algebra a field K, which is equivalent f finite dimensin as a vectr space ver K. Onemay,justasfrfields,definethetracetr A/k (x) f an element x in A as the trace f the multiplicatin map x : A! A. Andwecandefinethetrace frm as tr A/K (xy). We knw frm cmmutative algebra that A is a prduct f lcal algebras, i.e., A = Q i A i. Clearly this is an rthgnal decmpsitin f A with respect t the trace 12

13 frm, i.e., tr(xy) =0if x 2 A i and y 2 A j and i 6= j. Indeed,itisallreadyrthgnal fr the prduct! The trace frm is therefre nn-degenerate if and nly if the trace frm tr Ai /k f each factr is nn-degenerate. Prpsitin 9 The trace frm tr A/K (xy) is nndegenerate if and nly if A is ismrphic t a prduct f fields L i,allbeingseparableverk. Prf: After what we just said, we assume that A is a lcal algebra, and must shw that tr A/K (xy) is nn-degenerate if and nly if A is a separable field extensin f K. Assume that the trace frm is n-degenerate. Then A has n nilptents, since if x is nilptent, xy is nilptent fr all y, hencetr A/K (xy) =0fr all y. Therefr A is a field and by xxx it is separable. The discriminant f a field extensin Let K L be a finite, separable field extensin. T every K-basis v 1,...,v n, wich we call B, weasssiatethedeterminant B =det(tr L/K (v i v j )). It is called the discriminant f the basis B. Of curse, the dicriminant depends n the basis B. Assumethatv 0 1,...,v 0 n is anther basis, and let V dente the transitin matrix between the tw basises. As with any quadratic frm, the matrix f the quadratic frm transfrmss as (tr L/K (v i v j )) = V (tr L/K (v 0 iv 0 j))v t.hencetakingdeterminants,weseethat B =(detv ) 2 B 0 Relatin t the Dedekind determinant The setting is as usual witk K L a separable field extensin and asufficientlybigextensinfieldfk, mrespesifically, we assume that the number f embeddings f L in is equal t n, the different embeddings being 1,..., n. FranyK-basis v 1,...,v n fr L, wemaycnciderthematrix ( j (v i )), Dedekindshwedthismatrixtbeinvertibleanditsdeterminantisclsely related t the discriminant: Prpsitin 10 B =det( j (v i )) 2 Prf: One calculates the matrix prduct: MM t =( j (v i ))( i (v j )) = ( k k(v i ) k (v j )) = ( k k(v i v j )) = (tr K/L (v i v j )) k( ) by therem xxx. The prp- whetre the last equality hlds since tr K/L ( ) = P k sitin fllws. 13

14 The case f a primitive element If is a generatr fr the field L ver K, the discriminat is clsely related t the van der Mnde determinant at least t the discriminant f the cannical basis A cnsisting f the pwers 1,,..., n 1 and thus t the minimal plynmial f. Infact,clselyrelatedisanunderstatement,the discriminant f the basis A is equal t the discriminant f the minimal plynmial f. And they are bth equal t the square f the van der Mnde. We let j be the different embeddings f K( ) in and let j = j ( ). Frthevan der Mnde matrix V we have V =( i 1 j ) 1applei,japplen.HencecmputingVV t,weget VV t =( k i 1 k j 1 k )=( k i+j 2 k )=(tr L/K ( i 1 j 1 )),andhence Prpsitin 11 Assume that L = K( ) is f degree n, andleta be the natural basis 1,,..., n 1.ThenifV dentes the van der Mnde matrix ( i 1 j 1 ),then A =(detv ) 2. If f(x) dentes the minimal plynmial f, thenthediscriminantff and f the basis A cincide: f = A. Versjn: Tuesday, September 24, :28:24 AM 14