Separability MAT4250 — Høst 2013

Separability

Version 1.1 — 24. september 2013 klokken 10:28

Multiple roots and separable polynomials Let as usual K be a field, and let a polynomial f K[t] of degree m is given. It 2 is convenient—and we loose nothing—to suppose that f is monic. Let ⌦ be a field containing K such that ! ⌦ is a root of f. We may then write 2 f(t)=(x !)rg(x) where g(x) ⌦[t] is a polynomial not vanishing at ! and r 1. The root is asimple 2 root if r =1,andinthecontrarycase,i.e., if r 2,wecall! amultipleroot.One has the well known and elementary, but usfull lemma

Lemma 1 The root ! is multiple if and only if f 0(!)=0. Proof: The product rule for the derivative gives

r 1 r f 0(x)=r(x !) g(x)+(x !) g0(x), which shows that f 0(!)=0if and only if r 2. o If the polynomial f splits completely over ⌦,thereisafactorization f(x)=(x ! ) (x ! ) 1 ····· m with all the roots !i in ⌦.Whentheroots!i are distinct, we say f splits simply over ⌦ and call f(x) a separable polynomial.Ifnot,thatis,ifatleastonerootoccursat least twice, the polynomial is said to be inseparable. Problem 1. Verify that if f splits simply in one extension ⌦ of K it splits simply in any other extension ⌦0 where it splits. Hint: Chose a common algebraic closure of ⌦ and ⌦0. X It is of course an easy matter to find polynomials with multiple roots, but to find inseparable, irreducible polynomial is a lot more subtle1. As the proposition below will show, one has to search in positive characteristic to find such species. Among the many new and contra-intuitive phenomenon that can happen in positive characteristic, is that the derivative of a polynomial may vanishes identically without the polynomial being constant. The simplest example of a polynomial with this behavior is tp a. More m generally, if f(x)=g(xp ),byapplyingthechainrule,weseethat

m pm 1 pm f 0(x)=p x g0(x )=0 · since the characteristic of K is p. Now, these examples are in fact the only examples. We have:

—1— Separability MAT4250 — Høst 2013

PolyWithVanishDer Lemma 2 Let K be a field of characteristic p and assume f(x) is a polynomial in n K[x] whose derivative vanishes identically. Then f(x)=g(xp ) for some polynomial g(x) in K[x] whose derivative does not vanish identically.

i i 1 Proof: Let f(t)= 0 i r aix .Sincethederivativef 0(x)= 0 i r iait vanishes     identically, iai =0for 0 i r. This implies that either ai =0or i =0,the P   P latter means that i has p as factor. Hence the only non-vanishing terms of f are those pj whose degree is divisible by p, and we may write f(x)= 0 j r/p apjt .Or,phrased   differently, f(x)=g(xp) for some polynomial g.Ifg has a non-vanishing derivative, we P are happy, if not, we repeat the process until happiness. o

Proposition 1 Let K be a field and f(x) K[x] an irreducible polynomial. Then f(x) 2 is inseparable if and only if f 0(x) vanishes identically, i.e., f 0(x)=0as a polynomial. In particular, f(x) is separable if one of the following conditions is fulfilled:

⇤ The characteristic of K is zero.

⇤ The degree of f(x) is prime to the characteristic of K.

Proof: Let E be the stem field of f(x),thatisE = K[x]/(f(x)). Then f has a root ↵ in E,andf(x) is the minimal polynomial to ↵. The degree of f 0(x) being one less than he degree of f(x), it follows that if f 0(↵)=0,thenf 0(x)=0identically. In characteristic zero this can of course not not happen since f(x) is not constant, and lemma 2 shows that in case it happens in positive characteristic, the degree of f must be divisible by the characteristic of K. o

Problem 2. Assume that f(x) is an inseparable, irreducible polynomial. Show that n one may write f(x)=g(xp ) where g(x) is separable and irreducible. Conclude that all roots of f have the same multiplicity pn. X

Problem 3. Let K be a field of characteristic p and let a K be an element. Show n 2 that xp a is irreducible over K if and only if a is not pn-th power in K. Hint: n n n Use that the Frobenius map x xp is additive to show that if ap = bp in a field of 7! characteristic p,thena = b.Hencealltherootsoff in a splitting field ⌦ are equal. X

Separable field extensions

Recall that giving an embedding over K of the stem field Ef of an irreducible polyno- mial f(x) K[x] into a field ⌦,isequivalenttogivingarootoff in ⌦. 2 Assume that ⌦ is a splitting field for f.Forseparablepolynomials—thatis,poly- nomials that have as many different roots in ⌦ as the degree n indicates—this means that there are exactly n emddings into ⌦,andfortheinseparablepolynomialsthere are less. Hence

—2— Separability MAT4250 — Høst 2013

StemFieldSeparable Proposition 2 Suppose that K be a field and let f(x) K[x] be an irreducible polyno- 2 mial of degree n.Let⌦ be a splitting field for f.Thenf is separable if and only there are n different embeddings over K of the stem field Ef into ⌦. A small, but useful, extension of this result is to the case when K is not contained in ⌦, but just embedded. That is, there is given an embedding : K ⌦. Then the result ! above reads as f(x) is separable if and only if there are exactly n extensions of to L. The proof is not difficult and left as an exercise. Proposition 2 above may be generalized to fields not necessarily being stem fields (but in the end of the day they will be, due to what is called the “Theorem of the primitive element”). Assume that L is a field extension of K.Anelement↵ L is 2 said to be a separable element if the minimal polynomial is separable. The extension is called a separable extension if any element in L is separable. All this should be understood relatively to the field K.However,onehasthethe lemma SepInTowers Lemma 3 Assume that K E L is a tower of finite field extensions. If an element ✓ ✓ ↵ of L is separable over K,itisseparableoverE. Proof: Let f(x) be the minimal polynomial of ↵ over K. Then of course f(x) E[x], 2 but it is not necessarily irreducible there. We may therefore factor f(x)=g(x)h(x) over E, where g(x) is the minimal polynomial of ↵ over E and where h(↵) =0.Applying 6 the product rule, we find

f 0(x)=g0(x)h(x)+g(x)h0(x).

Hence if f 0(↵) =0,theng0(↵) =0(since h(↵) =0and g(↵)=0). o 6 6 6 Problem 4. Show by a (stupid) example that the converse of the lemma is not true Hint: Take E = L. X

Proposition 3 Let L be a finite extension of the field K.Assumethat⌦ is an algebraic closed field and that is an embedding of K in ⌦.ThenL is a separable extension if and only if the number of different embeddings of L in ⌦ extending equals the degree [L: K] of L over K. Proof: Pick an element ↵ L.IfL = K(↵) we are done, since then L is isomorphic to 2 the stem field Ef and ↵ is separable over K.Ifnot,wehavethetowerK K(↵) L. The ✓ ✓ number of embeddings of K(↵) in ⌦ equals the degree [K(↵): K],andbyinduction2, each of these can be extended to L in [L: K(↵)] ways since L is separable over K by lemma 3,andwearethroughsincethedegreeismultiplicativeintowers. o 1Just for this reason, many authors apply the terms separable and inseparable only to irreducible polynomials. 2In fact, we need to prove a slightly more general statement to make the induction work: Any embedding of K into ⌦ can be extended to L in exactly [L: K] ways.

—3— Separability MAT4250 — Høst 2013

Corollary 1 In a tower K E L of field extensions, L is separable over K if and ✓ ✓ only if both E is separable over K and L is separable over L. Proof: We count extensions of embedding of the fields into some algebraically closed field ⌦:AssumeL separable over E and E separable over K. Then by the proposition, an embedding of K into ⌦ can be extended to L in [E : L] different ways, and each of these can be further extended to L in [L: E] ways. Hence the total number of extensions to L equals [E : K][L: E],andthisisequalto[L: K]. The other way round is just lemma 3 above. o

Perfect fields A perfect field is a field K having the property that any extension L of K is separable. All fields of characteristic zero are perfect, and the good news for number theorists is that all finite fields are perfect. One has the following characterisation of perfect fields relating to existence of p- roots in K: Proposition 4 Assume that K is a field of characteristic p.ThenK is perfect if and only if every one of its elements is a p-th power, i.e., every element in K has a p-th root in K. Proof: Assume that every element has a p-th root, and suppose that L is an insepa- rable extension of K. Then there is an element ↵ L,notinK,satisfyingaminimal 2 relation nq (n 1)q q ↵ + an 1↵ + ...a1↵ + a0 =0 (K) where q = pm,andtheai’s are in K.SinceeveryelementinK has a p-root, they all MinRelInSep q have a q-th root as well, and we may find bi’s in K with bi = ai. The relation then takes the form nq q (n 1)q q q q ↵ + bn 1↵ + ...b1↵ + b0 =0, but, as rising elements to q-th power is additive in characteristic p,thisgives n n 1 q (↵ + bn 1↵ + ...b1↵ + b0) =0, contradicting the minimality of the relation K. The other way around, if a K is not a p-th power, the polynomial xp a is 2 p irreducible and the extension Ef = K[x]/(x a) of K is inseparable. o Finite fields are perfect The fields of finite characteristic we shall meet in this course, are are almost exclusively the finite fields. They are all perfect, so we very seldom will have to cope with inseparable fields extensions—however w’ll meet inseparable algebras, but that is another story we come back to later. n To see that a finite field Fq—where q = p is a prime power—is separable. We check that every element is a p-th power. The Frobenius map x xp is additive (this is true 7! in characteristic p)andinjective—xp =0obviously implies that x =0.Henceitmust be bijective since Fq is finite.

—4— Separability MAT4250 — Høst 2013

All this fuzz about nothing? The reader deserves at least one example of a extension that is not separable. The easiest is probably the following. Let K be any field of characteristic p and let x be a variable. Then K(xp) K(x) is not separable, ✓ since xp is not a p-th power in K(xp) (any p-th power in K(xp) is of degree p2 in K(x)).

Multiple roots and the discriminant We know since childhood (at least mathematically speaking) that a quadratic polyno- mial ax2 + bx + c has a double root if and only if the discriminant b2 4ac vanishes. Indeed, the discriminant is the square of the difference of the two roots. The discrimi- nant is a very convenient tool; without actually knowing the roots, one can decide if they are equal or not. In general, if the monic polynomial f(x) of degree n has the n roots !1,...,!n in splitting field ⌦,onedefinesthediscriminantoff as the product

= (! ! )2, f i j 1 i

2n 2 2 = a (! ! ) . f i j 1 i

Discriminant and derivatives Since f(x)= (x !j),oneeasilyseesbyapplying j the product rule, that f 0(!i)= j, j=i(!i !j).Hence 6 Q Q 2 f 0(! )= (! ! ) = ✏ (! ! ) i i j i j i i j, j=i ! i

f(x) a In case is not monic, but has leading coefficient ,wefind DiskProduktAvDerivert

n(n 1)/2 n 2 =( 1) a f 0(! ). f i i Y This relation between the discriminant and the derivative of f has nice formulation in terms of the norm. Recall that if K L is a separable extension, the norm N (↵) ✓ L/K of ↵ equals the product i i(↵) of the value at ↵ of the different embeddings i of L in some sufficiently big field ⌦.Let⌦ be a splitting field for f and let ! be one of the Q roots. Then if L = K(!),thevaluesi(!) are just the different roots !i of f in ⌦ and i(f 0(!)) = f 0(!i). Combining this with equation (H)above,weobtain

Proposition 5 Let f(x) be a monic polynomial in K[x] and let !1,...,!n be the roots of f in some splitting field ⌦.Then n(n 1)/2 =( 1) N (f 0(!)). f K(!)/K The case of the trinome xn + ax + b Computing discriminants by hand can be achallenge,butluckily,todayonehassoftwarethatcancandothedirtyjobsforus. However, in the special case of the the trinomes xn + ax + b, the computations—with some cleverness—are not horrible at all, and is a classic. So we shall do it. This furnishes us we many examples, and it reminds us that cumputation by hand sometimes is an option, and probably most importantly, you have a chance to become wiser by doing the calculatuons your self and not only pushing the button. So let a and b be elements of the field K and let f(x)=xn + ax + b.Let! be a n 1 n 1 1 root of f. The derivative is given as f 0(!)=n! + a.Now! = a b! ,so 1 renaming the derivative x and using x = (n 1)a nb! ,wecansolvefor! and obtain nb ! = . x +(n 1)a This shows that nb f( )=0. x +(n 1)a Clearing the denominator, we get n n 1 n n 1 n (x +(n 1)a) na(x +(n 1)a) +( 1) b n =0. Interpreting x as a variable, the left side is the minimal polynomial of f 0(!),andthe n norm of f 0(!), which we are looking for, is the constant term with the factor ( 1) . The binomial theorem and some cleaning up gives n(n 1)/2 n n 1 n 1 n 1 n =( 1) (n b +( 1) (n 1) a ) f Specializing to case of a cubic x3 + ax + b,onegetstheubiquiousformulaforthe cubic discrimant = 27b2 4a3 f

—6— Separability MAT4250 — Høst 2013

Problem 5. Compute the discriminant of the two cubic polynomials x3 + x +1and x3 x 1. X 3 2 2 3 Problem 6. Let g(x)=x + ax + b show that g = 27b 4a b. Hint: Relate 3 1 g to the discriminat of f(x)=x g(x ) X

Problem 7. Compute the discriminat of x3 2x2 +2. X Problem 8. The aim of this exercise is to show that if x3 + ax + b is cubic polnomial whose discriminant is negative, then it has a unique real root given by

! = 3 ( b + /3p 3)/2+ 3 ( b /3p 3)/2 (J) q q where is a square root of the discrimant 4a3 27b2. RealRoot a) Let ⇠ be a primitive third root of unity. Show that there are real numbers s and t such that s⇠ + t⇠¯ and s⇠¯+ t⇠ are the two complex roots of f(x). b) Show that the real root equals s + t. Hint: The sum of the roots is zero. c) Show that b = s3 + t3. Hint: The product of the roots is b. d) Show that the product of the differences of the roots (i.e., )is 3p 3(s3 t3). ± e) Show the formula (J). Hint: Solve for s and t.

X

3 2 2 2 3 3 2 2 Problem 9. If f(x)=ax +bx +cx+d,showthatf = b c 4ac 4b d 27a d + 18abcd. X The Van der Monde determinant To treat the van der Monde determinant we do it generically, as we did with the discriminant. We use n independent variables w1,...,wn be variables, so the formulas we obtain are valid in the polynomial ring Z[w1,...,wn].Subsequentlyonemaysubstituteforthewi whatever elements one wants, as long as they are elements of a commutative ring, hence the formulas are universally valid.

11...... 1 w w ...... w 1 2 n w2 w2 ...... w2 Vn = Vn(w1,...,wn)= 1 2 n . . . . . . . . . . n 1 n 1 n 1 w w ...... w 1 1 n Setting wi = wj kills the determinant two columns then being equal. This means that wi wj divides V (and this is where we use the ring Z[w1,...,wn] and that it is a ufd ). It follows that the product Pn = (wi wj) is a factor of V . i>j Q —7— Separability MAT4250 — Høst 2013

This product and the van der Monde are both polynomials of the same degree. n 1 Indeed, every term of the van der Monde is of degree i=0 i = n(n +1)/2,andthe product has as many (linear) factors as there are pairs with i>j,thatisn(n +1)/2. P Hence the two expressions agree up to a scalar factor, say ✏n.Settingwn =0and n 1 developing Vn along the last column gives Vn wn=0 =( 1) w1 wn 1Vn 1 and n 1 | ····· similarly Pn wn=0 =( 1) w1 wn 1Vn 1.Hence✏n = ✏n 1,andofcourse✏2 =1, | ····· so the scalars ✏n all reduce to 1. We have established the following:

Proposition 6 Assume that f(t) is a polynomial of degree n whose roots are !1,...,!n in some extension field ⌦ of K.Thenwehavethefollowingrelationbetweenhediscri- minant of f and the van der Monde determinant made out of the roots: 2 f = Vn (!1,...,!n)

Nilpotency and trace There is a close connection between nilpotency of a linear operator A on a vector space and the tracelessness of all the powers of A.Infact,overafieldofcharacteristiczero,A being nilpotent is equivalent to the vanishing of the traces tr Ai for i 1.Iftheground field is of positive characteristic, however , the situation is a little more complicated due to the fact that there are non-constant polynomials with a vanishing derivative. Even if our main objects of study—rings of algebraic integers in number fields—all live in characteristic zero, we have a great interest in considering the case of positive characteristic. The reason is as follows. If A B is an extension of say number rings— ✓ for example could B be the integtral closure of A in some finite field extension of the fraction fielsd of A—and p A aprimeideal,wearecertainlyinterestedingettinga ✓ hand on the set of prime ideals q in B with q A = p—that is, the fibre of the map \ spec B spec A over p. This fibre is in one-to-one-correspondence with the primes in ! the ring B/pB,andthisringisanalgebraoverthefinitefieldA/p. The lesson is: We are also interested in finite dimensional algebras over finite fields! Example ￿. An illustrative example is when B is generated by one element over A; to be concrete, say A = Z and B = Z[↵] where ↵ belongs to some finite extension L of Q,and↵ is integral over Z. Then B = Z[x]/(F (x)) where F (x) is the minimal polynomial of ↵ over Q,andthispolynomialhasintegralcoefficients,↵ being integral. If now p Z is a rational prime, we can perform the following small computation 2 B/pB = Z[x]/(F (x),p)=Fp[x]/(f(x)), where f(x) Fp[x] denotes the polynomial obtained from F by reducing all coefficients 2 mod p.Ofcourse,f(x) is not necessarily irreducible (even if F is). Some times it is and some times not. v1 vr If f(x)=g1(x) gr(x) is a factorization of f where the gi(x)’s are different ····· irreducible poly’s in Fp[x], then by some well known chinese theorem

v1 vr vi B/pB = Fp[x]/(g1(x) gr(x) )= Fp[x]/(gi(x) ) ····· Y —8— Separability MAT4250 — Høst 2013

vi Each factor Fp[x]/(gi(x) ) is an algebra over Fp,anditisafield if and only if vi =1, that is, if and only if it is without nilpotent elements. So we see that nilpotency is couple to multiple factors of f,orincasethefactorsarelinear,tomultiplerootsof f. Poly’s with multiple roots in some extension of the base field, are the inseparable poly’s,sothelessonis:Nilpotencyiscoupledtoinseparability. e

Problem 10. Show that x2 +1is irreducible mod p if and only if p 1mod4. X ⌘ Linear operators Coming back to linear operators, we fix a field K and a linear operator A on the finite dimensional vector space V over K. The characteristic polynomial of A is, as we know, defined by PA(t) = det(t I A).OversomeextensionL of K the characteristic · polynomial splits into a product P (t)= (t i) of linear factors. The i’s are the i eigenvalues of A in . Q In addition to the characteristic polynomial we shall make use of the following associated polynomial:

n 1 Q (t)= (1 t)=t P (t ) A i A Y It still has coefficients in K. The degree of QA may drop compared to the degree of PA.Itisequaltothenumberofnon-zeroeigenvaluesofA.Infact,therootsofQ are the inverses of the non-vanishing eigenvalues of A. A consequence of Cayley-Hamilton is that the operator A is nilpotent if and only if QA is constant. Our next lemma is completely elementary, on the level of a first course in calculus, but will be very usual:

FundFormel Lemma 4 Assume that Q(t)= 1 i n(1 it) is a polynomial where the i’s are in   some commutative ring A.Thentheformulabelowholdstrueinthepowerseriesring Q A[[t]]: Q0(t) k k 1 = t Q(t) i 1 i n,k 1  X  The identity in the lemma is completely formal. That is, we may assume that i’s are independent variable, and the identity holds in the ring of formal power series3 Z[1,...,n][[t]]. It is therefore valid whenever the two sides have a meaning.

Proof: The product rule gives Q(t)0 = i( i) j=i (1 jt). Then 6 Q0(t) P Q = i Q(t) 1 t i i X and the lemma follows from the the ubiquitous formula for the sum of a geometric series. o 3Admittedly, this formulation is not very much first-calculus-course-ish.

—9— Separability MAT4250 — Høst 2013

Lemma 5 If A is an operator on a finite dimensional vector space V of dimension n over the field K,then QA0 (t) k k 1 = tr A t QA(t) 1 k X n 1 holds in the power series ring K[[t]],whereQA(t)=t PA(t ) and PA(t) denotes the characteristic polynomial of A.

Proof: If the i’s with 1 i n are the eigenvalues of A,thenonehasQA(t)= k k  (1 it) and tr A = . o i i i Q P Proposition 7 For the operator A,thetwofollowingconditionsareequivalent

For all k 1,thetraceofAk vanishes, i.e., tr Ak =0. ⇤  ⇤ QA(t)0 =0

The trace form and separability We have now come to main theorem of this section. Let as usual K be a field and L afiniteextensionofK. The trace gives us a bilinear and symmetric form on L with values in K ,namelytheform(x, y)=tr (xy) for x, y L.Indeedtr (xy) is L/K 2 L/K clearly symmetric, and the product being linear in each variable, trL/K (xy) is bilinear. This form is very useful since it detects inseparability; more precisely the trace form is non-degenerate if and only if the extension L of K is separable:

TraceFormNondegerate Theorem 1 Let K L be a finite separable field extension. Then the bilinear form ✓ trL/K (xy) is non degenerate if and only if the extension L of K is separable.

Proof: Assume first that L is separable over K. We use induction on n =[L: K],andletL be a counter example of least degree over K. The field L being a counter example, there is an element x L not in K with 2 trL/K (xy)=0for all y.HenceL = K(x) since L was the smallest bad guy; indeed K(x) is separable over K,and,ofcourse,trK/L xy =0for all y K(x). k 1 k 2 By putting y = x we see that trL/K (x )=0for all k.Since

n n 1 x = a0 + a1x + + an 1x ···

with the ai’s in K,anda0 =0,thisgivestr a0 = na0 =0,andn must be divisible by 6 the characteristic p of K. n 1 Let now Q(t)=t P (t ) where P (t) denotes the characteristic polynomial of the multiplication map ⇢x. Then by lemma 4 we have

Q0(t) k k 1 = tr (x )t =0, Q(t) L/K k 1 X —10— Separability MAT4250 — Høst 2013

and Q0 vanishes identically. Now n 1 1 n 2 1 n 2 1 0=Q(t)0 = nt P (t )+t P 0(t )=t P 0(t ) as p divides n.HenceP 0 vanishes identically. But the characteristic and the minimal polynomial of x coincide (since L = K(x)), and by consequence x is not separable over K. For the proof of the implication in the other direction, assume that L is not separab- le, and let x L be inseparable over K.WefirstlookattheextensionK(x) generated 2 by x.IfP (t) is the characteristic polynomial (which is equal to the minimal polynomial of x over K)ofmultiplicationbyx in K(x),thenP 0 =0.HenceQ0 =0, and by what we just did, trK(x)/x(y)=0for all y K(x). It follows that trL/K =trK(x)/L trL/K(x) =0. 2 o Just one remark about this proof. Most of it is there to deal with the characteristic k p case. In characteristic zero, one concludes immediately from the fact that trL/K (x )= for all k 1,thatQ0 =0,henceQ(t) constant, i.e., that x is nilpotent, which means that x =0since K is a field. And thus the trace form is non-degenerate.

The dual basis and finiteness of the integral closure

Assume that v1,...,vn is a K basis for the sperable extension L of K. Then by the general theory of non-degenerate quadratic forms, there exsists a dual basis. This is a basis v10 ,...,vn0 such that trL/K (vivj0 )=ij. The proof of the following result illustrates the usefulness of the dual basis. The result is really the basis for the whole algebraic number theory. It is fundamental that the integral closure of a Dedekind ring in a finite extension of the field of fraction, still is Dedekind. The main ingredient in the proof of this result, and the only subtle one, is that the integral closure remains noetherian. This follows from our next proposition.

Proposition 8 Let A be noetherian domain, integrally and closed in its fields of frac- tions K.AssumethatL is a finite extension of K such that the bilinear form trL/K (xy) is non-degenerate. Then integral closure B of A in L is a finitely generated module over A.

Proof: Let v1,...,vn be a basis for L over K.Wemayassumethatthevi’s are contained in B (multipliedbyacommondenominator,theystillconstituteabasis). Let v10 ,...,vn0 be the dual basis with respect to the bilinear form trL/K (xy). Then we claim that B is contained in the A-submodule of L generated by the vi’s. This submodule is by definition finitely generated, and A being notherian, it follows that B is finitely generated. Since v0 ,...,v0 is a K-basis for L,any B may be written as = a1v0 + +anv0 1 n 2 1 ··· n with ai K.Ourtaskistoverifythattheai’s are in A.Now 2

trL/K (vj)= ai trL/K (vi0vj)=aj X —11— Separability MAT4250 — Høst 2013

and since both and vi are in B their product is, and hence tr (vi) A by L/K 2 proposition 7 on 6 in the section Norms and Traces.. o

Theorem 2 Let A K be a dedekind ring and its field of fractions. Let L be a finite, ✓ separable extension of K and denote by B the integral closure of A in B.ThenB is a Dedekind ring.

Proof: We know that B is integrally closed and finitely generated as an A-module. Hence it is noetherian and of the same Krull-dimension as A (It is generally true that if B is an A-algebra which is finitely generated as an A-module, then if one of A and B is noteherian, the other is, and they have the same Krull-dimension ). o Without the hypothesis that L be separable over K,thedomainB need not be finite over A, but it will be noetherian and of Krull dimension one. This is called the Krull- Akizuki-theorem, and we shall not prove it. So, the theorem remains true even if we skip the assumption that L is separable over K.

Etale or separable algebras One of the main problems in algebraic number theory is to describe how a rational prime p factorizes in the ring A of integers in an algebraic number field K.Aspart of this, it is of great importance to decide which primes p have a multiple factor in A. Such primes are said to ramify in A. One way of attacking this problem, is to study the Fp-algebra A/pA. Lemma 6 The prime p ramifies in A if and only if A/pA has non-zero nilpotent elements

⌫i Proof: If pA = i p the chinese theorem gives Q A/pA = A/p⌫i i Y which clearly is without nilpotents if and only if all the non-zero ⌫i’s are equal to one. o In the case that A is geneated by one element over Z,thatis,oftheformZ[x]/(f(x)) we could test nilpotency in A/pA by testing f(x) for multiple roots mod p. This hap- pends if and only if the discriminant of f has p has a factor. However not all rings of integers are generated by one element! Hence we something more general than just the discriminant of a polynomial. Let A be an algebra a field K, which is equivalent of finite dimension as a vector space over K.Onemay,justasforfields,definethetracetrA/k(x) of an element x in A as the trace of the multiplication map ⇢x : A A.Andwecandefinethetrace form ! as trA/K (xy). We know from commutative algebra that A is a product of local algebras, i.e.,

A = i Ai. Clearly this is an orthogonal decomposition of A with respect to the trace Q —12— Separability MAT4250 — Høst 2013

form, i.e., tr(xy)=0if x Ai and y Aj and i = j.Indeed,itisallreadyorthogonal 2 2 6 for the product! The trace form is therefore non-degenerate if and only if the trace form trAi/k of each factor is non-degenerate.

Proposition 9 The trace form trA/K (xy) is nondegenerate if and only if A is isomorp- hic to a product of fields Li,allbeingseparableoverK.

Proof: After what we just said, we assume that A is a local algebra, and must show that trA/K (xy) is non-degenerate if and only if A is a separable field extension of K. Assume that the trace form is no-degenerate. Then A has no nilpotents, since if x is nilpotent, xy is nilpotent for all y,hencetrA/K (xy)=0for all y. Therefor A is a field and by xxx it is separable. o

The discriminant of a field extension

Let K L be a finite, separable field extension. To every K-basis v1,...,vn, wich we ✓ call ,weassosiatethedeterminant B

=det(trL/K (vivj)). B It is called the discriminant of the basis . B Of course, the dicriminant depends on the basis .Assumethatv0 ,...,v0 is anot- B 1 n her basis, and let V denote the transition matrix between the two basises. As with any quadratic form, the matrix of the quadratic form transformss as (trL/K (vivj)) = t V (trL/K (vi0vj0 ))V .Hencetakingdeterminants,weseethat

=(detV )2 B B0 Relation to the Dedekind determinant The setting is as usual witk K L a ✓ separable field extension and ⌦ asufficientlybigextensionfieldofK,morespesifically, we assume that the number of embeddings of L in ⌦ is equal to n, the different embed- dings being 1,...,n.ForanyK-basis v1,...,vn for L,wemayconciderthematrix (j(vi)),Dedekindshowedthismatrixtobeinvertibleanditsdeterminantisclosely related to the discriminant:

2 Proposition 10 =det(j(vi)) B Proof: One calculates the matrix product:

t MM =(j(vi))(i(vj)) = ( k(vi)k(vj)) = ( k(vivj)) = (trK/L(vivj)) Xk Xk whetre the last equality holds since trK/L(↵)= k k(↵) by theorem xxx. The propo- sition follows. o P

—13— Separability MAT4250 — Høst 2013

The case of a primitive element If ↵ is a generator for the field L over K, the discriminat is closely related to the van der Monde determinant—at least to the n 1 discriminant of the canonical basis consisting of the powers 1,↵,...,↵ —and thus A to the minimal polynomial of ↵.Infact,closelyrelatedisanunderstatement,the discriminant of the basis is equal to the discriminant of the minimal polynomial of A ↵. And they are both equal to the square of the van der Monde. We let j be the different embeddings of K(↵) in ⌦ and let ↵j = j(↵).Forthevan i 1 t der Monde matrix V we have V =(↵j )1 i,j n.HencecomputingVV ,weget  

t i 1 j 1 i+j 2 i 1 j 1 VV =( ↵k ↵k )=( ↵k )=(trL/K (↵ ↵ )) Xk Xk ,andhence

Proposition 11 Assume that L = K(↵) is of degree n,andlet be the natural basis n 1 i A1 j 1 1,↵,...,↵ .ThenifV denotes the van der Monde matrix (↵ ↵ ),then

=(detV )2. A If f(x) denotes the minimal polynomial of ↵,thenthediscriminantoff and of the basis coincide: A f = . A

Versjon: Tuesday, September 24, 2013 10:28:24 AM

—14—