3 Extensions of local fields
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1 3 Extensions of local fields ocal field = field complete wrt an AV. (Sometimes people are more restrictive e.g. some people require the field to be locally compact.) We re going to study extensions of such things. Recall standard field theory: /K finite extension of degree n, the K n as K-vs and so for x, the K-linear transformation [ x]: has a char.poly f x,/k (X) = X n + n 1 i=0 a ix i K[X], and tr /K (x) := tr[ x] = a n 1 K, N /K (x) = det[ x] = ( 1) n a 0 K. If /K is a finite Galois extension with group G then tr /K (x) = σ G σx, N /K (x) = σ G σx. For any finite /K we define the trace form: t /K : K, t /K (x, y) = tr /K (xy). which is symmetric and K-bilinear. Basic fact: t /K is nondegenerate (i.e. tr /K is not identically 0) iff /K is separable. 1 In the cases of interest to us this is simple to prove: if K has characteristic 0 this is trivial, since tr /K (1) = n 0. The other case we need is K = F q finite. Then = F q n and tr /K (x) = x+x q + +x qn 1 is a polynomial of degree q n 1 so can t vanish identically on. So if /K is separable and e 1,..., e n is an (ordered) basis for /K then disc(e 1,..., e n ) := det(tr /K (e i e j )) K is non-zero. Theorem 3.1 (Finiteness of IC). et R be an integrally closed domain, K = Frac R. Assume R is Noetherian domain. et be a finite separable extension of K = Frac(R), S = integral closure of R in. Then S is a finite R-algebra. Proof. et {e i } be a basis for /K with all e i S. et {f i } be the dual basis for t /K, so that tr /K (e i f j ) = δ ij. et x S, and write x = a i f i. Then for all i, a i = tr /K (e i x) R (since e i x S, hence is integral over R). So S Rf i is a submodule of a finite R-module, hence (since R is Noetherian) is itself finite. ecture 7 Theorem 3.2. et K be complete wrt an AV, and /K an algebraic extension. i) There exists a unique AV on whose restriction to K is. ii) If [ : K] = n < and x then x = N /K (x) 1/n. iii) Assume K is nonarchimedean, with valuation ring R. The valuation ring R of equals the integral closure of R in. We only need (and will prove) this in certain cases. If K = R or C, it s trivial. We assume now that K is NA, complete with respect to a discrete valuation, with valuation ring R, uniformiser π, residue field k. (The proof works for arbitrary NA fields, given a suitable version of Hensel s lemma.) emma 3.3. (i) f K[T ] monic, irreducible, f(0) R. Then f R[T ]. (ii) If /K is finite and z with N /K (z) R then z is integral over R. 1 Non-degenerate means that for all nonzero x there exists y with t /K (x, y) 0. This is equivalent to saying that tr /K is surjective, since if tr /K (a) = 1 then t /K (x, x 1 a) = 1. 11
2 Proof. (i) et d = deg(f) and let m be minimal such that f (T ) = π m f(t ) = d i=0 a it i has R-coefficients. Assume m > 0, and let j be the largest integer with a j R. By hypothesis, 0 < j < d, hence we can write f = ḡ h = (ā j T j + + ā 0 )(0.T d j T + ā j ), a factorisation in k[t ] with (ḡ, h) = 1. So by Hensel s lemma it lifts to a factorisation f = gh in R[T ], contradicting irreducibility of f. For (ii), apply (i) with f the minimal polynomial of z. Proof of theorem. First do the case [ : K] = n <. Existence in (i), (ii): define by the formula given. Clearly satisfies all the axioms except possibly (AV3N). Suppose x, y with x y ; STP x + y y. Equivalently, STP that if z 1 then z et f = min.poly of z/k, m its degree. Then z = f(0) 1/m so f(0) 1 i.e. f(0) R. By the emma, this forces f R[X], so as f(x 1) is the min.poly of 1 + z, 1 + z = f( 1) 1/m 1. For the remainder, let z R. Then z is integral over R by emma 3.3(ii). As R is integrally closed (being a valuation ring), it equals the integral closure of R in. This proves (iii) Now if is any other AV on extending its valuation ring R is integrally closed, hence contains R, so by Thm 1.3(iii) is equivalent to. In general is the union of its subfields finite over K, and the extensions of to therefore define an extension to all of. So there is a unique extension of to the algebraic closure of K. In particular we can uniquely extend the normalised p-adic absolute value to Q p. The value group is v p (Q p) equals Q, since clearly v p (p a/b ) = a/b. Important fact: Q p is not complete. [Warning: This is nothing to do with the fact that the value group of Q p is not complete.] See ex. sheet 2. Proposition 3.4. et K be complete wrt a discrete valuation, /K a finite separable extension. Then the AV on is discrete, and is complete. Moreover R R n as R-modules. Proof. Clearly K 1/n so is discretely valued. By finiteness of IC, R is finitely generated as an R-module. As R is a PID and R is obviously torsion-free as R-module, we have R R n. Now π K R = π e R for e = v(π K )/v(π ), and so lim R /π m R l = lim m m R /π me R = lim so R = lim R /π m R, hence is complete. m R /π m KR lim (R/π m KR K ) n = R n Note: if K is not discretely valued, a finite extension will still be complete. But in general R will not be a free R-module (R is no longer a PID). ecture 8 Remark: the proof of 3.2 has a gap: given a NA AV on K, we have proved it extends to a unique NA AV on. But could there be an archimedean AV extending? The answer is NO, because of the following fact: an AV on a field K is non-archimedean iff for every n Z, n.1 K 1. [Proof: = by 12
3 strong triangle inequality. Other way: by binomial theorem, see that x + y r (r + 1) max( x, y ) r for every r 1, and letting r get that is NA.] Until the end of this we assume all valuations are discrete. Common and convenient shorthand: cdvf (complete discretely valued field). et K be such a field. Notation: o K = valuation ring of K, π K a uniformiser, v K the normalised valuation (with v K (π K ) = 1). k K = o K /π K o K the residue field. et /K be a finite separable extension of degree n. Since π K π o, the inclusion o K o induces a homorphism k K k, which is therefore a field extension. Definition. The residue class degree of /K is the integer f = f(/k) = [k : k K ]. The ramification degree is e = e(/k) = v (π K ) Note that by definition, π e(/k) o = π K o. Proposition 3.5. et /K be a finite separable extension of cdvfs. Then: (i) e(/k)f(/k) = [ : K]. (ii) K [:K] as topological K-vector spaces. Proof. (i) By emma 1.5, π i o /π i+1 o k, and so by the sequence of inclusions πo e π e 1 l o π o o we have dim kk o /π K o = e dim k/k k = ef. But by 3.4, o o n K so o /π K o has dimension n. (ii) This follows from the proof of 3.4. Definition. We say a finite extension /K is unramified if (i) e(/k) = 1 and (ii) the extension k /k K is separable. The condition e = 1 is equivalent to saying that π K is also a uniformiser of. (In applications, k K will be finite so (ii) is automatic.) Unramified extensions are easy to classify. Proposition 3.6. Suppose /K is finite. TFAE: i) /K is unramified; ii) = K(x) for some x o K for which f x,/k k K [T ] is separable. If so then o = o K [x] for any x as in (ii). Proof. Suppose /K is unramified, and let x k be any element with k = k K (x). (It exists by separability.) Then pick any x o lifting x, and let g be its minimal polynomial; it is in o K [T ] since x is integral over o K. Then ḡ( x) = 0, and since f(/k) = n, this forces ḡ to be the minimal polynomial of x. Conversely, suppose x is as in (ii). Claim f x,/k is irreducible. If not, as it is separable it factors into 2 coprime polynomials in k K [T ]. So by Hensel s emma f x,/k is reducible: contradiction. Therefore k k ( x)/k K is separable of degree n, so k = k k ( x) and /K is unramified. Finally, if o K [x] o, there exists y o with π K y o K [x] but y o K [x]. Write π K y = n 1 i=0 a ix i. As 1, x,..., x n 1 is a basis for k /k K, y o implies all a i π K o K, hence y o K [x], contradiction. 13
4 If /, M/K are finite separable extensions then any K-algebra homomorphism M maps o to o M, hence induces a map k k K. So k is a functor {finite separable extensions of K} {finite extensions of k K } Theorem 3.7. (i) et /K be unramified, and M/K any field extension. Then the natural map Hom K-algebras (, M) Hom kk -algebras(k, k M ) is a bijection. (ii) et k /k K be a finite separable extension. There exists /K unramified with k = k, and it is unique up to isomorphism. Proof. (i) Write = K(x) for x, g as in the propn. Then by Hensel Hom K-algebras (, M) {y M g(y) = 0} = {y o M g(y) = 0} {ȳ k M ḡ(ȳ) = 0} = Hom kk -algebras(k, k M ) (ii) Can write k = k K ( x), ḡ( x) = 0 for some irreducible ḡ k K [T ]. So barg and ḡ are coprime. et = K(x) where g is any monic lift of ḡ. Then g(x) / m so by propn above /K is unramified and k = k. Part (i) with M = shows uniqueness. ecture 9 Recall Thm.3.7. categories: It implies that the functor k defines an equivalence of (finite unramified extensions of K) (finite separable extensions of k K ) Remark. et K be a cdvf, /K a separable algebraic extension /K. Extend the normalised valuation v K of K to. We say /K is unramified if v K ( ) = Z and k /k K is separable. Equivalently, /K is unramified if all its finite subextensions are unramified. The conclusions of the theorem apply equally in this case. Corollary 3.8. Suppose k K = F q is finite. Then K has a unique unramified extension of degree n, for every n 1, namely the splitting field of T qn 1 1. Proof. Follows from the corresponding statement for extensions of F q. Corollary 3.9. (i) et /K be unramified. Then /K is Galois iff k /k K is, and the Galois groups are canonically isomorphic. (ii) Suppose that k K = F q is finite. Then every finite unramified extension /K is Galois. There exists a unique element σ /K Gal(/K), called the arithmetic Frobenius such that for every x o, σ /K (x) x q (mod π ). It generates Gal(/K). Proof. (i) Take M = in (i). (ii) Every extension of finite fields F q n/f q is Galois, with cyclic Galois group generated by x x q. Take σ /K = corresponding element of Gal(/K) under (i). 14
5 The inverse F /K = σ 1 /K is called the geometric Frobenius of /K. Remark. Recall that F q = n 1 F q n = (m,p)=1 F q(µ m ). et Q p Q p with K/Q p finite. Then K nr = (m,p)=1 K(µ m) is the union of all the unramified finite extensions /K inside Q p. It is called the maximal unramified extension of K. It is Galois and we have Gal(K nr /K) Gal(F q /F q ) = lim Gal(F q n/f q ) lim Z/nZ = Ẑ n 1 n 1 et φ K Gal(K nr /K) be the automorphism corresponding to φ q Gal(F q /F q ). Then φ K is an infinite cyclic subgroup of Gal(K nr /K), which is dense in it. Ramification For (considerable) simplicity we now only consider extensions /K for which k /k K is separable. Theorem /K finite separable, k /k K is separable. Then unique intermediate field K 0 such that 0 /K is unramified and / 0 is totally ramified (i.e. f /0 = 1). If K F then F 0 iff F/K is unramified. 0 is called the maximal unramified subfield of /K. Proof. By 3.7(ii) there exists K /K unramified with residue field k, and by (i) the identity map on k defines a unique embedding K. et 0 be its image. Then 0 /K is unramified of residue degree f(/k) so / 0 is totally ramified. Obviously F 0 = F/K unramified. Conversely, if F/K unramified then k F k = k 0 so applying 3.7(i) gives unique maps F 0 lifting the maps on residue fields, hence F 0. So a finite extension can be broken up into an unramified extension, followed by a totally ramified one. We now look at the latter. Definition. A monic polynomial g = T n + n 1 i=0 a it i o K [T ] is Eisenstein if for all 0 i n 1, v K (a i ) > 0, and v K (a 0 ) = 1. Eisenstein s criterion then says that g is irreducible over K. Theorem (i) If g is an Eisenstein polynomial over K and x is a root of g, then = K(x) is totally ramified, x is a uniformiser of and o = o K [x]. (ii) Conversely, if /K is totally ramified, and π is a uniformiser, then the min-poly of π is Eisenstein and = K(π ). Example: let = Q p (µ q ), q = p r. Then ζ q is a root of Φ q (T ) = (T q 1)/(T q/p 1), and the usual argument shows that Φ q (T + 1) is an Eisenstein polynomial. So o = Z p [ζ q ], and π = ζ q 1 is a uniformiser of. ecture 10 Proof. (i) Say g = T n +a n 1 T n 1 +a 1 T +a 0. et v K be the normalised valuation on K, extended to = K(x). Then n 1 x n = a i x i i=0 15
6 implies that v K (x) > 0. But then for all i 0, v K (a i x i ) > 1 = v K (a 0 ), hence v K (RHS) = 1. Therefore v K (x) = 1/n, and ef = n implies that e = n and v (x) = 1, i.e. x = π is a uniformiser of. Now consider y = n 1 i=0 b iπ i K[π ]. Then v (b i π i ) = nv K(b i ) + i, so all the terms have different valuations (as they belong to different residue classes mod n). Therefore v (y) = min{nv K (b i ) + i}, by triangle inequality. In particular, y o iff for each i, nv K (b i ) i i.e. v K (b i ) i/n. As i/n < 1 this means y o iff all b i o K. (ii) et [ : K] = n, and let g = T m + m 1 i=0 a it i be the min.poly of π. Then m n and v (a i ) = nv K (a i ), and from the equation π m = and the same argument as above, we have m 1 i=0 a i π i m = v (π m ) = min{v (a i π i )} = min{i + nv K (a i ) 0 i m 1} and this can only be satisfied if v K (a i ) 1 for all i and v K (a 0 ) = 1, which means m = n, so = K(π ); then o = o K [π ] as in (i). Remark. Suppose K/Q p is finite, with q = #k K. The normalised AV (or modulus) is often defined to be x K = q vk(x) ; thus K = [K:Qp] p. We explain which. K is a locally compact topological group, hence has (up to scalar) a unique Haar measure µ (translation-invariant measure, for which every compact set is measurable). It is easy to describe µ (without any fancy measure theory). Every compact subset of K has a compact open neighbourhood, so we need to specify the values µ(x + πk n o K), for x K and n Z. Translation invariance says µ(x + πk n o K) = µ(πk n o K), and as (πk n o K : π n+1 K ) = q we deduce µ(πn K o K) = o K), hecne qµ(π n+1 K µ(x + π n Ko K ) = µ(π n Ko K ) = q n µ(o K ). for all x K, n Z. So fixing µ(o K ) determines µ completely. In particular, for any open compact = U K and x K, the quotient µ(xu)/µ(u) is just x K. For a general l.c. top.field K the map x µ(xu)/µ(u) is a homomorphism K R which doesn t depend on U (measurable with non-0 measure). For K = R it is x, and for K = C it is x 2. 4 Ramification theory /K finite separable extension of cdvf, with separable residue field extension. Consider the trace form t /K : K. If {x i } is an o K -basis for o, and {y i } is the dual basis wrt t /K, then tr /K (x i y j ) = δ ij. So X = o K y i is the o -submodule of given by {x o tr /K (xy) o K y o } called the inverse different of /K, written D 1 /K. Obviously D 1 /K o, and since π nd 1 /K o when n = min{ v (y i )} we have D 1 /K = π δ(/k) o for some δ(/k) N. 16
7 Definition. D /K = π δ/k o is the different of /K, and δ(/k) is the differential exponent. Theorem 4.1. (i) M//K = D M/K = D M/ D /K. (ii) If o = o K [x] with g = min.poly of x, then D /K = (g (x)). (iii) δ(/k) e(/k) 1, with equality iff e 0 (mod p). In particular, /K is unramified iff D /K = o. (Here p is the residue characteristic of K.) Proof. (i) follow definition. (ii) et x = x 1,..., x n be the roots of g. Then since x i x j we have partial fractions decomposition 1 g(t ) = n i=1 1 (T x i )g (x i ) Expanding both sides as power series in 1/T we have n T n a n 1 T n = g (x i ) 1 (T 1 + x i T 2 + x 2 i T ) = i=1 tr /K g (x) 1 x r T r 1 and equating coefficients gives = 0 if 0 r < n 1 tr /K (x r g (x) 1 ) = 1 if r = n 1 o K for all r r=0 This implies that {g (x) 1 x i D /K = (g (x)). 0 i n 1} is an o K -basis for D 1 /K, hence ecture 11 (iii) Applying this with /K unramified gives by Propn. 3.6 that D /K = o. So by (i) D /K = D /0 where 0 is the maximal unramified subfield, so it is enough to consider the case /K totally ramified. In this case [ : K] = e and we may take x = π, a root of an Eisenstein polynomial g = T e + a i T i. Then e 1 g (π ) = eπ e 1 + ia i π i 1 v (a i ) e. So v (g (π )) = e 1. But if e 0 (mod p) then each term on the RHS has v e. and if e 0 (mod p) then the term eπ e 1 i=1 has v = e 1, whereas v (ia i π i 1 Definition. /K is tamely ramified if p e /K. Otherwise /K is wildly ramified. If k has characteristic zero, then any extension of K is at most tamely ramified. We henceforth assume k K has characteristic p > 0. Example: K n = Q p (ζ p n), p > 2. Know [K n : Q p ] = p n 1 (p 1) and K n /Q p is totally ramified, uniformiser π n = ζ p n 1. So K 1 /Q p is tamely ramified and D K1 /Q p = (π p 2 1 ). For n > 1, K n /K n 1 has degree p and o Kn = o Kn 1 [ζ p n], min.poly of ζ p n over K n 1 is g(t ) = T p ζ p n 1. So D Kn/Kn 1 = (pζ p 1 p ) = (p) and δ(k n/k n n 1 ) = p n 2 (p 1). Therefore D Kn/Qp = (p n 1 π p 2 1 ). 17
8 The case of /K Galois et G = Gal(/K). Then σ G, v σ = v so σ(o ) = o and σ(m ) = m. So G actos on o and on the quotients o /m i+1, i 0. Definition. G i = G i (/K) = ker(g Aut(o /m i+1 )) (i 0) are the ramification groups of /K. It s convenient to set G 1 = G. Obviously G i G and G i G i+1. Also G i = ker(g Aut(o /m i+1 )) = ker(g Aut o ) = {1} i so G i = {1} for i 0. Definition. I = I(/K) = G 0, the inertia subgroup of /K; P = P (/K) = G 1, the wild ramification subgroup of /K. If 0 is the maximal unramified subfield, obviously I = ker(g Gal(k /k K )) = ker(g Gal(/ 0 )) so I = Gal(/ 0 ). In particular, /K is unramified iff I = {1}, and G/I Gal(k /k K ). Also, for i G i (/K) = G i (/ 0 ). Proposition 4.2. Assume /K is totally ramified, π a uniformiser of. Then: (i) G i (/K) = {σ Gal(/K) v (σ(π ) π ) i + 1}. (ii) Define maps { k for i = 0 θ i : G i m i /mi+1 for i 1 σ σ(π ) π mod m (i = 0) σ(π ) π 1 mod m i+1 (i 1) (well-defined by (i)). Then θ i is a homomorphism, independent of the choice of π, and ker(θ i ) = G i+1, for all i 0. Proof. et σ G i. mod m i+1. Therefore Then if u o, σ(u) u mod mi+1 and so σ(u)/u 1 σ(uπ ) uπ = σ(u) σ(π ) = σ(π ) mod m i+1 u π π so θ i (σ) is independent of the choice of π. So for any τ G i, θ i (σ) = σ(τ(π ))/τ(π ) 1. If i = 0 and σ, τ G 0 then θ 0 (σ)θ 0 (τ) = σ(τ(π )) τ(π ) = στ(π ) τ(π ) π π = θ 0 (στ) and θ 0 is a homomorphism. ikewise, if i 1 then θ i (σ)θ i (τ) = 0 and so θ i (στ) = σ(τ(π )) 1 = σ(π ) π τ(π ) By definition of G i, ker θ i = G i+1. τ(π ) 1 = (θ i (σ)+1)(θ I (τ)+1) 1 = θ i (σ)+θ i (τ). π 18
9 ecture 12 Corollary 4.3. (i) G 0 /G 1 is cyclic of order prime to p, and for all i 1, G i /G i+1 is an elementary abelian p-group. (ii) P = G 1 is the unique Sylow p-subgroup of I, and is normal in G. Moreover P = {1} iff /K is tamely ramified. (iii) If k K is finite, G is solvable. Proof. (i) We have G 0 /G 1 k and G i/g i+1 m i /mi+1 k. Every finite subgroup of a field is cyclic. (ii) From (i) P is a Sylow p-subgroup of I and is normal; so it is the unique Sylow p, hence is normal in G. (iii) We have I/P cyclic, P a p-group and since k K finite, G/I cyclic. Example: K n = Q p (ζ p n). Then K n /Q p totally ramified = G = G 0, and G = Gal(K n /Q p ) (Z/p n Z) (σ a : ζ ζ a ) a Now π n = ζ p n 1 is a uniformiser of K n. et (a Z/p n Z), a 1 p n m b with 0 < m n and (p, b) = 1. Then v Kn (σ a (π n ) π n ) = v Kn (σ a (ζ p n) ζ p n) = v Kn (ζ a p n ζ p n) = v K n (ζ a 1 p n 1) = v Kn (ζ b p m 1) = v K n (ζ p m 1) = v Kn (π m ) = [K n : K m ] = p n m. and therefore by 4.2(i) (putting r = n m) G i = ker((z/p n Z) (Z/p r Z) ) if p r 1 i p r 1. 19
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