Dieudonne

Dieudonne

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  • Discrete valuation (results) #ALNT-LB 1.8.R Chapter 1 Section 8 #Algebraic Number Theory # Lecture note Benois

    Definition

    Let K be a field, then a discrete valuation on K is a map v: K*-->Z such that 

    (1) v(xy)=v(x)+v(y) for any x, y in K*

    (2) v(x+y) geq min{v(x), v(y)}

    (3) v(K*) neq 0.

    Remark

    (1) From (3), v(K*)=nZ for some n geq 1 (hence dividing by n we can assume that v is surjective).

    (2) Put v(0)=+infty

    (3) Given v, we can define ||x||_v=r^{v(x)} (so discrete valuation gives a non-Archimedean norm) for some 0non-Archimedean normed field.

    Definition

    Let A be a commutative integral ring, then we call A a DVR if it is both

    (1) local 

    (2) PID

    Remark

    For a DVR A, there exists an element pi in A such that the unique maximal m=(pi). Notice that A - m = U(A) (invertible elements of A), as m+x=(1) for any x in A-m. Hence for any x in A, x = upi^n for some u in U(A) and n in N (if x is not in U(A), then it is in m, hence equals pi^n up to multiplying by a unit).

    Definition

    Any generator of m is called a uniformiser. 

    Theorem 8.1 ("equivalence" between DVR and discrete valuation)

    (1) Let K be a field, v be a discrete valuation on K.

    Let Av={ x in K, v(x) geq 0 }. Then Av is a DVR. 

    Namely, assume that v(K*)=Z, then we have

    (1.1) m_v={ x in K; v(x) geq 1 } is the unique nonzero prime.

    (1.2) pi in Av is a uniformizer if and only if v(pi)=1.

    (1.3) U(Av)={ x in K*; v(x)=0 } 

    (1.4) Frac(Av)=K.

    (2) Conversely, let A be a DVR and K=Frac A, then 

    (2.1) Any x in K* can be written x=u pi^n with u in U(A) and n in Z.

    (2.2) The map v: K* --> Z; v(x)=n (assume x=u pi^n) is a discrete valuation on K. (moreover as in (1), the discrete valuation induces an Av and m_v, with no surprise indeed Av=A and m_v=m).

    Notation

    Let K be a field with v a discrete valuation on K. Then we have defined Av as preceding. Now put K_v = the completion of K w.r.t ||-||_v, and O_v its valuation ring (defined in section 7 for any field endowed with a non-Archimedean norm).

    Proposition 8.2 (the relation of discrete on a field and on its completion w.r.t the corresponding norm)

    (1) v can be extended to a discrete valuation on K_v (v was in priori only a discrete valuation on K: for example v on Q can extend to v on Q_p).

    (2) O_v is the topological closure of Av in K_v and the maximal ideal m_v subset O_v is the topological closure of mAv (for example Z_p is the closure of Z_(p) in Q_p). 

    (3)  The residue field 

    k_v=O_v/m_v and k_{A_v}=A_v/mA_v coincide.

    Basic Examples: (application of Prop 8.2)

    (1) p-adic valuation

    In Q, we can define v_p: Q* --> Z as imagined, Z_(p) is the valuation ring with p a uniformiser and pZ_(p) the unique maximal ideal. 

    One can extend v_p to Q_p and denote Z_p the ring of integers of Q_p, which is the closure of Z_(p) in Q_p.

    One has Z_p/pZ_p =Z/pZ=F_p.

    (2) K=k(X).



    18 min
  • Non-Archimedean norm (results) #ALNT-LB 1.7.R Chapter 1 Section 7 #Algebraic Number Theory # Lecture note Benois

    Let (K, ||-||_K) be a normed field equipped with a non-Archimedean norm ||-||_K.

    Proposition 7.1

    (1) The set O={ x in K;  ||x||_K leq 1 }  is a commutative ring.

    (2) m={ x in K; ||x||_K<1 } is the unique maximal ideal of O. (hence O is a local ring)

    and U(K)={ x in K; ||x||_K=1 } is the group of units of O.

    (3) K = Frac O, and O is integrally closed in K.

    Proposition 7.2 

    Sum_{k=1}^infty a_k, a_k in K, converges in K if and only if lim_{k --> +infty} a_k = 0.

    Proposition 7.3 (Hensel's Lemma)

    (Some sufficient conditions for existence of root of polynomial with coefficient in a complete field with non-Archimedean norm)

    Let (K, ||-||_K) be a normed field equipped with a non-Archimedean norm ||-||_K. Let f(X) in O_K[X] and  a_0 in K be an element such that 

    ||f(a_0)||_K < || f'(a_0)||_K^2.

    Then the sequence

    a_{i+1}=a_i - f(a_i)/f'(a_i), i geq 0

    converges in K to a root of f(X).

    (We can't guarantee, in general, that Newton's approximation always converges. But Hensel's lemma tells us that it always converge for non-Archimedean case. So in some sense, non-Archimedean is easier.)

    Proposition 7.4

    Let f^bar(X)in k_K[X], non-zero, and let alpha^bar in k_K be a simple root of f^bar(X). Let f(X) in O_K[X] be such that f^bar(X) congruent f(X) mod m_K. Then there exists a unique root alpha in O_K of f(X) such that alpha^bar congruent alpha mod m_K.

    10 min
  • Extension of norms (results) #ALNT-LB 1.6.R Chapter 1 Section 6 #Algebraic Number Theory #Lecture note Benois

    Theorem 6.1 

    (1) ||-||_w1, ... , ||_||_wn are exactly all extensions of ||-||_w to L.

    (2) Sum_{i=1}^m[ L_wi : K_v ] = [ L : K ].

    Corollary 6.2

    For any w|v, we have a map 

    Hom_{Kv}(Lw, Kv^bar) ---> Hom_K(L, K^bar)

    and the following map is a bijection:

    U_{w|v} Hom_{Kv}(Lw, Kv^bar) ---> Hom_K(L, K^bar).

    Corollary 6.3

    N_{L/K}(x)=Prod_{w|v} N_{Lw/Kv}(x).

    Tr_{L/K}(x)=Sum_{w|v} Tr_{Lw/Kv}(x).

    Corollary 6.4

    For any x in L, ||N_{L/K}(x)||_v=Prod_{w|v} ||x||_w^[Lw : Kv].

    43 min
  • Descriminants; Traces and norms of field extensions #Appendix

    Corollary 1.10.5 Assume L/K is not separable. Then Tr_{L/K}=0.

    proof: Let x in L with x_i the roots (in some algebraic closure of K, counting multiplicities) of its minimal polynomial P over K.

    case (1) If x is SEPARABLE over K.

    case (2) If x is not separable over K.

    Example 1.10.6

    (1) Let K be a field, x algebraic over K and P(X)=X^n+a_1X+...+a_n in K[X] its minimal polynomial, then

    Tr_{K(x)/K}(x)=-a_1

    N_{K(x)/K}(x)=(-1)^na_n

    chi_{x,L/K}=P.

    (2) If L/K is a SEPARABLE finite extension, K^bar an algebraic closure of K and Hom_{K-alg}(L,K^bar)={sigma_1, ... , sigma_d}, we have d=[L : K], and 

    Tr_{L/K}(x)=Sum sigma_i(x)

    N_{L/K}(x)=Sum sigma_i(x)


    Corollary 1.10.7

    Let A be an integrally closed domain, K = FracA, L/K a finite extension and B the integral closure of A in L. If b in B, then Tr_{L/K}(b), N_{L/K}(b) in A and chi_{b, L/K} in A[X] and we have b in B* if and only if N_{L/K}(x) in A*. 

    39 min
  • Trace and norms (definition part) #Appendix

    Definition 1.10.2  Let A be a ring and M a free A-module of finite rank and f in End_A(M). If B^tilde is an A-basis of M, we can describe f by its matrix (a_{i,j}) in B^tilde. The trace, the determinant and the characteristic polynomial of f are 

    Tr(f)=  Sum_i a_{i,i} in A,

    det(f) = det (a_{i,j}) in A,

    chi_f(X) = det (XI_n-(a_{i,j})) in  A[X].

    Proposition 1.10.4 Let L/K be a finite field extension, x in L and x_1, ..., x_n the roots (in some algebraic closure K^bar of K, counted with multiplicities) of the minimal polynomial P of x over K. Then

     Tr_{L/K}(x)= [L : K(x)] Sum_i x_i,

    N_{L/K}(x) = (Prod x_i)^[L:K(x)],

    chi_{x, L/K} = P^^[L:K(x)].

    59 min
  • completion and extension of norms (detailed proof) #ALNT-LB 1.4-1.5.DP Chapter 1 Section 4-5 #Algebraic Number Theory # Lecture note Benois

    Definition A normed field (K, ||-||_K) is complete if every Cauchy sequence converges in K.

    Theorem 4.1 Let (K, ||-||_K) be a normed field, then there exists a unique field K' (up to isometry) containing K that has a norm ||-||_K' extending ||-||_K and being complete for this norm.

    Theorem 5.1 Let (K, ||-||_K) be a complete normed field and L/K an algebraic extension of dimension n, then there exists a unique norm extension over L, given by ||x||_L=||N_{L/K}(x)||_K^{1/n} for any x in L.

    Theorem 5.2 Let (K, ||-||_K) be a complete normed field and let V be a finite vector space over K. Then

    (1) Any two norms on V are equivalent.

    (2) V is complete for any norm.

    Corollary1 Let (K, ||-||_K) be a complete normed field and let L/K be a finite Galois extension, then Gal(L/K) acts continuously on L and it preserves  norm.

    Corollary 2 Let (K, ||-||_K) be a complete normed field and let L/K be a separable extension, then N_{L/K}: L --> K and  Tr_{L/K}: L --> K are both continuous.

    Remark that N_{L/K}(x)=Prod sigma_i(x) (for all possible sigma_i L/K embeds into K^bar/K) and Tr_{L/K}(x)=Prod sigma_i(x) (for all possible sigma_i L/K embeds into K^bar/K).

    45 min
  • Completion and extension of norms (results) #ALNT-LB 1.4-1.5.R Chapter 1 Section 4-5 #Algebraic Number Theory # Lecture note Benois

    Definition A normed field (K, ||-||_K) is complete if every Cauchy sequence converges in K.

    Theorem 4.1 Let (K, ||-||_K) be a normed field, then there exists a unique field K' (up to isometry) containing K that has a norm ||-||_K' extending ||-||_K and being complete for this norm. 

    Theorem 5.1 Let (K, ||-||_K) be a complete normed field and L/K an algebraic extension of dimension n, then there exists a unique norm extension over L, given by ||x||_L=||N_{L/K}(x)||_K^{1/n} for any x in L.

    Theorem 5.2 Let (K, ||-||_K) be a complete normed field and let V be a finite vector space over K. Then

    (1) Any two norms on V are equivalent.

    (2) V is complete for any norm.

    Corollary1 Let (K, ||-||_K) be a complete normed field and let L/K be a finite Galois extension, then Gal(L/K) acts continuously on L and it preserves  norm. 

    Corollary 2 Let (K, ||-||_K) be a complete normed field and let L/K be a separable extension, then N_{L/K}: L --> K and  Tr_{L/K}: L --> K are both continuous.

    20 min
  • The approximation theorem (detailed proof) #ALNT-LB 1.3.DP Chapter 1 Section 3 #Algebraic Number Theory # Lecture note Benois

    Theorem 3.1 

    Let ||-||_1, ..., ||-||_n be inequivalent norms on K, and let a_1, ..., a_n be arbitrary elements of K. Then for any epsilon>0, there exists a in K such that ||a-a_i||_i < epsilon. 

    Lemma

    Let ||-||_1, ..., ||-||_n be inequivalent norms on K, then there exists t in K such that ||t||_1>1 and ||t||_i < 1 for i in {2, ..., n}. 

    Question:

    1. Easy proof of the lemma for n=2 case?

    2. A rigorous  proof that ||t^N/(1+t^N)|| tends to 1 when ||t||>1 ? (should use triangular inequality)

    3. Relation of the theorem with the Chinese remainder theorem?

    27 min

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