On the Rational Approximations to the Powers of an Algebraic Number: Solution of Two Problems of Mahler and Mend S France

Total Page:16

File Type:pdf, Size:1020Kb

On the Rational Approximations to the Powers of an Algebraic Number: Solution of Two Problems of Mahler and Mend S France Acta Math., 193 (2004), 175 191 (~) 2004 by Institut Mittag-Leffier. All rights reserved On the rational approximations to the powers of an algebraic number: Solution of two problems of Mahler and Mend s France by PIETRO CORVAJA and UMBERTO ZANNIER Universith di Udine Scuola Normale Superiore Udine, Italy Pisa, Italy 1. Introduction About fifty years ago Mahler [Ma] proved that if ~> 1 is rational but not an integer and if 0<l<l, then the fractional part of (~n is larger than l n except for a finite set of integers n depending on ~ and I. His proof used a p-adic version of Roth's theorem, as in previous work by Mahler and especially by Ridout. At the end of that paper Mahler pointed out that the conclusion does not hold if c~ is a suitable algebraic number, as e.g. 1 (1 + x/~ ) ; of course, a counterexample is provided by any Pisot number, i.e. a real algebraic integer c~>l all of whose conjugates different from cr have absolute value less than 1 (note that rational integers larger than 1 are Pisot numbers according to our definition). Mahler also added that "It would be of some interest to know which algebraic numbers have the same property as [the rationals in the theorem]". Now, it seems that even replacing Ridout's theorem with the modern versions of Roth's theorem, valid for several valuations and approximations in any given number field, the method of Mahler does not lead to a complete solution to his question. One of the objects of the present paper is to answer Mahler's question completely; our methods will involve a suitable version of the Schmidt subspace theorem, which may be considered as a multi-dimensional extension of the results mentioned by Roth, Mahler and Ridout. We state at once our first theorem, where as usual we denote by IIx]] the distance of the complex number x from the nearest integer in Z, i.e. Ilxll := min{Jx-mJ : mCZ}. 176 P. CORVAJA AND U. ZANNIER THEOREM 1. Let a>l be a real algebraic number and let 0<l<1. Suppose that IIanll < In for infinitely many natural numbers n. Then there is a positive integer d such that a d is a Pisot number. In particular, a is an algebraic integer. We remark that the conclusion is best possible. For assume that for some positive integer d, /3:--a d is a Pisot number. Then for every positive multiple n of d we have Ilanll << I n, where l is the dth root of the maximum absolute value of the conjugates of/3 different from/3. Here, Mahler's example with the golden ratio is typical. Also, the conclusion is not generally true without the assumption that a is algebraic; for this see the appendix. The present application of the subspace theorem seems different from previous ones, and occurs in Lemma 3 below. Related methods actually enable us to answer as well a question raised by Mend~s France about the length of the periods of the continued fractions for a n, where a is now a quadratic irrational; this appears as Problem 6 in [Me]. We shall prove, more generally, the following result: THEOREM 2. Let a>0 be a real quadratic irrational. If a is neither the square root of a rational number, nor a unit in the ring of integers of Q(a), then the period length of the continued fraction for a n tends to infinity with n. If a is the square root of a rational number, the period length of the continued fraction for a 2n+1 tends to infinity. If a is a unit, the period length of the continued fraction for a n is bounded. Clearly, if a is the square root of a rational number, then the continued frac- tion for a ~n is finite,(1) so Theorem 2 gives a complete answer to the problem of Mendbs France. The main tool in the proof of both theorems is the following new lower bound for the fractional parts of S-units in algebraic number fields. We first need a definition: Definition. We call a (complex) algebraic number a a pseudo-Pisot number if (i) lal > 1 and all its conjugates have (complex) absolute value strictly less than 1; (ii) a has an integral trace: WrQ(a)/Q(a)EZ. Of course, pseudo-Pisot numbers are "well approximated" by their trace, hence are good candidates for having a small fractional part compared to their height. The algebraic integers among the pseudo-Pisot numbers are just the usual Pisot numbers. We shall prove the following result: (1) Its length tends to infinity by a result of Pourchet, proved in greater generality in [CZ]. ON THE RATIONAL APPROXIMATIONS TO THE POWERS OF AN ALGEBRAIC NUMBER 177 MAIN THEOREM. Let FcQ • be a finitely generated multiplicative group of algebraic numbers, let 5cQ • be a non-zero algebraic number and let ~>0 be fixed. Then there are only finitely many pairs (q,u)~z• with d=[Q(u):Q] such that 1Saul>l, 5qu is not a pseudo-Pisot number and o < 115q ll < H(u)-*q -d-*. (1.1) Note again that, conversely, starting with a Pisot number a and taking q=l and u=(f ~ for n=l, 2, ... produces an infinite sequence of solutions to 0< Ilqull <H(u) -~ for a suitable s>0. The above main theorem can be viewed as a Thue-Roth inequality with "moving target", as the theorem in [CZ], where we considered quotients of power sums with in- tegral roots instead of elements of a finitely generated multiplicative group. The main application of the theorem in [CZ] also concerned continued fractions, as for our Theo- ram 2. 2. Proofs We shall use the following notation: Let K be a number field, embedded in C and Galois over Q. We denote by M/( (rasp. M~) the set of places (resp. archimedean places) of K; for each place v we denote by 1 Iv the absolute value corresponding to v, normalized with respect to K; by this we mean that if vEM~ then there exists an automorphism aEGal(K/Q) of K such that, for all xEK, Ixlv = I (x)l where d(a)=l if a(K)=KcR, and d(a)=2 otherwise (note that d(a) is now constant since K/Q is Galois). Non-archimedean absolute values are normalized accordingly, so that the product formula holds and the absolute Weit height reads H(x) = H max{l, Ixlv}. vEMK For a vector x=(xl, ...,Xn)CK n and a place vEMK we shall denote by I]xl]~ the v-norm of x, namely,(2) HxHv := max{ Ix1 Iv,..., Ixn M, (2) We believe that there will be no confusion with the previously recalled symbol Ilxll for the distance to the nearest integer. 178 P. CORVAJA AND U. ZANNIER and by H(x) the projective height H(x) = H max{lxlIv,..., Ixnl~}. vEMK We begin by proving the main theorem. First of all notice that, by enlarging if necessary the multiplicative group F, we can reduce to the situation where FcK • is the group of S-units: F= O~ = {uEK:iu[~ = 1 for all v~S} of a suitable number field K, Galois over Q, with respect to a suitable finite set of places S, containing the arehimedean ones and stable under Gatois conjugation. Our first lemma can be easily deduced from a theorem of Evertse, which in turn was obtained as an application of the already mentioned subspace theorem; we prefer to give a proof for completeness. LEMMA 1. Let K and S be as before, al, ...,an be distinct automorphisms of K, /kl,...,An be non-zero elements of K, ~>0 be a positive real number and wCS be a distinguished place. Let =~C O~ be an infinite set of solutions uEO~ of the inequality [.~lO'l (U)T... T /~n(Tn(U)lw < max{]crl (u)iw, ..., [(rn(U)[w}H(u) -~. Then there exists a non-trivial linear relation of the form alal(u)+...+anan(u)=O, aiEK, satisfied by infinitely many elements of =. Proof. Let ~ be as in the lemma. Going to an infinite subset of ~ we may assume that lal(U)lw=max{tal(U)l~, ..., lan(u)l~,} for all the involved u's. Let us consider, for each yES, n linear forms Lv,1, ..., L~,n in n variables x=(xl, ..., xn) as follows: Put L~,I (x) := Alxl +...+Anxn while for (v,i)ES• with (v,i)#(w, l), put L,,~(x)=xi. Note that the linear forms L~,I, ..., Lv,n are indeed linearly independent. Now put x = (ol(u), :.., ~d(u)) e (o~) d and consider the double product H l~ ILv,dx)lv ON THE RATIONAL APPROXIMATIONS TO THE POWERS OF AN ALGEBRAIC NUMBER 179 By multiplying and dividing by Ix~l~=l~(u)l~, and using the fact that the coordinates of x are S-units, we obtain the equality H let In~,~(x)l~ _ iAl~l(U)+...+Anan(U)lw i~(u)l;~H(x)_n. ~s~=~ Ilxll~ Since uC~, we have I~l(U)+...+ ~na~(U)lw I~I(U)I~ x < H(u)-~, and then v~s~=lII ILv,I)xllv (x)lv < The height of the point x=(al(u), ..., ad(U)) is easily compared with the height of u by the estimate H(x) ~< H(u) [~;:q] . Hence the above upper bound for the double product also gives H fl IL~"(x)lv H(x)-n-~/[K:Q]" Now, an application of the subspace theorem in the form given in e.g.
Recommended publications
  • MATH 304: CONSTRUCTING the REAL NUMBERS† Peter Kahn
    MATH 304: CONSTRUCTING THE REAL NUMBERSy Peter Kahn Spring 2007 Contents 4 The Real Numbers 1 4.1 Sequences of rational numbers . 2 4.1.1 De¯nitions . 2 4.1.2 Examples . 6 4.1.3 Cauchy sequences . 7 4.2 Algebraic operations on sequences . 12 4.2.1 The ring S and its subrings . 12 4.2.2 Proving that C is a subring of S . 14 4.2.3 Comparing the rings with Q ................... 17 4.2.4 Taking stock . 17 4.3 The ¯eld of real numbers . 19 4.4 Ordering the reals . 26 4.4.1 The basic properties . 26 4.4.2 Density properties of the reals . 30 4.4.3 Convergent and Cauchy sequences of reals . 32 4.4.4 A brief postscript . 37 4.4.5 Optional: Exercises on the foundations of calculus . 39 4 The Real Numbers In the last section, we saw that the rational number system is incomplete inasmuch as it has gaps or \holes" where some limits of sequences should be. In this section we show how these holes can be ¯lled and show, as well, that no further gaps of this kind can occur. This leads immediately to the standard picture of the reals with which a real analysis course begins. Time permitting, we shall also discuss material y°c May 21, 2007 1 in the ¯fth and ¯nal section which shows that the rational numbers form only a tiny fragment of the set of all real numbers, of which the greatest part by far consists of the mysterious transcendentals.
    [Show full text]
  • Algebraic Number Theory
    Algebraic Number Theory William B. Hart Warwick Mathematics Institute Abstract. We give a short introduction to algebraic number theory. Algebraic number theory is the study of extension fields Q(α1; α2; : : : ; αn) of the rational numbers, known as algebraic number fields (sometimes number fields for short), in which each of the adjoined complex numbers αi is algebraic, i.e. the root of a polynomial with rational coefficients. Throughout this set of notes we use the notation Z[α1; α2; : : : ; αn] to denote the ring generated by the values αi. It is the smallest ring containing the integers Z and each of the αi. It can be described as the ring of all polynomial expressions in the αi with integer coefficients, i.e. the ring of all expressions built up from elements of Z and the complex numbers αi by finitely many applications of the arithmetic operations of addition and multiplication. The notation Q(α1; α2; : : : ; αn) denotes the field of all quotients of elements of Z[α1; α2; : : : ; αn] with nonzero denominator, i.e. the field of rational functions in the αi, with rational coefficients. It is the smallest field containing the rational numbers Q and all of the αi. It can be thought of as the field of all expressions built up from elements of Z and the numbers αi by finitely many applications of the arithmetic operations of addition, multiplication and division (excepting of course, divide by zero). 1 Algebraic numbers and integers A number α 2 C is called algebraic if it is the root of a monic polynomial n n−1 n−2 f(x) = x + an−1x + an−2x + ::: + a1x + a0 = 0 with rational coefficients ai.
    [Show full text]
  • Approximation to Real Numbers by Algebraic Numbers of Bounded Degree
    Approximation to real numbers by algebraic numbers of bounded degree (Review of existing results) Vladislav Frank University Bordeaux 1 ALGANT Master program May,2007 Contents 1 Approximation by rational numbers. 2 2 Wirsing conjecture and Wirsing theorem 4 3 Mahler and Koksma functions and original Wirsing idea 6 4 Davenport-Schmidt method for the case d = 2 8 5 Linear forms and the subspace theorem 10 6 Hopeless approach and Schmidt counterexample 12 7 Modern approach to Wirsing conjecture 12 8 Integral approximation 18 9 Extremal numbers due to Damien Roy 22 10 Exactness of Schmidt result 27 1 1 Approximation by rational numbers. It seems, that the problem of approximation of given number by numbers of given class was firstly stated by Dirichlet. So, we may call his theorem as ”the beginning of diophantine approximation”. Theorem 1.1. (Dirichlet, 1842) For every irrational number ζ there are in- p finetely many rational numbers q , such that p 1 0 < ζ − < . q q2 Proof. Take a natural number N and consider numbers {qζ} for all q, 1 ≤ q ≤ N. They all are in the interval (0, 1), hence, there are two of them with distance 1 not exceeding q . Denote the corresponding q’s as q1 and q2. So, we know, that 1 there are integers p1, p2 ≤ N such that |(q2ζ − p2) − (q1ζ − p1)| < N . Hence, 1 for q = q2 − q1 and p = p2 − p1 we have |qζ − p| < . Division by q gives N ζ − p < 1 ≤ 1 . So, for every N we have an approximation with precision q qN q2 1 1 qN < N .
    [Show full text]
  • Mop 2018: Algebraic Conjugates and Number Theory (06/15, Bk)
    MOP 2018: ALGEBRAIC CONJUGATES AND NUMBER THEORY (06/15, BK) VICTOR WANG 1. Arithmetic properties of polynomials Problem 1.1. If P; Q 2 Z[x] share no (complex) roots, show that there exists a finite set of primes S such that p - gcd(P (n);Q(n)) holds for all primes p2 = S and integers n 2 Z. Problem 1.2. If P; Q 2 C[t; x] share no common factors, show that there exists a finite set S ⊂ C such that (P (t; z);Q(t; z)) 6= (0; 0) holds for all complex numbers t2 = S and z 2 C. Problem 1.3 (USA TST 2010/1). Let P 2 Z[x] be such that P (0) = 0 and gcd(P (0);P (1);P (2);::: ) = 1: Show there are infinitely many n such that gcd(P (n) − P (0);P (n + 1) − P (1);P (n + 2) − P (2);::: ) = n: Problem 1.4 (Calvin Deng). Is R[x]=(x2 + 1)2 isomorphic to C[y]=y2 as an R-algebra? Problem 1.5 (ELMO 2013, Andre Arslan, one-dimensional version). For what polynomials P 2 Z[x] can a positive integer be assigned to every integer so that for every integer n ≥ 1, the sum of the n1 integers assigned to any n consecutive integers is divisible by P (n)? 2. Algebraic conjugates and symmetry 2.1. General theory. Let Q denote the set of algebraic numbers (over Q), i.e. roots of polynomials in Q[x]. Proposition-Definition 2.1. If α 2 Q, then there is a unique monic polynomial M 2 Q[x] of lowest degree with M(α) = 0, called the minimal polynomial of α over Q.
    [Show full text]
  • Algebraic Numbers, Algebraic Integers
    Algebraic Numbers, Algebraic Integers October 7, 2012 1 Reading Assignment: 1. Read [I-R]Chapter 13, section 1. 2. Suggested reading: Davenport IV sections 1,2. 2 Homework set due October 11 1. Let p be a prime number. Show that the polynomial f(X) = Xp−1 + Xp−2 + Xp−3 + ::: + 1 is irreducible over Q. Hint: Use Eisenstein's Criterion that you have proved in the Homework set due today; and find a change of variables, making just the right substitution for X to turn f(X) into a polynomial for which you can apply Eisenstein' criterion. 2. Show that any unit not equal to ±1 in the ring of integers of a real quadratic field is of infinite order in the group of units of that ring. 3. [I-R] Page 201, Exercises 4,5,7,10. 4. Solve Problems 1,2 in these notes. 3 Recall: Algebraic integers; rings of algebraic integers Definition 1 An algebraic integer is a root of a monic polynomial with rational integer coefficients. Proposition 1 Let θ 2 C be an algebraic integer. The minimal1 monic polynomial f(X) 2 Q[X] having θ as a root, has integral coefficients; that is, f(X) 2 Z[X] ⊂ Q[X]. 1i.e., minimal degree 1 Proof: This follows from the fact that a product of primitive polynomials is again primitive. Discuss. Define content. Note that this means that there is no ambiguity in the meaning of, say, quadratic algebraic integer. It means, equivalently, a quadratic number that is an algebraic integer, or number that satisfies a quadratic monic polynomial relation with integral coefficients, or: Corollary 1 A quadratic number is a quadratic integer if and only if its trace and norm are integers.
    [Show full text]
  • Solution Set for the Homework Problems
    SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5. Problem 8. Prove that if x and y are real numbers, then 2xy x2 + y2. ≤ Proof. First we prove that if x is a real number, then x2 0. The product of two positive numbers is always positive, i.e., if x 0≥ and y 0, then ≥ ≥ xy 0. In particular if x 0 then x2 = x x 0. If x is negative, then x is positive,≥ hence ( x)2 ≥0. But we can conduct· ≥ the following computation− by the associativity− and≥ the commutativity of the product of real numbers: 0 ( x)2 = ( x)( x) = (( 1)x)(( 1)x) = ((( 1)x))( 1))x ≥ − − − − − − − = ((( 1)(x( 1)))x = ((( 1)( 1))x)x = (1x)x = xx = x2. − − − − The above change in bracketting can be done in many ways. At any rate, this shows that the square of any real number is non-negaitive. Now if x and y are real numbers, then so is the difference, x y which is defined to be x + ( y). Therefore we conclude that 0 (x + (−y))2 and compute: − ≤ − 0 (x + ( y))2 = (x + ( y))(x + ( y)) = x(x + ( y)) + ( y)(x + ( y)) ≤ − − − − − − = x2 + x( y) + ( y)x + ( y)2 = x2 + y2 + ( xy) + ( xy) − − − − − = x2 + y2 + 2( xy); − adding 2xy to the both sides, 2xy = 0 + 2xy (x2 + y2 + 2( xy)) + 2xy = (x2 + y2) + (2( xy) + 2xy) ≤ − − = (x2 + y2) + 0 = x2 + y2. Therefore, we conclude the inequality: 2xy x2 + y2 ≤ for every pair of real numbers x and y. ♥ 1 2SOLUTION SET FOR THE HOMEWORK PROBLEMS Page 5.
    [Show full text]
  • Proof Methods and Strategy
    Proof methods and Strategy Niloufar Shafiei Proof methods (review) pq Direct technique Premise: p Conclusion: q Proof by contraposition Premise: ¬q Conclusion: ¬p Proof by contradiction Premise: p ¬q Conclusion: a contradiction 1 Prove a theorem (review) How to prove a theorem? 1. Choose a proof method 2. Construct argument steps Argument: premises conclusion 2 Proof by cases Prove a theorem by considering different cases seperately To prove q it is sufficient to prove p1 p2 … pn p1q p2q … pnq 3 Exhaustive proof Exhaustive proof Number of possible cases is relatively small. A special type of proof by cases Prove by checking a relatively small number of cases 4 Exhaustive proof (example) Show that n2 2n if n is positive integer with n <3. Proof (exhaustive proof): Check possible cases n=1 1 2 n=2 4 4 5 Exhaustive proof (example) Prove that the only consecutive positive integers not exceeding 50 that are perfect powers are 8 and 9. Proof (exhaustive proof): Check possible cases Definition: a=2 1,4,9,16,25,36,49 An integer is a perfect power if it a=3 1,8,27 equals na, where a=4 1,16 a in an integer a=5 1,32 greater than 1. a=6 1 The only consecutive numbers that are perfect powers are 8 and 9. 6 Proof by cases Proof by cases must cover all possible cases. 7 Proof by cases (example) Prove that if n is an integer, then n2 n. Proof (proof by cases): Break the theorem into some cases 1.
    [Show full text]
  • Lie Group Definition. a Collection of Elements G Together with a Binary
    Lie group Definition. A collection of elements G together with a binary operation, called ·, is a group if it satisfies the axiom: (1) Associativity: x · (y · z)=(x · y) · z,ifx, y and Z are in G. (2) Right identity: G contains an element e such that, ∀x ∈ G, x · e = x. (3) Right inverse: ∀x ∈ G, ∃ an element called x−1, also in G, for which x · x−1 = e. Definition. A group is abelian (commutative) of in addtion x · y = y · x, ∀x, y ∈ G. Definition. A topological group is a topological space with a group structure such that the multiplication and inversion maps are contiunuous. Definition. A Lie group is a smooth manifold G that is also a group in the algebraic sense, with the property that the multiplication map m : G × G → G and inversion map i : G → G, given by m(g, h)=gh, i(g)=g−1, are both smooth. • A Lie group is, in particular, a topological group. • It is traditional to denote the identity element of an arbitrary Lie group by the symbol e. Example 2.7 (Lie Groups). (a) The general linear group GL(n, R) is the set of invertible n × n matrices with real entries. — It is a group under matrix multiplication, and it is an open submanifold of the vector space M(n, R). — Multiplication is smooth because the matrix entries of a product matrix AB are polynomials in the entries of A and B. — Inversion is smooth because Cramer’s rule expresses the entries of A−1 as rational functions of the entries of A.
    [Show full text]
  • MAT 240 - Algebra I Fields Definition
    MAT 240 - Algebra I Fields Definition. A field is a set F , containing at least two elements, on which two operations + and · (called addition and multiplication, respectively) are defined so that for each pair of elements x, y in F there are unique elements x + y and x · y (often written xy) in F for which the following conditions hold for all elements x, y, z in F : (i) x + y = y + x (commutativity of addition) (ii) (x + y) + z = x + (y + z) (associativity of addition) (iii) There is an element 0 ∈ F , called zero, such that x+0 = x. (existence of an additive identity) (iv) For each x, there is an element −x ∈ F such that x+(−x) = 0. (existence of additive inverses) (v) xy = yx (commutativity of multiplication) (vi) (x · y) · z = x · (y · z) (associativity of multiplication) (vii) (x + y) · z = x · z + y · z and x · (y + z) = x · y + x · z (distributivity) (viii) There is an element 1 ∈ F , such that 1 6= 0 and x·1 = x. (existence of a multiplicative identity) (ix) If x 6= 0, then there is an element x−1 ∈ F such that x · x−1 = 1. (existence of multiplicative inverses) Remark: The axioms (F1)–(F-5) listed in the appendix to Friedberg, Insel and Spence are the same as those above, but are listed in a different way. Axiom (F1) is (i) and (v), (F2) is (ii) and (vi), (F3) is (iii) and (vii), (F4) is (iv) and (ix), and (F5) is (vii). Proposition. Let F be a field.
    [Show full text]
  • Chapter 9: Roots and Irrational Numbers
    Chapter 9: Roots and Irrational Numbers Index: A: Square Roots B: Irrational Numbers C: Square Root Functions & Shifting D: Finding Zeros by Completing the Square E: The Quadratic Formula F: Quadratic Equations G: Cube Roots Name: _____________________________________________________ Date: ____________________ Period: ______ Algebra Square Roots 9A Square roots, cube roots, and higher level roots are important mathematical tools because they are the inverse operations to the operations of squaring and cubing. In this unit we will study these operations, as well as numbers that come from using them. First, some basic review of what you’ve seen before. Exercise #1: Find the value of each of the following principal square roots. Write a reason for your answer in terms of a multiplication equation. (a) (b) (c) (d) (e) (f) It is generally agreed upon that all positive, real numbers have two square roots, a positive one and a negative one. We simply designate which one we want by either including a negative sign or leaving it off. Exercise #2: Give all square roots of each of the following numbers. (a) 4 (b) 36 (c) Exercise #3: Given the function , which of the following is the value of ? (1) 22 (3) 16 (2) 5 (4) 7 Square roots have an interesting property when it comes to multiplication. Lets discover that property. Exercise #4: Find the value of each of the following products. (a) (b) (c) (d) What you should notice in the last exercise is the following important property of square roots. MULTIPLICATON PROPERTY OF SQUARE ROOTS 1. LIKEWISE 2. One1. obviousa b use for a b this is to multiply two “unfriendlylikewise” square roots to get a nice2.
    [Show full text]
  • Totally Ramified Primes and Eisenstein Polynomials
    TOTALLY RAMIFIED PRIMES AND EISENSTEIN POLYNOMIALS KEITH CONRAD 1. Introduction A (monic) polynomial in Z[T ], n n−1 f(T ) = T + cn−1T + ··· + c1T + c0; is Eisenstein at a prime p when each coefficient ci is divisible by p and the constant term c0 is not divisible by p2. Such polynomials are irreducible in Q[T ], and this Eisenstein criterion for irreducibility is the way essentially everyone first meets Eisenstein polynomials. Here we will show Eisenstein polynomials at a prime p give us useful information about p-divisibility of coefficients of a power basis and ramification of p in a number field. Let K be a number field, with degree n over Q. A primep number p is said to be totally n 2 ramified in Kpwhen pOK = p . For example, in Z[i] and Z[ p−5] we have (2) = (1 + i) and (2) = (2; 1 + −5)2, so 2 is totally ramified in Q(i) and Q( −5). 2. Eisenstein polynomials and p-divisibility of coefficients Our first use of Eisenstein polynomials will be to extract information about coefficients for algebraic integers in the power basis generated by the root of an Eisenstein polynomial. Lemma 2.1. Let K=Q be a number field with degree n. Assume K = Q(α), where α 2 OK and its minimal polynomial over Q is Eisenstein at p. For a0; a1; : : : ; an−1 2 Z, if n−1 (2.1) a0 + a1α + ··· + an−1α ≡ 0 mod pOK ; then ai ≡ 0 mod pZ for all i. Proof. Assume for some j 2 f0; 1; : : : ; n − 1g that ai ≡ 0 mod pZ for i < j (this is an empty condition for j = 0).
    [Show full text]
  • Direct Proof  Proof by Contraposition  Proof by Contradiction
    Mathematical theorems are often stated in the form of an implication Example: If x > y, where x and y are positive real numbers, then x 2 > y 2. ∀ x,y [(x > 0) ∧∧∧ (y > 0) ∧∧∧ (x > y) → (x 2 > y 2)] ∀ x,y P(x,y) → Q(x,y) We will first discuss three applicable proof methods (Section 1.6): Direct proof Proof by contraposition Proof by contradiction Direct proof In a direct proof , we prove p → q by showing that if p is true , then q must necessarily be true Example: Prove that if n is an odd integer, then n 2 is an odd integer. Proof: Assume that n is odd. That is n = (2k + 1) for some integer k. Note that n 2 = (2k+1) 2 = (4k 2 + 4k + 1) We can factor the above to get 2(2k 2 + 2k) + 1 Since the above quantity is one more than even number, we know that n 2 is odd. ☐ Direct proofs are not always the easiest way to prove a given conjecture. In this case, we can try proof by contraposition How does this work? Recall that p → q ≡ ¬q → ¬p Therefore, a proof of ¬q → ¬p is also a proof of p → q Proof by contraposition is an indirect proof technique since we don’t prove p → q directly. Let’s take a look at an example… Prove: If n is an integer and 3n + 2 is odd, then n is odd. First, attempt a direct proof: Assume that 3n + 2 is odd, thus 3n + 2 = 2k + 1 for some k Can solve to find that n = (2k – 1)/3 Where do we go from here?!? Now, try proof by contraposition: Assume n is even, so n = 2k for some k 3(2k) + 2 = 6k + 2 = 2(3k + 1) So, 3n + 2 is also even.
    [Show full text]