The Radicans. James B. Wilson. April 13, 2002

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1 The Radicans James B. Wilson April 13, Radices of Integers Computational mathematics has brought awareness to the use of various bases in representing integers. The standard for most number systems is decimal, which is base 10, or has radix Common examples inclue the radix for: binary, 2; octal, 8; and hexidecimal, 16. Definition 1.1 A positive integer r is a radix. A sequence α = (a k,..., a 0 ) is a radix r sequence if 0 a i < r when r > 1 and a i = 1 if r = 1, for all i = 1,..., k. The set of all radix r sequences of integers is defined as N r and is called a primitive radican of r. 2 The translation of a positive integer to a specific radix in computer science can be done recursively as follows: a i = (n k j=i+1 a j )/r i, where k = log r n. This method ensures the proper allocation of space and takes advantage of the implicit flooring done with integer division in computers. However this recursion runs backwards and so is potentially confussing. 3 The following direct and more number theoretic recursion offers a greater insight into the mechinisms behind a radix. j 1 a j r j n a i r i (mod r j+1 ) where each a j is restricted to the least residue class. Notice n = m r j + j 1 a ir i and (r j, r j+1 ) = r j so the explicit form can be given through cancelation as a j m (mod r). 1 The use of the term base is more common than radix. However due to the use of the the terms basis and bases later, base will be avoided in favor of radix. 2 The set N includes 0. 3 Running backwards allows for a program to convert the radix to a reduced precision thus is ideal for floating point situations as well. 1

2 The process terminates when j a ir i = n, which it must at some point. Regardless of the algorithm used, the output is well-defined and unique and is denoted as n r. Note the length of n r is always log r n and 1 when n = 0. Definition 1.2 An non-negative integer n is said to be expressed (in the radix r) by n r = (a k,..., a 0 ) if for all i = 1,..., k, 0 a i < r when r > 1 and a i = 1 when r = 1 (called a tally). Proposition 1.3 The map (r) : N r N, defined as α(r) = a k r k + + a 1 r + a 0, is well-defined. Furthermore r : N N r defined by n r = (a k,..., a 0 ) is the inverse map of (r), so n r(r) = n and α(r) r = α. 4 Typcially the sequence is denoted with the most significant figures to the left and the least significant to the right, just as with decimal notation. The individual coefficients a i are called r-dits, or simply radits when r is clear from the context. Definition 1.4 Given a set A N, define A r = {n r n A}. Also define n = n so that A = A and in particlar N = N. Before moving on it is important to note several techniques exist to extend the radix representation of a positive integer to its negative: simply fix a new symbol (the minus sign) in front of the number, or for the case of more complex systems, such as the p-adic integers, use infinite sequences. Most of these methods can be extended to the following construction but for simplicity negative integers are ignored. 2 Radix Algorithm The construction of a radix makes sense to other number systems as well. But as with other principles, the development of radix does not in general follow the integer model. The restrictions that make an integer like radix to exist are considered but the certain useful results can be derived even with radix representations that do not exhibit all the necessary properties. Lemma 2.1 Given a right Z-module M and a submodule S, and an integer r 0, a b (mod S) if and only if ar i br i (mod Sr i ) for all i 0. Proof: a + S = b + S if and only if (a + S)r i = (b + S)r i. Furthermore (x + S)r i = {(x + s)r i s S} = {xr i + sr i s S} = xr i + Sr i. Therefore a + S = b + S if and only if ar i + Sr i = br i + Sr i. 4 The notation mimicks the equation a/b b = a as a mnemonic. 2

3 Theorem 2.2 (Radix Algorithm) Let M be a right Z-module and S a submodule of M with transversal T r(s) and of finite index r. For every element a M there exists a unique sequence a S = {a i i N, a i T r(s)} such that for all j N. a a i r i (mod Sr j ) (1) Proof: Proceed with induction. Given an element a M there exists a unique a 0 T such that a + S = a 0 + S by the definition of T, thus verifying a a 0 (mod Sr 0 ). Presume the subsequence a 0,..., a j satisfies Equation (1), which requires a j a ir i Sr j. Thus there exists an s S such that a j a ir i = sr j. However clearly the order of sr j in M/Sr j+1 divides r. Therefore there exists a t M such that s tr (mod Sr). Once again there exists a unique a j+1 T r(s) such that a j+1 t (mod S). Applying Lemma-2.1 and substituting yields a j+1 r j+1 tr j+1 = sr j = a a i r i (mod Sr j+1 ). Therefore a S is uniquely defined. Proposition 2.3 If an element is expressible then its expression is unique. Proof: Suppose a is expressed by (a i ) i N and (b i ) i N. The transitivity of mod equivalence infers j a ir i j b ir i (mod Sr j ). If j = 0 clearly a 0 b 0 (mod S) and since both are in the transversal it follows a 0 = b 0. Proceeding with induction, assume a i = b i for all i j for some non-negative integer j. Canceling the common sum leaves a j+1 r j+1 b j+1 r j+1 (mod Sr j+1 ), and applying Lemma-2.1 reduces the expression to a j+1 b j+1 (mod S). By the same argument as above a j+1 = b j+1, so in fact (a i ) i N = (b i ) i N. When context allows, we write a S as a r and M S as M r since the integer radix is generally of greatest interest. Traditionally the sequence derived from the radix algorithm is written from left to write starting with the most significant radit. Some confussion may arrise of the radix algorithm is not used to compute the sequence. Example 2.4 In the ring Z/4, 0 = 1 4. Therefore 0 = However 0 2 (1, 0, 0) because = 4 / 0 ( ) (mod 2 3 ). 3

4 Example 2.5 Consider the radix algorithum with the chain 0 Z/r 0 Z/r Z/r where r > 1. The element (1, 0) (0, 0) (mod Z/r 0) so a 0 = 0. Next (1, 0) a 1 r + 0 = a 1 r (mod 0). Therefore (1, 0) = a 1 r yet every element of x Z/r Z/r has order dividing r so (1, 0) a 1 r for any a 1 in any transversal. Therefore Z/r Z/r is not expressible. Proposition 2.6 Every cyclic group is expressible by any subgroup of finite index greater than one. Proof: Consider the arbitrary cyclic group C = g and the subgroup S of finite index r > 1 in C. Take an arbitrary element a C. By the definition of the transversal T r(s) there exists a unique a 0 a (mod S). Assume a j a ir i (mod Sr j ) for all j k, with a i T r(s), for some k and proceed by induction. Both a and k a ir i are in C so a = gn and k a ir i = gm. Therefore g(n m) 0 (mod Sr k ) and since g is a generator of C it follows gs = B for some s Z. Therefore r k n m... hmm, requires some more work, r j could be greater than the group size. Definition 2.7 (Radix Algorithm) Let H = {H i i N} be an r-regular descending normal series so that [H i, H i+1 ] = r for all H i 0 and H 0 = G. Let φ i : {0,..., r} H i be a map such that φ i (m)h i+1 = φ i (n)h i+1 implies m = n and for each ah i+1 there exists an n such that φ i (n)h i+1 = ah. G γ0 γ G/H 1 γ 1 H 1 /H 2 γ 2 i γ H i /H i+1 i+1 where γ i (ah i ) = a φ i (a i ) where φ(a i )H i = ah i. The radix expression of x G by H is the sequence (x i ) i N where for all j N. xφ 0 (x 0 ) 1 φ j (x j ) 1 H j+1, Proposition 2.8 The radix algorithm is well-defined. Proof: When j = 0 there exists a unique x 0 such that φ 0 (x 0 )H 1 = xh 1 which verifies xφ 0 (x 0 ) 1 H 1. Assume xφ 0 (x 0 ) 1 φ j (x j ) 1 H j+1 for some set of unique x 0 through x j and proceed with induction. Once again the definition of φ j+1 provides for a unique x j+1 such that φ j+1 (x j+1 )H j+2 = x 1 φ 0 (x 0 ) 1 φ j (x j ) 1 H j+2. Therefore x j+1 φ i(x i ) 1 H j+2. By induction (x i ) i N is uniquely determined for all x G so the radix algorithm is well-defined. Definition 2.9 A regular descending normal series is a collection H = {H i i I} where H i+1 H i and [H i, H i+1 ] = [H j, H j+1 ] where H i and H j are not 0. A strongly regular descending normal series requires further that H i /H i+1 = H j /H j+1 when once again H i and H j are non-trivial. The analog exists for ascending normal series. 4

5 Definition 2.10 Given an r-regular descending normal series H with H 0 = G, the expression of G by H is called a radix expression and r is called the radix. If H is strongly regular then G is expressed base r. Theorem 2.11 If H is a strongly regular descending normal series then G is isomorphic to I G/H 1 with radix multiplication. Proof: blah. Example 2.12 Express D 4 base 2. First locate a regular descending normal series: 0 a 2 a D 4. Next define the φ i functions: φ 0 (0) = e and φ 0 (1) = b, φ 1 (0) = e and φ 1 (1) = a, and finally φ 2 (0) = e and φ 2 (1) = a 2. Now apply the radix algorithm to each element. For instance: a e = φ 0 (0) (mod a ), aφ 0 (0) 1 = ae 1 = a a = φ 1 (1) (mod a 2 ), aφ 0 (0) 1 φ 1 (1) 1 = ae 1 a 1 = e e = φ 2 (0) (mod 0), so a 2 = 010. The same works with ab: ab b = φ 0 (1) (mod a ), abφ 0 (1) 1 = ab 1 = a a = φ 1 (1) (mod a 2 ), abφ 0 (01 1 φ 1 (1) 1 = ab 1 a 1 = e e = φ 2 (0) (mod 0), and therefore ab 2 = 011. Notice that in the second step abφ 0 (1) 1 is in a so in fact the evaluation is in a / a 2 not D 4 / a 2 as required. When completed the algorithm generates the following table uniquely identifying every element. e a a 2 a 3 b ab a 2 b a 3 b D/4/ a φ 0 (0) φ 0 (0) φ 0 (0) φ 0 (0) φ 0 (1) φ 0 (1) φ 0 (1) φ 0 (1) a / a 2 φ 1 (0) φ 1 (1) φ 1 (0) φ 1 (1) φ 1 (0) φ 1 (1) φ 1 (0) φ 1 (1) a 2 /0 φ 2 (0) φ 2 (0) φ 2 (1) φ 2 (1) φ 2 (0) φ 2 (0) φ 2 (1) φ 2 (1) Notice the representation need not be unique. For instance take instead the regular descending normal series: H 3 = 0 H 2 = a 2 H 1 = {e, a 2, b, a 2 b} H 0 = D 4. 5

6 Now define: φ 0 (1) = a, φ 1 (1) = b, and φ 2 (1) = a 2. The radix algorithm now generates: e a a 2 a 3 b ab a 2 b a 3 b H 0 /H 1 φ 0 (0) φ 0 (1) φ 0 (0) φ 0 (1) φ 0 (0) φ 0 (1) φ 0 (0) φ 0 (1) H 1 /H 2 φ 1 (0) φ 1 (0) φ 1 (0) φ 1 (0) φ 1 (1) φ 1 (1) φ 1 (1) φ 1 (1) H 2 /H 3 φ 2 (0) φ 2 (0) φ 2 (1) φ 2 (1) φ 2 (0) φ 2 (0) φ 2 (1) φ 2 (1) Notice the results are simply permutations of each other: Interchanging the first two rows in the table creates the other. Proposition 2.13 If an element is expressible then its expression is unique. Proof: Suppose a is expressed by (a i ) i N and (b i ) i N. The transitivity of modular equivalence infers j φ i(a i ) j φ i(b i ) (mod H j+1 ) for all j. If j = 0 clearly φ 0 (a 0 ) φ 0 (b 0 ) (mod H 1 ) implies φ 0 (a 0 )H 1 = φ 0 (b 0 )H 1 so a 0 = b 0 by the definition of φ 0. Proceeding with induction, assume a i = b i for all i j for some non-negative integer j. Using cancelation j+1 φ i(a i ) j+1 φ i(b i ) (mod H j+2 ) reduces to φ j+1 (a j+1 ) φ j+1 (b j+1 ) (mod H j+2 ) where once again φ j+1 (a j+1 ) φ j+1 (b j+1 ) implies a j+1 = b j+1 by the definition of φ j+1. Therefore the expression of a is unique by induction. If G is of finite order n then the maximum length required to express any element in G by the radix r is log r n. Note clearly r n. Example 2.14 Take r i Z rz Z for any r > 1. Define φ i (n) = nr i. Then every element of Z is expressible. Proposition 2.15 Given a right Z-module M and a submodule S of finite index r, and take a M such that a r = {a i i N}. Let s T r(s) be the element such that s + S = S. The following are true: 1. if a = j a ir i for some j then a k = s for all k > j; 2. 0 r = s; 3. if Sr j = 0 for some j then a i = s for all i > j. Proof: Take a = j a ir i and let k > j. Clearly a a i r i = 0 = 0r k sr k = a k r k (mod Sr k ). By Lemma-2.1, a k = s. 6

7 Clearly 0 = 0 1 so condition 1 is satisfied leaving 0 S = s. Suppose Sr j = 0 for some j. Given that a j a ir i (mod Sr j ) it follows a j a ir i (mod 0) so a = j a ir i in M/0 = M. Applying condition 1 it follows a i = s for all i > j. Selecting various representatives for the transversal generate different associations of numbers. Example 2.16 Takeing T r(2z) to equal {1, 3}, instead of {0, 1}, changes 0 2 from 0 to 1 and 1 2 from 01 to 033. Additionally 3 2 becomes simply 03 instead of the usual 011. Thus the order of the integers are generally not preserved, with respect the corresponding lexicographic order, by the radix algorithm. The goal of a radix is to provide a normal form for the elements expressed according to the radix. This requires that the representation be unique. In most situations the ideal conditions would be a finit sequence, but sense the design of the radix algorithm functions for all positive integers, the best condition is to produce a sequence of finite type. Definition 2.17 An element a is expressed by S when for some j, a i = a j for i j and is furthermore primitive if a j = 0. M is [primitively]expressed by S when every element of M is [primitively]expressible by S. When M is expressed by S, the index [M : S] = r is a radix expressing M and is said to be expressible by r. Notice Sr j 1 Sr S R and at each step [R : Sr j 1 ] = r j. Proposition 2.18 Given a family of right Z-modules {A i i = 1,..., n}, expressible by r, the direct product n i=1 A i is expressible by r. Proof: Note A i / S i = Ai /S i. Therefore (a 1,..., a n ) r = (a 1 r,..., a n r). Corollary 2.19 All finitely generated abelian groups are expressible. Proof: The Fundamental Theorem of Finitely Generated Abelian Groups together with Proposition Example = 010 and 2 2 = 010. First 2 0 (mod 2), and next 2 0 = 2 2 = 1 2 (mod 4). Now suppose 2 j i=1 2i (mod 2 j+1 ) for some j. Then 2 j i=1 2i 0 (mod 2 j+1 ) and so 2 j i=1 2i 2 j+1 (mod 2 j+2 ) 2 (2 + 0) 4 4 (mod 8) =-8 The rational numbers are not expressible. 7

8 Notice all primitive expressions are weak direct product (or equivalently a polynomial when M is a ring) with coefficients in T r(s). Since T r(s) is a traversal for M/T r(s) it follows a S maps a into N M/S. In particular if M is primitively expressed by S then R is embedded in N R/S. Considering a S as a polynomial, even if multiplication is not defined, it is possible to consider substituting r for x in the polynomial. Thus if a S = (a 0,..., a k, 0,... ) then However a a S(r) = k a i r i a i r i 1 (mod Sr j 1 ) i=1 for all j Z + so for all j > k, a j 0 so a j = 0 since 0 is the unique representative for the coset S, in T. In this way a for all j > k, so a S(r) = a. k a i r i = a S(r) (mod Sr j ) i=1 Proposition 2.21 R is primitively expressed by S if and only if R = N R/S = (R/S)[x]. The substitution map allows for a way to sign the elements of M. Take the positive elements to be P = M S(r). Definition 2.22 (Algebraic Log) Suppose some subset A of M is expressible by r. The algebraic log base r of a A is defined as the length minus 1 of a r and is denoted alog r a. alog r a n = φ(a i ) r j i where φ(a i ) = 0 when a i = 0 and 1 everywhere else. Therefore alog r 100 = 2, alog r 101 = 2 + 1/r 2, etc. Usually log b x is defined as the unique y such that b y = x. Here alog r is similar but can best be thought of as a map from M to the rationals. Unlike log, alog r 0 is artificially assigned the value 0. For any elements that cannot be expressed by r the alog is not defined, recall the largest set of unexpressables are called negatives. 8

Ring Theory Problems. A σ

Ring Theory Problems. A σ Ring Theory Problems 1. Given the commutative diagram α A σ B β A σ B show that α: ker σ ker σ and that β : coker σ coker σ. Here coker σ = B/σ(A). 2. Let K be a field, let V be an infinite dimensional