Topological conjugacy and symbolic dynamics
... One way to describe orbits of the tent map (and hence also orbits of the logistic map f4 (x) = 4x(1 − x)) is by their ‘left-right’ itineraries: to each x0 ∈ [0, 1] we assign a sequence of L’s and R’s by writing ‘L’ if the nth point of the orbit of x0 is in [0, 1/2] and ‘R’ if it is in [1/2, 1]. This ...
... One way to describe orbits of the tent map (and hence also orbits of the logistic map f4 (x) = 4x(1 − x)) is by their ‘left-right’ itineraries: to each x0 ∈ [0, 1] we assign a sequence of L’s and R’s by writing ‘L’ if the nth point of the orbit of x0 is in [0, 1/2] and ‘R’ if it is in [1/2, 1]. This ...
Example sheet 4
... 12. Show that a complex number α is an algebraic integer if and only if the additive group of the ring Z[α] is finitely generated (i.e. Z[α] is a finitely generated Z-module). Furthermore if α and β are algebraic integers show that the subring Z[α, β] of C generated by α and β also has a finitely ge ...
... 12. Show that a complex number α is an algebraic integer if and only if the additive group of the ring Z[α] is finitely generated (i.e. Z[α] is a finitely generated Z-module). Furthermore if α and β are algebraic integers show that the subring Z[α, β] of C generated by α and β also has a finitely ge ...
NAP PROBLEM SET #1, SOLUTIONS 1. We follow the procedure in
... integers. If m is even, say, m = 2`, then g = (x` )2 , contrary to our assumption. Thus m is odd, and similarly n is odd. Therefore m + n is even, say, m + n = 2q. Now we have gh = xm xn = xm+n = x2q = (xq )2 . This shows that gh is a square. (Note that the proof did not use the hypothesis that |G| ...
... integers. If m is even, say, m = 2`, then g = (x` )2 , contrary to our assumption. Thus m is odd, and similarly n is odd. Therefore m + n is even, say, m + n = 2q. Now we have gh = xm xn = xm+n = x2q = (xq )2 . This shows that gh is a square. (Note that the proof did not use the hypothesis that |G| ...
MMExternalRepresentations
... 3) gives another representation in the plane of the same rectangle. The family of rectangles of different sizes may given by parameters, two real numbers representing the lengths of two adjacent sides. (If you give the length of two adjacent sides, the other two sides are determined by the fact that ...
... 3) gives another representation in the plane of the same rectangle. The family of rectangles of different sizes may given by parameters, two real numbers representing the lengths of two adjacent sides. (If you give the length of two adjacent sides, the other two sides are determined by the fact that ...
Some proofs about finite fields, Frobenius, irreducibles
... coefficients in k, then Φ(β) is also a root. So any polynomial with coefficients in k of which α is a zero must have factors x − Φi (α) as well, for 1 ≤ i < d. By unique factorization, this is the unique such polynomial. P must be irreducible in k[x], because if it factored in k[x] as P = P1 P2 then ...
... coefficients in k, then Φ(β) is also a root. So any polynomial with coefficients in k of which α is a zero must have factors x − Φi (α) as well, for 1 ≤ i < d. By unique factorization, this is the unique such polynomial. P must be irreducible in k[x], because if it factored in k[x] as P = P1 P2 then ...
Problem Score 1 2 3 4 or 5 Total - Mathematics
... Solution: We turn the attention to the possibility of finding an element √ of Z[ −5] with non-unique factorization. We search for possible candidates among elements of Z[ −5] with small norm, the ...
... Solution: We turn the attention to the possibility of finding an element √ of Z[ −5] with non-unique factorization. We search for possible candidates among elements of Z[ −5] with small norm, the ...
MATH 361: NUMBER THEORY — TENTH LECTURE The subject of
... Thus K is a superfield of k containing an element r such that f (r) = 0. For example, since the polynomial f (X) = X 3 − 2 is irreducible over Q, the corresponding quotient ring R = Q[X]/hX 3 − 2i = {a + bX + cX 2 + hX 3 − 2i : a, b, c ∈ Q} is a field. And from R we construct a field (denoted Q(r) o ...
... Thus K is a superfield of k containing an element r such that f (r) = 0. For example, since the polynomial f (X) = X 3 − 2 is irreducible over Q, the corresponding quotient ring R = Q[X]/hX 3 − 2i = {a + bX + cX 2 + hX 3 − 2i : a, b, c ∈ Q} is a field. And from R we construct a field (denoted Q(r) o ...
Algebra 2: Harjoitukset 2. A. Definition: Two fields are isomorphic if
... polynomial of α over K has degree p. [Hint: Review lecture notes from 12.9.] D. Find the minimal polynomial for i + 1 over Q. E. Let K = Q(S) where S is the set of all the complex p-th roots of unity, where p is prime. Prove that [K : Q] = p − 1. You may assume that the polynomial xp−1 + xp−2 + · · ...
... polynomial of α over K has degree p. [Hint: Review lecture notes from 12.9.] D. Find the minimal polynomial for i + 1 over Q. E. Let K = Q(S) where S is the set of all the complex p-th roots of unity, where p is prime. Prove that [K : Q] = p − 1. You may assume that the polynomial xp−1 + xp−2 + · · ...
Math 113 Final Exam Solutions
... so this set in this case is not closed under addition and therefore not an ideal. c) The commutator subgroup of a simple group is trivial. False: Consider the group A5 . Its commutator subgroup is not trivial since A5 /{σe } is not abelian. d) Q, + is a cyclic group. False: if Q were ha/bi, where a, ...
... so this set in this case is not closed under addition and therefore not an ideal. c) The commutator subgroup of a simple group is trivial. False: Consider the group A5 . Its commutator subgroup is not trivial since A5 /{σe } is not abelian. d) Q, + is a cyclic group. False: if Q were ha/bi, where a, ...
January 2008
... Do two problems from each of the three sections, for a total of six problems. If you have doubts about the wording of a problem, please ask for clarification. In no case should you interpret a problem in such a way that it becomes trivial. A. Groups and Character Theory 1. Let G be a nonabelian grou ...
... Do two problems from each of the three sections, for a total of six problems. If you have doubts about the wording of a problem, please ask for clarification. In no case should you interpret a problem in such a way that it becomes trivial. A. Groups and Character Theory 1. Let G be a nonabelian grou ...
Solutions to Exercises for Section 6
... c = d. So there are 10 reducible monic polynomials of degree 2, leaving 6 monic irreducible polynomials of degree 2. To find one of these irreducible polynomials, you could compute the 10 reducible ones and then take a polynomial not on the list, but that looks a little tedious, so you could just ch ...
... c = d. So there are 10 reducible monic polynomials of degree 2, leaving 6 monic irreducible polynomials of degree 2. To find one of these irreducible polynomials, you could compute the 10 reducible ones and then take a polynomial not on the list, but that looks a little tedious, so you could just ch ...
MATH 521A: Abstract Algebra Homework 7 Solutions 1. Consider
... This is similar to the previous problem. We have x2 = (x)2 , x2 + 2 = (x + 1)(x + 2), x2 + x = x(x + 1), x2 + x + 1 = (x + 2)2 , x2 + 2x = x(x + 2), x2 + 2x + 1 = (x + 1)2 . The others, namely x2 + 1, x2 + x + 2, x2 + 2x + 2, are all irreducible. 5. Find some f (x) ∈ Z5 [x] that is monic, of degree ...
... This is similar to the previous problem. We have x2 = (x)2 , x2 + 2 = (x + 1)(x + 2), x2 + x = x(x + 1), x2 + x + 1 = (x + 2)2 , x2 + 2x = x(x + 2), x2 + 2x + 1 = (x + 1)2 . The others, namely x2 + 1, x2 + x + 2, x2 + 2x + 2, are all irreducible. 5. Find some f (x) ∈ Z5 [x] that is monic, of degree ...
Math 323. Midterm Exam. February 27, 2014. Time: 75 minutes. (1
... (2) [5] Let R be a commutative ring with 1. Prove that if there exists a prime ideal P of R that contains no zero divisors, then R is an integral domain. Solution: Suppose R had zero divisors, say ab = 0 with a, b 6= 0. Then consider R/P . We should have āb̄ = 0̄ in R/P , but since P is prime, R/P ...
... (2) [5] Let R be a commutative ring with 1. Prove that if there exists a prime ideal P of R that contains no zero divisors, then R is an integral domain. Solution: Suppose R had zero divisors, say ab = 0 with a, b 6= 0. Then consider R/P . We should have āb̄ = 0̄ in R/P , but since P is prime, R/P ...
Math 296. Homework 4 (due Feb 11) Book Problems (Hoffman
... its inverse map ψ : W → V is a linear transformation. Conclude that vector space homomorphism φ : V → W is an isomorphism if and only if it surjective with trivial kernel. 3. A “bad version” of the complex numbers. (1) Let B = R2 . Let the operation + : B × B → B be componentwise addition. Invent a ...
... its inverse map ψ : W → V is a linear transformation. Conclude that vector space homomorphism φ : V → W is an isomorphism if and only if it surjective with trivial kernel. 3. A “bad version” of the complex numbers. (1) Let B = R2 . Let the operation + : B × B → B be componentwise addition. Invent a ...
Homework 10 April 13, 2006 Math 522 Direction: This homework is
... 2. Describe the finite field GF (24 ) using polynomial p(x) = x16 −x. What are the irreducible factors of p(x)? Based on these irreducible factors find the finite fields that are isomorphic to GF (24 ). Answer: The polynomial p(x) = x16 − x has 16 distinct roots since p(x) = x16 − x and p0 (x) = −1 ...
... 2. Describe the finite field GF (24 ) using polynomial p(x) = x16 −x. What are the irreducible factors of p(x)? Based on these irreducible factors find the finite fields that are isomorphic to GF (24 ). Answer: The polynomial p(x) = x16 − x has 16 distinct roots since p(x) = x16 − x and p0 (x) = −1 ...
Solutions to selected problems from Chapter 2
... Thus, we have to check if there exists a polynomial of the form p(x) = p2 x2 + p1 x + p0 over GF (2), which divides a(x). (a) As a(0) = 1 we can conclude that x − 0 = x is not a factor of a(x). (b) As a(1) = 1 we can conclude that x − 1 = x + 1 is not a factor of a(x). If there exists a polynomial p ...
... Thus, we have to check if there exists a polynomial of the form p(x) = p2 x2 + p1 x + p0 over GF (2), which divides a(x). (a) As a(0) = 1 we can conclude that x − 0 = x is not a factor of a(x). (b) As a(1) = 1 we can conclude that x − 1 = x + 1 is not a factor of a(x). If there exists a polynomial p ...
FINAL EXAM
... (3) (9 points) Let f (x) = x3 + 2x2 + x + 1, viewed as a polynomial in Zp [x]. Determine whether f is irreducible when: (a) p = 2 ...
... (3) (9 points) Let f (x) = x3 + 2x2 + x + 1, viewed as a polynomial in Zp [x]. Determine whether f is irreducible when: (a) p = 2 ...
Math 403A assignment 7. Due Friday, March 8, 2013. Chapter 12
... Gauss’s lemma says that the primitive polynomials form a multiplicative subset of Q[x]. We can write each polynomial k(x) ∈ Z[x] as af (x), where f (x) is primitive and a is the g.c.d. of the coefficients of k(x). Let J be an ideal in Z[x] such that J ∩ Z = (0) and such that if k(x) lies in J, then ...
... Gauss’s lemma says that the primitive polynomials form a multiplicative subset of Q[x]. We can write each polynomial k(x) ∈ Z[x] as af (x), where f (x) is primitive and a is the g.c.d. of the coefficients of k(x). Let J be an ideal in Z[x] such that J ∩ Z = (0) and such that if k(x) lies in J, then ...
June 2007 901-902
... Do two problems from each of the three sections, for a total of six problems. If you have doubts about the wording of a problem, please ask for clarification. In no case should you interpret a problem in such a way that it becomes trivial. A. Groups and Character Theory 1. Consider the collection of ...
... Do two problems from each of the three sections, for a total of six problems. If you have doubts about the wording of a problem, please ask for clarification. In no case should you interpret a problem in such a way that it becomes trivial. A. Groups and Character Theory 1. Consider the collection of ...
1 PROBLEM SET 9 DUE: May 5 Problem 1(algebraic integers) Let K
... Let f (x), g(x) ∈ Z[x]. The content of a polynomial is defined to be the greatest common divisors of its coefficients, and denoted by c(f ). A polynomial is said to be primitive if it has content 1. (1). Show that c(f g) = c(f )c(g).(Hint: use the morphism in Problem 1). (2). Then prove that a polyn ...
... Let f (x), g(x) ∈ Z[x]. The content of a polynomial is defined to be the greatest common divisors of its coefficients, and denoted by c(f ). A polynomial is said to be primitive if it has content 1. (1). Show that c(f g) = c(f )c(g).(Hint: use the morphism in Problem 1). (2). Then prove that a polyn ...
Algebra 1 : Fourth homework — due Monday, October 24 Do the
... (a) Show that the natural action of GLn (k) on k n induces a transitive action of GLn (k) on Pn−1 (k), and compute the stabilizer of the line k × 0 × · · · × 0 under this action. (b) Taking k to be a finite field Fq , use the result of part (a) to inductively compute the order of GLn (Fq ). 2. (This ...
... (a) Show that the natural action of GLn (k) on k n induces a transitive action of GLn (k) on Pn−1 (k), and compute the stabilizer of the line k × 0 × · · · × 0 under this action. (b) Taking k to be a finite field Fq , use the result of part (a) to inductively compute the order of GLn (Fq ). 2. (This ...
Math 331: hw 7 Solutions 5.1.4 Show that, under congruence
... 5.2.6 Each element of the given congruence-class ring can be written in the form [ax + b] (Why?). Determine the rules for addition and multiplication of congruence classes. (In other words, if the product [ax + b][cx + d] is the class [rx + s], describe how to find r and s from a, b, c, d, and simil ...
... 5.2.6 Each element of the given congruence-class ring can be written in the form [ax + b] (Why?). Determine the rules for addition and multiplication of congruence classes. (In other words, if the product [ax + b][cx + d] is the class [rx + s], describe how to find r and s from a, b, c, d, and simil ...
H8
... 1. Let a(x) = (x − 2)2 and b(x) = (x − 3)2 in R[x]. (a) Find polynomials u(x) and v(x) in R[x] so that a(x)u(x) + b(x)v(x) = 1. (b) Find reconstruction polynomials c1 (x), c2 (x) ∈ R[x] so that given any f1 (x) and f2 (x) in R[x] the polynomial f (x) = c1 (x)f1 (x) + c2 (x)f2 (x) satisfies f (x) ≡ f ...
... 1. Let a(x) = (x − 2)2 and b(x) = (x − 3)2 in R[x]. (a) Find polynomials u(x) and v(x) in R[x] so that a(x)u(x) + b(x)v(x) = 1. (b) Find reconstruction polynomials c1 (x), c2 (x) ∈ R[x] so that given any f1 (x) and f2 (x) in R[x] the polynomial f (x) = c1 (x)f1 (x) + c2 (x)f2 (x) satisfies f (x) ≡ f ...
Solutions - UBC Math
... where g ′ is a polynomial with integer coefficients. Now, if the gcd of all the coefficients of g ′ is some integer m > 1, we get: m|D, since g was a monic polynomial, and so the leading coefficient of g ′ is D. Then we can cancel m on the both sides of equation (1) (and replace D with D/m). Thus, w ...
... where g ′ is a polynomial with integer coefficients. Now, if the gcd of all the coefficients of g ′ is some integer m > 1, we get: m|D, since g was a monic polynomial, and so the leading coefficient of g ′ is D. Then we can cancel m on the both sides of equation (1) (and replace D with D/m). Thus, w ...
Math 210B. Homework 4 1. (i) If X is a topological space and a
... and 1 in R = k[X, Y ]/(X(X − 1)(X − λ)) where λ ∈ k − {0, 1}, and determine the associated decomposition of R as a direct product in each case. Draw pictures. (iii) If Z ⊂ k n is an affine algebraic set, prove every point has a connected neighborhood (so all connected components are open) and interp ...
... and 1 in R = k[X, Y ]/(X(X − 1)(X − λ)) where λ ∈ k − {0, 1}, and determine the associated decomposition of R as a direct product in each case. Draw pictures. (iii) If Z ⊂ k n is an affine algebraic set, prove every point has a connected neighborhood (so all connected components are open) and interp ...