MATH1022 ANSWERS TO TUTORIAL EXERCISES III 1. G is closed
... order 12. They are non-abelian since in T ET , for example, the products of the rotation through 2π/3 fixing A, and which takes B to C to D and back to B, and that through 2π/3 fixing B, and which takes A to C to D and back to A, are different half-turns depending on which order they are performed, ...
... order 12. They are non-abelian since in T ET , for example, the products of the rotation through 2π/3 fixing A, and which takes B to C to D and back to B, and that through 2π/3 fixing B, and which takes A to C to D and back to A, are different half-turns depending on which order they are performed, ...
Polynomials over finite fields
... Size of a finite field Theorem 1.1 The cardinality of F is pn where n = [F : Fp] and Fp denotes the prime subfield of F. Proof. The prime subfield Fp of F is isomorphic to the field Z/pZ of integers mod p. Since the field F is an n-dimensional vector space over Fp for some finite n, it is set-isomo ...
... Size of a finite field Theorem 1.1 The cardinality of F is pn where n = [F : Fp] and Fp denotes the prime subfield of F. Proof. The prime subfield Fp of F is isomorphic to the field Z/pZ of integers mod p. Since the field F is an n-dimensional vector space over Fp for some finite n, it is set-isomo ...
Math 323. Midterm Exam. February 27, 2014. Time: 75 minutes. (1
... so c = d = 0. Similarly, since x commutes with j, we get b = 0. (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/ ...
... so c = d = 0. Similarly, since x commutes with j, we get b = 0. (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/ ...
Explanation via surplus structure.
... • what is lost by choosing a representation? • what kind of epistemic commitment is it to choose a ...
... • what is lost by choosing a representation? • what kind of epistemic commitment is it to choose a ...
Math 322, Fall Term 2011 Final Exam
... (b) Let R = Z[i] be the ring of Euclidean integers. In the polynomial ring R[X] the prime elements are the same as irreducible elements. (c) If a group of order 25 acts on a set X, then there is an element x ∈ X such that its orbit has 8 elements. (d) The number of conjugacy classes in Sn is equal t ...
... (b) Let R = Z[i] be the ring of Euclidean integers. In the polynomial ring R[X] the prime elements are the same as irreducible elements. (c) If a group of order 25 acts on a set X, then there is an element x ∈ X such that its orbit has 8 elements. (d) The number of conjugacy classes in Sn is equal t ...
F08 Exam 1
... [1] Define a ring homomorphism. Define an ideal. Prove that the kernel of a ring homomorphism is an ideal. ...
... [1] Define a ring homomorphism. Define an ideal. Prove that the kernel of a ring homomorphism is an ideal. ...
Groups
... • Associative: a(bc) = (ab)c • Identity: 1 G, 1a = a1 = a, aG • Inverse: a-1 G, a-1a = aa-1 = 1, a G ...
... • Associative: a(bc) = (ab)c • Identity: 1 G, 1a = a1 = a, aG • Inverse: a-1 G, a-1a = aa-1 = 1, a G ...
ALGEBRAIC NUMBER THEORY
... there must√be a sign change in between. For the second, we can write z = reiθ and then reiθ/2 is a square root. Now if K is an extension of R of finite odd degree, it must equal R (for there is a primitive element α, and its minimal polynomial is of odd degree and so has a root in R, so R[α] ∼ = R. L ...
... there must√be a sign change in between. For the second, we can write z = reiθ and then reiθ/2 is a square root. Now if K is an extension of R of finite odd degree, it must equal R (for there is a primitive element α, and its minimal polynomial is of odd degree and so has a root in R, so R[α] ∼ = R. L ...
Galois` Theorem on Finite Fields
... a 0 for addition, a 1 for multiplication, you can divide consistently, etc. When the number of elements of the field is n, a non-negative integer, it is called finite. If n is a prime then addition and multiplication modulo p defines a field, denoted GF(p). Thus in GF(3), 2 × 2 = 2 + 2 = 1 since 4 ( ...
... a 0 for addition, a 1 for multiplication, you can divide consistently, etc. When the number of elements of the field is n, a non-negative integer, it is called finite. If n is a prime then addition and multiplication modulo p defines a field, denoted GF(p). Thus in GF(3), 2 × 2 = 2 + 2 = 1 since 4 ( ...