07 some irreducible polynomials
... for d < 10. The fact that there are such primes can be verified in an ad hoc fashion by simply looking for them, and Dirichlet’s theorem on primes in arithmetic progressions assures that there are infinitely many such. The presence of primitive roots 2, 6, 7, 8 (that is, generators for the cyclic gr ...
... for d < 10. The fact that there are such primes can be verified in an ad hoc fashion by simply looking for them, and Dirichlet’s theorem on primes in arithmetic progressions assures that there are infinitely many such. The presence of primitive roots 2, 6, 7, 8 (that is, generators for the cyclic gr ...
TRUE/FALSE. Write `T` if the statement is true and `F` if the
... 39) Two integers are relatively _________ if their only common positive integer factor is 1. ...
... 39) Two integers are relatively _________ if their only common positive integer factor is 1. ...
Comments on Earlier Problems 76:60 Peter Weinberger Let jfj
... with a > b , 1, a even, b odd shows that if such an A exists then A(n; n) n2. Solution: Andrzej Schinzel writes that the answer to this problem is negative, and a simple counterexample is f = xab , 1, g = (xa , 1)(xb , 1), where jf j = 2, jgj = 4 and j(f; g)j can be arbitrarily large. The only di ...
... with a > b , 1, a even, b odd shows that if such an A exists then A(n; n) n2. Solution: Andrzej Schinzel writes that the answer to this problem is negative, and a simple counterexample is f = xab , 1, g = (xa , 1)(xb , 1), where jf j = 2, jgj = 4 and j(f; g)j can be arbitrarily large. The only di ...
m\\*b £«**,*( I) kl)
... Barnes [l] has constructed an example of a commutative semisimple normed annihilator algebra which is not a dual algebra. His example is not complete and when completed acquires a nonzero radical. In this paper we construct an example which is complete. The theory of annihilator algebras is develope ...
... Barnes [l] has constructed an example of a commutative semisimple normed annihilator algebra which is not a dual algebra. His example is not complete and when completed acquires a nonzero radical. In this paper we construct an example which is complete. The theory of annihilator algebras is develope ...
On the number of polynomials with coefficients in [n] Dorin Andrica
... For an elliptic curve (over a number field) it is known that the order of its TateShafarevich group is a square, provided it is finite. In higher dimensions this no longer holds true. We will present work in progress on the classification of all occurring non-square parts of orders of Tate-Shafarevi ...
... For an elliptic curve (over a number field) it is known that the order of its TateShafarevich group is a square, provided it is finite. In higher dimensions this no longer holds true. We will present work in progress on the classification of all occurring non-square parts of orders of Tate-Shafarevi ...
Math 261y: von Neumann Algebras (Lecture 1)
... to consider also infinite-dimensional representations V . For this to be sensible, we should assume that V is equipped with some sort of topology. Let us restrict our attention to the easiest case: assume that V is a (complex) Hilbert space and that the representation of G on V is unitary. In this c ...
... to consider also infinite-dimensional representations V . For this to be sensible, we should assume that V is equipped with some sort of topology. Let us restrict our attention to the easiest case: assume that V is a (complex) Hilbert space and that the representation of G on V is unitary. In this c ...
ON DENSITY OF PRIMITIVE ELEMENTS FOR FIELD EXTENSIONS
... bA = b1 a11 + b2 a12 + · · · + bn a1n = 0 is at most k n−1 . To see this, note that at least one of the entries a1j is nonzero which, without loss of generality, we assume is a11 . For each choice of (b2 , . . . , bn ) ∈ S n−1 , b1 is uniquely determined by the above equation, and this value of b1 m ...
... bA = b1 a11 + b2 a12 + · · · + bn a1n = 0 is at most k n−1 . To see this, note that at least one of the entries a1j is nonzero which, without loss of generality, we assume is a11 . For each choice of (b2 , . . . , bn ) ∈ S n−1 , b1 is uniquely determined by the above equation, and this value of b1 m ...
Two proofs of the infinitude of primes Ben Chastek
... clear that 5 3 + 7 is an algebraic number, but it is not obvious to which polynomial equation it is a root. A number that is not expressible as a root of a polynomial (with rational coefficients) is called transcendental. Some common examples are π and e (the base of the natural logarithms). An alge ...
... clear that 5 3 + 7 is an algebraic number, but it is not obvious to which polynomial equation it is a root. A number that is not expressible as a root of a polynomial (with rational coefficients) is called transcendental. Some common examples are π and e (the base of the natural logarithms). An alge ...
Solutions to Exercises for Section 6
... Is hX 2 + 1i a prime ideal? a maximal ideal? if neither, what is its radical? Solution: The possible remainders of polynomials when divided by X 2 + 1 are: 0, 1, X, X + 1. So the factor ring Z2 [X]/hX 2 + 1i has four elements - the images of these - which we may write as 0, 1, α, α + 1 (having writt ...
... Is hX 2 + 1i a prime ideal? a maximal ideal? if neither, what is its radical? Solution: The possible remainders of polynomials when divided by X 2 + 1 are: 0, 1, X, X + 1. So the factor ring Z2 [X]/hX 2 + 1i has four elements - the images of these - which we may write as 0, 1, α, α + 1 (having writt ...
Some definitions that may be useful
... • A 2-morphism is a cone on a bigon. Now again I’ll work over K = K-mod for K a ring. Pick algebras A, B and A = A-mod and B = B-mod, and take the forgetful maps as the fiber functors” Exercise: Any 1-morphism is exact, cocontinuous, faithful, etc. Corollary: To know E : A → B, it suffices to know E ...
... • A 2-morphism is a cone on a bigon. Now again I’ll work over K = K-mod for K a ring. Pick algebras A, B and A = A-mod and B = B-mod, and take the forgetful maps as the fiber functors” Exercise: Any 1-morphism is exact, cocontinuous, faithful, etc. Corollary: To know E : A → B, it suffices to know E ...
Some proofs about finite fields, Frobenius, irreducibles
... K by grouping them in d-tuples of roots of elements of irreducible monic polynomials with coefficients in k = Fq , where d runs over positive divisors of n including 1 and n. Let Nd be the number of irreducible monic polynomials of degree d with coefficients in k = Fq . Then this grouping and counti ...
... K by grouping them in d-tuples of roots of elements of irreducible monic polynomials with coefficients in k = Fq , where d runs over positive divisors of n including 1 and n. Let Nd be the number of irreducible monic polynomials of degree d with coefficients in k = Fq . Then this grouping and counti ...
Math 113 Final Exam Solutions
... those of the form (0, n)H and (1, n)H, where n can be any integer. Suppose (r, n)H = (s, m)H where s, r = 0 or 1. Then we have (r − s, n − m) ∈ H. Note that |r − s| = 0 or 1. At the same time, since H = h(2, 4)i, we have 2|r − s, so r − s = 0 necessarily. But then n − m = 0 · 4 = 0, and so we have t ...
... those of the form (0, n)H and (1, n)H, where n can be any integer. Suppose (r, n)H = (s, m)H where s, r = 0 or 1. Then we have (r − s, n − m) ∈ H. Note that |r − s| = 0 or 1. At the same time, since H = h(2, 4)i, we have 2|r − s, so r − s = 0 necessarily. But then n − m = 0 · 4 = 0, and so we have t ...
Groups, Rings and Fields
... • Let β be a nonzero element of GF(q) and let 1 be the multiplicative identity • Definition 2-12 The order of β is the smallest positive integer m such that βm = 1 • Theorem 2-10 If t = ord(β) then t | (q-1) • Definition 2-14 In any finite field, there are one or more elements of order q-1 called pr ...
... • Let β be a nonzero element of GF(q) and let 1 be the multiplicative identity • Definition 2-12 The order of β is the smallest positive integer m such that βm = 1 • Theorem 2-10 If t = ord(β) then t | (q-1) • Definition 2-14 In any finite field, there are one or more elements of order q-1 called pr ...
Universal Enveloping Algebras (and
... Under bracket multiplication, Lie algebras are non-associative. The idea behind the construction of the universal enveloping algebra of some Lie algebra g is to pass from this non-associative object to its more friendly unital associative counterpart U g (allowing for the use of asociative methods s ...
... Under bracket multiplication, Lie algebras are non-associative. The idea behind the construction of the universal enveloping algebra of some Lie algebra g is to pass from this non-associative object to its more friendly unital associative counterpart U g (allowing for the use of asociative methods s ...
1 Fields and vector spaces
... Multiplication by zero induces the zero endomorphism of F . Multiplication by any non-zero element induces an automorphism (whose inverse is multiplication by the inverse element). In particular, we see that the automorphism group of F acts transitively on its non-zero elements. So all no ...
... Multiplication by zero induces the zero endomorphism of F . Multiplication by any non-zero element induces an automorphism (whose inverse is multiplication by the inverse element). In particular, we see that the automorphism group of F acts transitively on its non-zero elements. So all no ...