
MA2215: Fields, rings, and modules
... with the leading coefficient 2n−1 and a polynomial gn (x) ∈ Z[x] of degree n−1 with the leading coefficient 2n−1 for which cos(nα) = fn (cos α) and sin(nα) = sin αgn (cos α). For n = 1 we take f1 (x) = x and g1 (x) = 1, and the statement is trivial. Let us assume that we know this statement for some ...
... with the leading coefficient 2n−1 and a polynomial gn (x) ∈ Z[x] of degree n−1 with the leading coefficient 2n−1 for which cos(nα) = fn (cos α) and sin(nα) = sin αgn (cos α). For n = 1 we take f1 (x) = x and g1 (x) = 1, and the statement is trivial. Let us assume that we know this statement for some ...
Math. 5363, exam 1, solutions 1. Prove that every finitely generated
... any element of order 6. Also, it can’t happen that every element other than 1 is of order 2. Therefore, there is element a ∈ G of order 3. This element generates the subgroup H = {1, a, a2 } ⊆ G of index 2. In particular, H is a normal subgroup. Since |G| = 2 × 3, there is a Sylow subgroup of G of o ...
... any element of order 6. Also, it can’t happen that every element other than 1 is of order 2. Therefore, there is element a ∈ G of order 3. This element generates the subgroup H = {1, a, a2 } ⊆ G of index 2. In particular, H is a normal subgroup. Since |G| = 2 × 3, there is a Sylow subgroup of G of o ...
Regular local rings
... Since B is regular,the middle vertical arrow is bijective. Then A is regular if and only if the map on the right is injective, which is true if and only if the map on the left is surjective. Since K is generated in degree one and the map is an isomorphism in degree one, A is regular if and only if G ...
... Since B is regular,the middle vertical arrow is bijective. Then A is regular if and only if the map on the right is injective, which is true if and only if the map on the left is surjective. Since K is generated in degree one and the map is an isomorphism in degree one, A is regular if and only if G ...
Computer-Generated Proofs of Mathematical Theorems
... process according to which it can be determined by a finite number of operations whether the equation is solvable in rational integers.” A well-known example of such a Diophantine equation is the Pythagorean equation, x2 + y 2 = z 2 , with the restriction that we only accept integer solutions such a ...
... process according to which it can be determined by a finite number of operations whether the equation is solvable in rational integers.” A well-known example of such a Diophantine equation is the Pythagorean equation, x2 + y 2 = z 2 , with the restriction that we only accept integer solutions such a ...
as a PDF
... calculator or a computer. Program 3.1 below is written in QBASIC and nicely demonstrates the algorithm . In line 120 we ask for double precision arithmetic to give us 16 digits of accuracy. In line 130 we ask for the square root of two (A = 2) and start with iterations with x0 = 1 (X = 1). In line 1 ...
... calculator or a computer. Program 3.1 below is written in QBASIC and nicely demonstrates the algorithm . In line 120 we ask for double precision arithmetic to give us 16 digits of accuracy. In line 130 we ask for the square root of two (A = 2) and start with iterations with x0 = 1 (X = 1). In line 1 ...
Faster Polynomial Multiplication via Discrete
... As our first contribution, for every field k, we present an algorithm Dk ∈ Ank , which is a generalization of Schönhage-Strassen’s construction that works over arbitrary fields and achieves the best known complexity upper bounds. In fact, we argue that the algorithm Dk stands for a generic polynomi ...
... As our first contribution, for every field k, we present an algorithm Dk ∈ Ank , which is a generalization of Schönhage-Strassen’s construction that works over arbitrary fields and achieves the best known complexity upper bounds. In fact, we argue that the algorithm Dk stands for a generic polynomi ...
Euler`s Formula and the Fundamental Theorem of Algebra
... addition and subtraction immediately from the definition. Multiplication, however, is not so easy to visualize with this representation. (1 + i) · (3 − 2i) looks like... well, what? A first time student only knows to multiply this out and then plot the result, 5 + i. There seems to be nothing anyone ...
... addition and subtraction immediately from the definition. Multiplication, however, is not so easy to visualize with this representation. (1 + i) · (3 − 2i) looks like... well, what? A first time student only knows to multiply this out and then plot the result, 5 + i. There seems to be nothing anyone ...
How to Hash into Elliptic Curves
... case, the discrete logarithm of H(m) with respect to G is known, which makes most protocols insecure. For example, it is easy to see that for Boneh-Franklin identity encryption scheme, the attacker can then decrypt any ciphertext. This remains true if we use H(m) = h1 (m).G1 + h2 (m).G2 or any such ...
... case, the discrete logarithm of H(m) with respect to G is known, which makes most protocols insecure. For example, it is easy to see that for Boneh-Franklin identity encryption scheme, the attacker can then decrypt any ciphertext. This remains true if we use H(m) = h1 (m).G1 + h2 (m).G2 or any such ...
GROUPS, RINGS AND FIELDS
... and multiplication. S is a commutative ring. The set of all n-square matrices over the real numbers is not a commutative ring. We define integral domain, which is commutative ring that obeys the following axioms: (M5) Multiplicative identity: There is an element 1 such that a1=1a=a for all a in R (M ...
... and multiplication. S is a commutative ring. The set of all n-square matrices over the real numbers is not a commutative ring. We define integral domain, which is commutative ring that obeys the following axioms: (M5) Multiplicative identity: There is an element 1 such that a1=1a=a for all a in R (M ...
WITT`S PROOF THAT EVERY FINITE DIVISION RING IS A FIELD
... finite dimension over z(D), say k(d). Thus |z(d)| = q k(d) . 3.5. The class equation for D× . The set D× of non-zero elements of D forms a finite group under multiplication. We write the class equation for D× . We have |D× | = |D| − 1 = q n − 1. The centre of D× is just z(D) \ {0} and so its cardina ...
... finite dimension over z(D), say k(d). Thus |z(d)| = q k(d) . 3.5. The class equation for D× . The set D× of non-zero elements of D forms a finite group under multiplication. We write the class equation for D× . We have |D× | = |D| − 1 = q n − 1. The centre of D× is just z(D) \ {0} and so its cardina ...