
SOME ALGEBRAIC DEFINITIONS AND CONSTRUCTIONS
... We have used a special case of a result of Kaplansky, and we will also need Kaplansky’s result in the proof of Munshi’s theorem. Theorem 29 (Kaplansky). Let R be an integral domain. Then the intersection I of the non-zero prime ideals in R[x] is zero. Let K be the field of fractions of R. We need a ...
... We have used a special case of a result of Kaplansky, and we will also need Kaplansky’s result in the proof of Munshi’s theorem. Theorem 29 (Kaplansky). Let R be an integral domain. Then the intersection I of the non-zero prime ideals in R[x] is zero. Let K be the field of fractions of R. We need a ...
Solutions Sheet 7
... x ∈ X possesses an affine open neighborhood U ⊂ X such that OX (U ) is noetherian. Then we already know that U has only finitely many irreducible components. It thus suffices to show that for any irreducible component Z of X the intersection Z ∩ U is either empty or an irreducbile component of U . S ...
... x ∈ X possesses an affine open neighborhood U ⊂ X such that OX (U ) is noetherian. Then we already know that U has only finitely many irreducible components. It thus suffices to show that for any irreducible component Z of X the intersection Z ∩ U is either empty or an irreducbile component of U . S ...
Noncommutative geometry on trees and buildings
... be expressed in terms of the inverse of the classical Dirac operator D, so that the Riemannian geometry is entirely encoded by the data (A, H, D) of the algebra of smooth functions (a dense subalgebra of the C ∗ -algebra of continuous functions), the Hilbert space of square integrable spinor section ...
... be expressed in terms of the inverse of the classical Dirac operator D, so that the Riemannian geometry is entirely encoded by the data (A, H, D) of the algebra of smooth functions (a dense subalgebra of the C ∗ -algebra of continuous functions), the Hilbert space of square integrable spinor section ...
security engineering - University of Sydney
... The strength of Diffie-Hellman is based upon two issues: – given p, g, ga, it is difficult to calculate a (the discrete logarithm problem) – given p, g, ga, gb it is difficult to calculate gab (the Diffie-Hellman problem) – we know that DL → DH but it is not known if DH → DL ...
... The strength of Diffie-Hellman is based upon two issues: – given p, g, ga, it is difficult to calculate a (the discrete logarithm problem) – given p, g, ga, gb it is difficult to calculate gab (the Diffie-Hellman problem) – we know that DL → DH but it is not known if DH → DL ...
Notes from a mini-course on Group Theory for
... theory, such groups are regarded as “the same”. An important class of problems in group theory is to classify groups satisfying given conditions up to isomorphism (“classification problems”). Exercise 2.11. Let G be a group, and g ∈ G. Show that the map G → G given by a 7→ gag −1 (called the conjuga ...
... theory, such groups are regarded as “the same”. An important class of problems in group theory is to classify groups satisfying given conditions up to isomorphism (“classification problems”). Exercise 2.11. Let G be a group, and g ∈ G. Show that the map G → G given by a 7→ gag −1 (called the conjuga ...
quotient rings of a ring and a subring which have a common right ideal
... Let S be a ring and let A be a right ideal of S. The idealizer of A in S, denoted 1(A), is the largest subring of S containing A as a two-sided ideal. Armendariz and Fisher [1] have shown that with various assumptions on A, S being semiprime right (left) Goldie is equivalent to 1(A) being semiprime ...
... Let S be a ring and let A be a right ideal of S. The idealizer of A in S, denoted 1(A), is the largest subring of S containing A as a two-sided ideal. Armendariz and Fisher [1] have shown that with various assumptions on A, S being semiprime right (left) Goldie is equivalent to 1(A) being semiprime ...
Finite Fields
... • the integers (Z, +, ∗) form an integral domain but not a field; • the rationals (Q, +, ∗), reals (R, +, ∗) and complex numbers (C, +, ∗) form fields; • the set of 2 × 2 matrices with real entries forms a non-commutative ring with identity w.r.t. matrix addition and multiplication. • the group Zn w ...
... • the integers (Z, +, ∗) form an integral domain but not a field; • the rationals (Q, +, ∗), reals (R, +, ∗) and complex numbers (C, +, ∗) form fields; • the set of 2 × 2 matrices with real entries forms a non-commutative ring with identity w.r.t. matrix addition and multiplication. • the group Zn w ...
lecture notes as PDF
... • the integers (Z, +, ∗) form an integral domain but not a field; • the rationals (Q, +, ∗), reals (R, +, ∗) and complex numbers (C, +, ∗) form fields; • the set of 2 × 2 matrices with real entries forms a non-commutative ring with identity w.r.t. matrix addition and multiplication. • the group Zn w ...
... • the integers (Z, +, ∗) form an integral domain but not a field; • the rationals (Q, +, ∗), reals (R, +, ∗) and complex numbers (C, +, ∗) form fields; • the set of 2 × 2 matrices with real entries forms a non-commutative ring with identity w.r.t. matrix addition and multiplication. • the group Zn w ...
POLYNOMIAL RINGS AND UNIQUE FACTORIZATION DOMAINS 1
... It’s possible that x has some irreducibles from R that we can factor out of it. If so, we do factor these irreducibles from R out, using Lemma 10. This leaves a primitive polynomial. Then a primitive polynomial factors only into polynomials of smaller positive degree. Thus, what we’ve done so far is ...
... It’s possible that x has some irreducibles from R that we can factor out of it. If so, we do factor these irreducibles from R out, using Lemma 10. This leaves a primitive polynomial. Then a primitive polynomial factors only into polynomials of smaller positive degree. Thus, what we’ve done so far is ...
Optimal strategies in the average consensus problem
... our matrices to be nonnegative, even though it will appear that the optimal matrices are. Observe now that the fact that the element in position i, j of the matrix I + K is different from zero, means that the system i needs to know exactly the state of the system j in order to compute its feedback a ...
... our matrices to be nonnegative, even though it will appear that the optimal matrices are. Observe now that the fact that the element in position i, j of the matrix I + K is different from zero, means that the system i needs to know exactly the state of the system j in order to compute its feedback a ...
Document
... She did a brilliant exhibition, first tapping it in 4, 4, then giving me a hasty glance and doing it in 2, 2, 2, 2, before coming for her nut. It is astonishing that Star learned to count up to 8 with no difficulty, and of her own accord ...
... She did a brilliant exhibition, first tapping it in 4, 4, then giving me a hasty glance and doing it in 2, 2, 2, 2, before coming for her nut. It is astonishing that Star learned to count up to 8 with no difficulty, and of her own accord ...
Sums of Fractions and Finiteness of Monodromy
... (we refer to this as the “dihedral case”), or else (λ, µ, ν) ∈ the finite list in Table 1 below. Remark. Again, though the statement of the theorem is purely (elementary) number theoretic, the proof uses the finiteness of a certain group Γ in GL2 (C). It would be interesting to find a purely number ...
... (we refer to this as the “dihedral case”), or else (λ, µ, ν) ∈ the finite list in Table 1 below. Remark. Again, though the statement of the theorem is purely (elementary) number theoretic, the proof uses the finiteness of a certain group Γ in GL2 (C). It would be interesting to find a purely number ...