EXERCISES IN MA 510 : COMMUTATIVE ALGEBRA
... (ii) Show that C ∗ and H ∗ are projectively equivalent to P ∗ . (43) Let φ : P1 → P2 be given by φ((x0 : x1 )) = (x20 : x0 x1 : x21 ). Show that C = φ(P1 ) and P1 are isomorphic as projective varieties but their homogeneous coordinate rings are not. (44) The variety defined by a linear form is calle ...
... (ii) Show that C ∗ and H ∗ are projectively equivalent to P ∗ . (43) Let φ : P1 → P2 be given by φ((x0 : x1 )) = (x20 : x0 x1 : x21 ). Show that C = φ(P1 ) and P1 are isomorphic as projective varieties but their homogeneous coordinate rings are not. (44) The variety defined by a linear form is calle ...
The Z-densities of the Fibonacci sequence
... divisibility properties of the Fibonacci sequence by considering the order α in G(Fp ). Hence we can relate Z(p) to the order of α = (3/2, 1/2) in G(Fp ), as is shown in Theorem 3.5. We define a n-th preimage of α under l to be an element β such that β multiplied by ln equal is to α. Then we may vie ...
... divisibility properties of the Fibonacci sequence by considering the order α in G(Fp ). Hence we can relate Z(p) to the order of α = (3/2, 1/2) in G(Fp ), as is shown in Theorem 3.5. We define a n-th preimage of α under l to be an element β such that β multiplied by ln equal is to α. Then we may vie ...
Soergel diagrammatics for dihedral groups
... Temperley-Lieb 2-category 2T L. Consider a crossingless matching, and color each region of the planar disk with one of two colors (say, red and blue) so that adjacent regions alternate colors. This is a 2-morphism in 2T L. Each crossingless matching can be colored in precisely 2 ways, giving two dif ...
... Temperley-Lieb 2-category 2T L. Consider a crossingless matching, and color each region of the planar disk with one of two colors (say, red and blue) so that adjacent regions alternate colors. This is a 2-morphism in 2T L. Each crossingless matching can be colored in precisely 2 ways, giving two dif ...
Simplifying Expressions Involving Radicals
... settings they do not necessarily find a solution to a given problem described in the easiest possible way. Simplification algorithms can be applied to express these solutions in a form that is more convenient for later use. For example, to determine whether the solution itself or the difference of t ...
... settings they do not necessarily find a solution to a given problem described in the easiest possible way. Simplification algorithms can be applied to express these solutions in a form that is more convenient for later use. For example, to determine whether the solution itself or the difference of t ...
Quadratic fields
... polynomial p, we may assume that p is irreducible; this means, in particular, that p is primitive, that is, the highest common factor of its coefficients is 1. I Multiplying by −1 if necessary, we may further assume that the leading coefficient of p is positive. We claim that the above two assumptio ...
... polynomial p, we may assume that p is irreducible; this means, in particular, that p is primitive, that is, the highest common factor of its coefficients is 1. I Multiplying by −1 if necessary, we may further assume that the leading coefficient of p is positive. We claim that the above two assumptio ...
19 Feb 2010
... multiplication by working in this modular arithmetic. This makes it easy for us to reject many possible factorizations before we start. In addition, the set {1, 2, . . . , 25} has many interesting properties under modular arithmetic that we can exploit further. CONCLUSION. Abstract algebra is a theo ...
... multiplication by working in this modular arithmetic. This makes it easy for us to reject many possible factorizations before we start. In addition, the set {1, 2, . . . , 25} has many interesting properties under modular arithmetic that we can exploit further. CONCLUSION. Abstract algebra is a theo ...
Public key principles, one
... Choose p and q: Test for primality Remember that when n = pq, we could factor n if we could find all four square roots of a second-degree equation Theorem: Suppose there exist integers x and y with x 2 = y 2 mod n but x 6= ±y mod n. Then n is composite, and gcd(x − y , n) gives a nontrivial factor ...
... Choose p and q: Test for primality Remember that when n = pq, we could factor n if we could find all four square roots of a second-degree equation Theorem: Suppose there exist integers x and y with x 2 = y 2 mod n but x 6= ±y mod n. Then n is composite, and gcd(x − y , n) gives a nontrivial factor ...