COMPUTING RAY CLASS GROUPS, CONDUCTORS AND
... chosen so that vp (π) = 1 (if this is not the case, then vp (p) = 1, i.e. p is unramified, and hence we replace π by π + p). Then for all m, if q = dm/ee = b(m + e − 1)/ec where e = e(p/p) is the ramification index of p, we have pm = pq ZK + π m ZK . Indeed, for any prime ideal q different from p, m ...
... chosen so that vp (π) = 1 (if this is not the case, then vp (p) = 1, i.e. p is unramified, and hence we replace π by π + p). Then for all m, if q = dm/ee = b(m + e − 1)/ec where e = e(p/p) is the ramification index of p, we have pm = pq ZK + π m ZK . Indeed, for any prime ideal q different from p, m ...
Notes from Unit 1
... We can always do the arithmetic to find products like this. The interesting question arises when we know the matrix A and the product vector b, and we want to find the vector x for which Ax = b — or even to know whether such a x exists. If we could “divide by A”, it would be an easy problem, just li ...
... We can always do the arithmetic to find products like this. The interesting question arises when we know the matrix A and the product vector b, and we want to find the vector x for which Ax = b — or even to know whether such a x exists. If we could “divide by A”, it would be an easy problem, just li ...
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 ...
1 Homework 1
... (2) If I is an ideal in R, show rad(I) = {x ∈ R|xn ∈ I for some n} is an ideal. SOLUTION: Suppose x, y ∈ rad(I) and xn , y m ∈ I. Then binomial expansion of (x + y)n+m−1 shows that each term is either of degree at least n in x or degree at least m in y, hence (x + y) ∈ rad(I). For the multiplicative ...
... (2) If I is an ideal in R, show rad(I) = {x ∈ R|xn ∈ I for some n} is an ideal. SOLUTION: Suppose x, y ∈ rad(I) and xn , y m ∈ I. Then binomial expansion of (x + y)n+m−1 shows that each term is either of degree at least n in x or degree at least m in y, hence (x + y) ∈ rad(I). For the multiplicative ...
Solving Systems of Linear Equations Substitution Elimination
... (2) Add a multiple of one equation to another. Again, this should not be taken literally. It really means to add a multiple of the left side of one equation to the left side of another and also add the same multiple of the right side of that equation to the right side of the other. (3) Interchange t ...
... (2) Add a multiple of one equation to another. Again, this should not be taken literally. It really means to add a multiple of the left side of one equation to the left side of another and also add the same multiple of the right side of that equation to the right side of the other. (3) Interchange t ...
October 17, 2011 THE ELGAMAL CRYPTOSYSTEM OVER
... Two of the most popular groups used in the discrete logarithm problem are the group of units of a finite field and the group of rational points of an elliptic curve over a finite field. The obvious question arises, are there any other groups? There are matrix groups out there, for example, the group ...
... Two of the most popular groups used in the discrete logarithm problem are the group of units of a finite field and the group of rational points of an elliptic curve over a finite field. The obvious question arises, are there any other groups? There are matrix groups out there, for example, the group ...
M1GLA: Geometry and Linear Algebra Lecture Notes
... In particular, x and y are perpendicular iff (x · y) = 0. Lines in R2 can be written as L = {u + λv | λ ∈ R}. This will be referred to as vector form. Lines can also be described by their Cartesian equation: px1 + qx2 + r = 0, where p1 , q1 ∈ R. Definition. Any vector perpendicular to the direction v ...
... In particular, x and y are perpendicular iff (x · y) = 0. Lines in R2 can be written as L = {u + λv | λ ∈ R}. This will be referred to as vector form. Lines can also be described by their Cartesian equation: px1 + qx2 + r = 0, where p1 , q1 ∈ R. Definition. Any vector perpendicular to the direction v ...