Introduction to Coding Theory
... Proof: E is a vector space over F , finite-dimensional since F is finite. Denote this dimension by n; then E has a basis over F consisting of n elements, say α1 , ..., αn . Every element of E can be uniquely represented in the form k1 α1 + ... + kn αn (where k1 , ..., kn ∈ F ). Since each ki ∈ F can ...
... Proof: E is a vector space over F , finite-dimensional since F is finite. Denote this dimension by n; then E has a basis over F consisting of n elements, say α1 , ..., αn . Every element of E can be uniquely represented in the form k1 α1 + ... + kn αn (where k1 , ..., kn ∈ F ). Since each ki ∈ F can ...
Proof techniques (section 2.1)
... Implicitly, we add new hypothesis which re ects basic facts of this subject. For example, we add the following hypothesis: • x is even if and only if there exists y such that x = 2y : (∀x) [P (x) ↔ (∃y) x = 2y] ...
... Implicitly, we add new hypothesis which re ects basic facts of this subject. For example, we add the following hypothesis: • x is even if and only if there exists y such that x = 2y : (∀x) [P (x) ↔ (∃y) x = 2y] ...
Algebraic Geometry
... Algebraic Schemes and Algebraic Spaces In this course, we have attached an affine algebraic variety to any algebra finitely generated over a field k. For many reasons, for example, in order to be able to study the reduction of varieties to characteristic p ¤ 0, Grothendieck realized that it is impor ...
... Algebraic Schemes and Algebraic Spaces In this course, we have attached an affine algebraic variety to any algebra finitely generated over a field k. For many reasons, for example, in order to be able to study the reduction of varieties to characteristic p ¤ 0, Grothendieck realized that it is impor ...
Homework assignments
... (You can use the following property of a compact Hausdorff space X. Let C be a closed subset of X and let p ∈ X, p ∈ / C. Then there is f ∈ A such that f (p) = 1 and f has value 0 at any point of S.) 2. Prove that for a subset S of X, S is closed if and only if there is an ideal I of A such that S = ...
... (You can use the following property of a compact Hausdorff space X. Let C be a closed subset of X and let p ∈ X, p ∈ / C. Then there is f ∈ A such that f (p) = 1 and f has value 0 at any point of S.) 2. Prove that for a subset S of X, S is closed if and only if there is an ideal I of A such that S = ...
analytic and combinatorial number theory ii
... pigeon-hole principle two distinct n-tuples are mapped by the forms to the same m-tuple. Their difference, which we denote (α1 , . . . , αn ), is mapped by the forms to the m-tuple of zeros, has |αi | ≤ r and not all αi are zero. We check that r = b(nA)m/(n−m) c satisfies the required inequality and ...
... pigeon-hole principle two distinct n-tuples are mapped by the forms to the same m-tuple. Their difference, which we denote (α1 , . . . , αn ), is mapped by the forms to the m-tuple of zeros, has |αi | ≤ r and not all αi are zero. We check that r = b(nA)m/(n−m) c satisfies the required inequality and ...
Basic Precalculus Toolbox Often the difficulties students run into in
... 12. How to solve an algebraic equation (No all equations can be solved in a closed form way, but when they can be solved, we follow the steps below): (a) Try to isolate the variable you want to solve for (say x) by using a sequence of inverses. (b) At each step cancel the ‘outside’ function on the x ...
... 12. How to solve an algebraic equation (No all equations can be solved in a closed form way, but when they can be solved, we follow the steps below): (a) Try to isolate the variable you want to solve for (say x) by using a sequence of inverses. (b) At each step cancel the ‘outside’ function on the x ...
Round 3 Solutions
... prime (in a, b, or c) must be exactly twice the smallest exponent of that same prime (in a, b, or c). In particular, if any prime divides one of a, b, and c, then it must divide all of them. We cannot have a, b, c all be a power of the same prime, since then clearly the smallest one would divide the ...
... prime (in a, b, or c) must be exactly twice the smallest exponent of that same prime (in a, b, or c). In particular, if any prime divides one of a, b, and c, then it must divide all of them. We cannot have a, b, c all be a power of the same prime, since then clearly the smallest one would divide the ...
