O I A
... ( i, j ) . But we know that A contains some matrix M whose entry ( i, j ) is not zero, ...
... ( i, j ) . But we know that A contains some matrix M whose entry ( i, j ) is not zero, ...
Homework #3
... Closure: suppose that a, b are two elements of µn . We want to prove that the product ab is an element of µn ; that is, we want to prove that (ab)n = 1. We raise ab to the nth power and use the fact that multiplication in a field is commutative to get (ab)n = an bn . Since a ∈ µn , we know that an = ...
... Closure: suppose that a, b are two elements of µn . We want to prove that the product ab is an element of µn ; that is, we want to prove that (ab)n = 1. We raise ab to the nth power and use the fact that multiplication in a field is commutative to get (ab)n = an bn . Since a ∈ µn , we know that an = ...
nnpc – fstp- maths_eng 1
... (1) The quadratic function has one turning value: a maximum for a < 0 and a minimum for a > 0. (2) The function is symmetrical about the line x = b/(2a). (3) The turning value of y is D/(4a) and occurs at the line of symmetry. (4) The curve of y(x) cuts the x-axis at 2 points if D > 0, touches the ...
... (1) The quadratic function has one turning value: a maximum for a < 0 and a minimum for a > 0. (2) The function is symmetrical about the line x = b/(2a). (3) The turning value of y is D/(4a) and occurs at the line of symmetry. (4) The curve of y(x) cuts the x-axis at 2 points if D > 0, touches the ...
Finite group schemes
... In other words, every morphism X → Y of k-schemes, with Y /k étale, factors uniquely via X → $0 (X). To understand what $0 (X) is, we use our description of étale k-schemes. Fix a separable closure k̄/k. Observe that Gal(k̄/k) acts on Spec(k̄), hence on, Xk̄ = X ×k Spec(k̄), hence on the topologic ...
... In other words, every morphism X → Y of k-schemes, with Y /k étale, factors uniquely via X → $0 (X). To understand what $0 (X) is, we use our description of étale k-schemes. Fix a separable closure k̄/k. Observe that Gal(k̄/k) acts on Spec(k̄), hence on, Xk̄ = X ×k Spec(k̄), hence on the topologic ...
