Universal enveloping algebra
... algebras. For this functor, f∗ = f for all F -algebra homomorphisms f : A → B. The reason that this works is elementary: f [a, b] = f (ab − ba) = f (a)f (b) − f (b)f (a) = [f (a), f (b)] ...
... algebras. For this functor, f∗ = f for all F -algebra homomorphisms f : A → B. The reason that this works is elementary: f [a, b] = f (ab − ba) = f (a)f (b) − f (b)f (a) = [f (a), f (b)] ...
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Division Algebras
... Theorem (Adams, Hopf Invariant One Problem, 1960). The only maps with Hopf invariant 1 are the Hopf fibrations in dimensions 1, 2, 4, 8. The original proof used delicate analysis of Steenrod operations. A shorter proof of Adams’ Theorem was given 1966 by Atiyah, using Adams operations and the (eight ...
... Theorem (Adams, Hopf Invariant One Problem, 1960). The only maps with Hopf invariant 1 are the Hopf fibrations in dimensions 1, 2, 4, 8. The original proof used delicate analysis of Steenrod operations. A shorter proof of Adams’ Theorem was given 1966 by Atiyah, using Adams operations and the (eight ...
Derived Representation Theory and the Algebraic K
... Remark: Throughout this paper, all Hom and smash product spectra will be computed using only cofibrant modules. If a module is not cofibrant by construction, we will always replace it with a weakly equivalent cofibrant model. We will sometimes do this without comment. An important method for constru ...
... Remark: Throughout this paper, all Hom and smash product spectra will be computed using only cofibrant modules. If a module is not cofibrant by construction, we will always replace it with a weakly equivalent cofibrant model. We will sometimes do this without comment. An important method for constru ...
LECTURES ON SYMPLECTIC REFLECTION ALGEBRAS 2. Algebras of Crawley-Boevey and Holland
... Pick a ∈ A /A6n−1 , b ∈ A6m /A6m−1 and lift them to elements ā ∈ A6n , b̄ ∈ A6m . Then āb̄ is an element in A6n+m . Moreover, since A6n−1 A6m , A6n A6m−1 ⊂ A6n+m−1 , the class of āb̄ in An+m = A6n+m /A6n+m−1 does not depend on the choice of ā, b̄. By definition, ab is that class. Exercise 2.3. Ch ...
... Pick a ∈ A /A6n−1 , b ∈ A6m /A6m−1 and lift them to elements ā ∈ A6n , b̄ ∈ A6m . Then āb̄ is an element in A6n+m . Moreover, since A6n−1 A6m , A6n A6m−1 ⊂ A6n+m−1 , the class of āb̄ in An+m = A6n+m /A6n+m−1 does not depend on the choice of ā, b̄. By definition, ab is that class. Exercise 2.3. Ch ...
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6.6. Unique Factorization Domains
... Proof. Suppose that f (x) ∈ R[x] is primitive in R[x] and irreducible in F [x]. If f (x) = a(x)b(x) in R[x], then one of a(x) and b(x) must be a unit in F [x], so of degree 0. Suppose without loss of generality that a(x) = a0 ∈ R. Then a0 divides all coefficients of f (x), and, because f (x) is prim ...
... Proof. Suppose that f (x) ∈ R[x] is primitive in R[x] and irreducible in F [x]. If f (x) = a(x)b(x) in R[x], then one of a(x) and b(x) must be a unit in F [x], so of degree 0. Suppose without loss of generality that a(x) = a0 ∈ R. Then a0 divides all coefficients of f (x), and, because f (x) is prim ...
Lecture 1: Lattice ideals and lattice basis ideals
... To see this, let F ∈ K [y1 , . . . , yP m ] be a polynomial with F (tb1 , . . . , tbm ) = 0. Say, F = c ac yc with ac ∈ K . Then ...
... To see this, let F ∈ K [y1 , . . . , yP m ] be a polynomial with F (tb1 , . . . , tbm ) = 0. Say, F = c ac yc with ac ∈ K . Then ...
