Dated 1/22/01
... by setting the parameters n and k equal to each other or nearly so, and whether there are other natural bases for F . ...
... by setting the parameters n and k equal to each other or nearly so, and whether there are other natural bases for F . ...
Abstract Algebra Prelim Jan. 2012
... factorization domain. 5. Let R be a commutative ring. Show a nonzero ideal I in R is a free R-module if and only I is a principal ideal with a generator that is not a zero divisor in R. (Hint: For the direction (⇒), show a basis of I can’t have more than one term in it.) 6. Give examples as requeste ...
... factorization domain. 5. Let R be a commutative ring. Show a nonzero ideal I in R is a free R-module if and only I is a principal ideal with a generator that is not a zero divisor in R. (Hint: For the direction (⇒), show a basis of I can’t have more than one term in it.) 6. Give examples as requeste ...
Week 10 Let X be a G-set. For x 1, x2 ∈ X, let x 1 ∼ x2 if and only if
... Let X be a G-set. For x1, x2 ∈ X, let x1 ∼ x2 if and only if ∃ g ∈ G s.t. gx1 = x2. The ∼ is an equivalence relation. Each cell in X/∼ is an orbit and the orbit contains x is denoted by Gx. Theorem Let X be a G-set and x ∈ X. Then |Gx| = (G : Gx ) where Gx = {g ∈ G | gx = x}. If |G| is finite, then ...
... Let X be a G-set. For x1, x2 ∈ X, let x1 ∼ x2 if and only if ∃ g ∈ G s.t. gx1 = x2. The ∼ is an equivalence relation. Each cell in X/∼ is an orbit and the orbit contains x is denoted by Gx. Theorem Let X be a G-set and x ∈ X. Then |Gx| = (G : Gx ) where Gx = {g ∈ G | gx = x}. If |G| is finite, then ...
Pre-Algebra - Duplin County Schools
... The reason (multiplication & division) and (add & subtract) are grouped is when those operations are next to each other you do the math from left to right. You do not necessarily do addition first if it is written next to subtraction. ...
... The reason (multiplication & division) and (add & subtract) are grouped is when those operations are next to each other you do the math from left to right. You do not necessarily do addition first if it is written next to subtraction. ...
OX(D) (or O(D)) for a Cartier divisor D on a scheme X (1) on
... Remark 1 : in this general de nition of OX (D), there is NO notion of degree involved. Remark 2 : there is an injective map OX → OX (D), which locally is the map A → f1 A, a 7→ a = f1 f a. In our example X = Pnk and the divisor D is a hyperplane H with equation a linear form `. (Let us not assume th ...
... Remark 1 : in this general de nition of OX (D), there is NO notion of degree involved. Remark 2 : there is an injective map OX → OX (D), which locally is the map A → f1 A, a 7→ a = f1 f a. In our example X = Pnk and the divisor D is a hyperplane H with equation a linear form `. (Let us not assume th ...
Solutions to assigned problems from Sections 3.1, page 142, and
... Solutions to assigned problems from Sections 3.1, page 142, and 3.2, page 150 In Exercises 3.1.2, 3.1.4, and 3.1.10, decide whether each of the given sets is a group with respect to the indicated operation. If it is not a group, state a condition in Definition 3.1 that fails to hold. 3.1.2 The set o ...
... Solutions to assigned problems from Sections 3.1, page 142, and 3.2, page 150 In Exercises 3.1.2, 3.1.4, and 3.1.10, decide whether each of the given sets is a group with respect to the indicated operation. If it is not a group, state a condition in Definition 3.1 that fails to hold. 3.1.2 The set o ...
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... and its subclass Rf consisting of those representations which are finite-dimensional as vector spaces. We consider a special representation F = (V, ·) where V is a fixed vector space with a basis B which is in bijective correspondence with G. If f : B → G is a required bijection, then we define ,,·” ...
... and its subclass Rf consisting of those representations which are finite-dimensional as vector spaces. We consider a special representation F = (V, ·) where V is a fixed vector space with a basis B which is in bijective correspondence with G. If f : B → G is a required bijection, then we define ,,·” ...
Solutions - Cal Poly
... n then |g | = 2 (we use the fact—shown many times in class—that |a| = n and k|n, then a k = k). Similarly, if G contains an element g of order 6, then |g 3 | = 2, and if G contains an element g of order 4, then |g 2 | = 2. So we conclude that G contains an element of order 2 unless all elemen ...
... n then |g | = 2 (we use the fact—shown many times in class—that |a| = n and k|n, then a k = k). Similarly, if G contains an element g of order 6, then |g 3 | = 2, and if G contains an element g of order 4, then |g 2 | = 2. So we conclude that G contains an element of order 2 unless all elemen ...
Garrett 03-30-2012 1 • Interlude: Calculus on spheres: invariant integrals, invariant
... especially easy case of Riesz-Fischer, there is Fx ∈ X such that f (x) = hf, Fx i for all f ∈ X. Since X is rotation-invariant and not {0}, the functional f → f (x) cannot be 0 on all of X, so ...
... especially easy case of Riesz-Fischer, there is Fx ∈ X such that f (x) = hf, Fx i for all f ∈ X. Since X is rotation-invariant and not {0}, the functional f → f (x) cannot be 0 on all of X, so ...
Quiz 1 Solutions, Math 309 (Vinroot) (1): The set of integers Z, with
... adding b to both sides and using properties of fields we would have a = b. So a + (−b) has some multiplicative inverse in F , say c. Multiplying the left side by c gives, using properties of vector spaces and fields, c(a + (−b))x = (c(a + (−b)))x = 1x = x, while the right side becomes c0 = 0. We thu ...
... adding b to both sides and using properties of fields we would have a = b. So a + (−b) has some multiplicative inverse in F , say c. Multiplying the left side by c gives, using properties of vector spaces and fields, c(a + (−b))x = (c(a + (−b)))x = 1x = x, while the right side becomes c0 = 0. We thu ...
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... the distributive axiom we cannot establish connections between addition and multiplication. Without scalar multiplication we cannot describe coefficients. With these equations we can define certain subalgebras, for example we see both axioms at work in Proposition 2. Given an algebra A, the set Z0 ( ...
... the distributive axiom we cannot establish connections between addition and multiplication. Without scalar multiplication we cannot describe coefficients. With these equations we can define certain subalgebras, for example we see both axioms at work in Proposition 2. Given an algebra A, the set Z0 ( ...
Constructions in linear algebra For all that follows, let k be the base
... k → End(V ) is still well-defined, while k → V ∗ ⊗ V is not.) 9. Define the evaluation map : V ∗ ⊗ V → k by (f1 ⊗ v1 + · · · + fr ⊗ vr ) = f1 (v1 ) + · · · fr (vr ) Show that this map is well-defined (for example, show that f (v ⊗w +v ⊗w0 ) = f (v ⊗(w +w0 )), and do the same for the other relatio ...
... k → End(V ) is still well-defined, while k → V ∗ ⊗ V is not.) 9. Define the evaluation map : V ∗ ⊗ V → k by (f1 ⊗ v1 + · · · + fr ⊗ vr ) = f1 (v1 ) + · · · fr (vr ) Show that this map is well-defined (for example, show that f (v ⊗w +v ⊗w0 ) = f (v ⊗(w +w0 )), and do the same for the other relatio ...
