By Joseph J. Rotman

E-book DescriptionThis book's organizing precept is the interaction among teams and jewelry, the place "rings" comprises the information of modules. It comprises simple definitions, entire and transparent theorems (the first with short sketches of proofs), and provides awareness to the subjects of algebraic geometry, pcs, homology, and representations. greater than only a succession of definition-theorem-proofs, this article placed effects and concepts in context in order that scholars can take pleasure in why a definite subject is being studied, and the place definitions originate. bankruptcy issues comprise teams; commutative earrings; modules; imperative excellent domain names; algebras; cohomology and representations; and homological algebra. for people attracted to a self-study consultant to studying complex algebra and its similar topics.Book information comprises uncomplicated definitions, whole and transparent theorems, and provides recognition to the themes of algebraic geometry, desktops, homology, and representations. for people drawn to a self-study advisor to studying complicated algebra and its comparable themes.

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Ir ) is the cycle (ir ir −1 . . i 1 ): (i 1 i 2 . . ir )−1 = (ir ir −1 . . i 1 ). (ii) If γ ∈ Sn and γ = β1 · · · βk , then γ −1 = βk−1 · · · β1−1 . Definition. Two permutations α, β ∈ Sn have the same cycle structure if their complete factorizations have the same number of r -cycles for each r . 4 on page 50, there are (1/r )[n(n − 1) · · · (n − r + 1)] r -cycles in Sn . This formula can be used to count the number of permutations having any given cycle structure if we are careful about factorizations having several cycles of the same length.

If f : X → Y and g : Y → Z are functions (note that the target of f is equal to the domain of g), then their composite, denoted by g ◦ f , is the function X → Z given by g ◦ f : x → g( f (x)); that is, first evaluate f on x, and then evaluate g on f (x). The chain rule in calculus is a formula for the derivative (g ◦ f ) in terms of g and f : (g ◦ f ) = g ◦ f · f . For example, (sin(ln x)) = cos(ln x) · 1 . x Given a set X , let F(X ) = {all functions X → X }. We have just seen that the composite of two functions in F(X ) is always defined; moreover, the composite is again a function in F(X ).

F (xm ) in Y , and so m ≤ n; using the bijection f −1 in place of f gives the reverse inequality n ≤ m. 58 If X and Y are finite sets with the same number of elements, show that the following conditions are equivalent for a function f : X → Y : (i) f is injective; (ii) f is bijective; (iii) f is surjective. Hint. If A ⊆ X and |A| = n = |X |, then A = X ; after all, how many elements are in X but not in A? 59 Let f : X → Y and g : Y → Z be functions. (i) If both f and g are injective, then g ◦ f is injective.