By Jerry R. Shipman

REA’s Algebra and Trigonometry challenge Solver

Each Problem Solver is an insightful and crucial research and answer consultant chock-full of transparent, concise problem-solving gem stones. solutions to your whole questions are available in a single handy resource from some of the most relied on names in reference resolution courses. extra precious, simpler, and extra informative, those research aids are the simplest evaluate books and textbook partners on hand. they are ideal for undergraduate and graduate studies.

This hugely beneficial reference is the best evaluate of algebra and trigonometry at the moment on hand, with hundreds and hundreds of algebra and trigonometry difficulties that conceal every little thing from algebraic legislation and absolute values to quadratic equations and analytic geometry. every one challenge is obviously solved with step by step precise strategies.

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3. Lying-Over, Incomparability, Going-Up, Going-Down 33 Let r ∈ P1 SQ2 ∩ R. Then r = xs for some x ∈ P1 S and s ∈ S \ Q2 . There exists a finitely generated R-subalgebra T of S such that x ∈ P1 T . Necessarily R ⊆ T is integral, so the finite generation implies that T is module-finite over R. Note that xT ⊆ P1 T T = P1 T . 8, x satisfies an equation xn + a1 xn−1 + · · · + an = 0 with ai ∈ P1 (actually in P1i , but we may ignore the powers). Set f (X) = X n +a1 X n−1 +· · ·+an . 19 it factors into monic polynomials over R[X].

Tc in T such that if νi = deg(ti ), then {ν1 , . . , νc } is a Z-basis of M . Since the νi are linearly independent over Q, the ti are algebraically independent over the field T0 . −1 Clearly T0 [t1 , . . , tc , t−1 1 , . . , tc ] ⊆ T . Let x ∈ T be a homogeneous elemc 1 ment. Write deg(x) = i mi νi for some mi ∈ Z. Then x and tm ∈T 1 · · · tc are homogeneous elements of the same degree, whence their quotient is a non−1 zero element of T0 . Hence x ∈ T0 [t1 , . . , tc , t−1 1 , . . , tc ].

In particular, T is (a localization) of a unique factorization domain, hence T is integrally closed. This proves (1). 6 Let R be a reduced N-graded ring, possibly non-Noetherian, such that the non-zero elements of R0 are non-zerodivisors in R. Then the integral closure of R is N-graded. Proof: Let α ∈ R. By writing α as a quotient of two elements of R and by collecting all the homogeneous parts of the two elements and of the coefficients of an equation of integral dependence, we see that there exist finitely many homogeneous elements x1 , .

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