Solutions to odd-numbered exercises
نویسنده
چکیده
which are just module axioms (MM2) and (MM3). [In the case of the second axiom, we have to show that (r1r2)θ = (r1θ)r2. We are thinking of r′ as an element of the free R-module R, and r2 as an element of the ring R acting on this module.] The kernel of θ is {r ∈ R : mr = 0}. Clearly any element r ∈ Ann(M) satisfies this condition. Conversely, since m generates M, every element of M has the form mr′ for some r′ ∈ R; then mr = 0 implies (mr′)r = (mr)r′ = 0, so r ∈Ann(M). [Remember that our rings are now commutative with identity!] So the Ker(θ) = Ann(M). Also, since m generates M, we have Im(θ) = M. So M ∼= R/Ann(M) (as R-modules).
منابع مشابه
Outlines of Solutions to Selected Homework Problems
This handout contains solutions and hints to solutions for many of the STA 6505 homework exercises from Categorical Data Analysis, second edition, by Alan Agresti (John Wiley, & Sons, 2002). It should not be distributed elsewhere without permission of the author. Additional solutions for odd-numbered exercises are available at the website for the text, http://www.stat.ufl.edu/∼aa/cda/cda.html. ...
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تاریخ انتشار 2007