Algebra Quals Fall 2012 1. This is an immediate consequence of the
... it’s a typo. so you need a product. in this case it’s trivial. c. Choose a basis. Notice that since it’s a sequence of free R-modules, it splits, so M = M 0 ⊕ M 00 .Then the isomorphism is obvious, writing it in this basis. For naturality, notice if we have an iso or exact sequences 0 → N 0 → N → N ...
... it’s a typo. so you need a product. in this case it’s trivial. c. Choose a basis. Notice that since it’s a sequence of free R-modules, it splits, so M = M 0 ⊕ M 00 .Then the isomorphism is obvious, writing it in this basis. For naturality, notice if we have an iso or exact sequences 0 → N 0 → N → N ...
Honors Algebra 4, MATH 371 Winter 2010
... as the images of the generators are a generating set. Since R is a PID, it is noetherian and any submodule of a finitely generated R-module is again finitely generated (proof?). That any quotient or submodule of a p-primary module is p-primary is obvious. (b) Let ann(Tp ) := {r ∈ R : rt = 0 for all ...
... as the images of the generators are a generating set. Since R is a PID, it is noetherian and any submodule of a finitely generated R-module is again finitely generated (proof?). That any quotient or submodule of a p-primary module is p-primary is obvious. (b) Let ann(Tp ) := {r ∈ R : rt = 0 for all ...
PRIMITIVE ELEMENTS FOR p-DIVISIBLE GROUPS 1. Introduction
... A0 -module M . For example, the A0 -module structure on A reviewed above is the one corresponding to the natural G-module structure on A. Given a G-module M , its submodule M G of G-invariants consists of all elements in M annihilated by the augmentation ideal I 0 in A0 . For any k-algebra R there i ...
... A0 -module M . For example, the A0 -module structure on A reviewed above is the one corresponding to the natural G-module structure on A. Given a G-module M , its submodule M G of G-invariants consists of all elements in M annihilated by the augmentation ideal I 0 in A0 . For any k-algebra R there i ...
