Structure theorem for abelian groups

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SUMMARY

The discussion centers on the structure theorem for abelian groups, specifically regarding the interpretation of diagonalized presentation matrices for modules over polynomial rings. The example provided illustrates that a diagonal matrix like diag[1, (t-1)²(t-2)] can be interpreted as a direct sum of cyclic groups, leading to the conclusion that the module M is isomorphic to k[X] × (k[X]/(X-1)²(X-2)). This highlights the relationship between polynomial rings and module theory in the context of abelian groups.

PREREQUISITES
  • Understanding of module theory over rings
  • Familiarity with polynomial rings, specifically k[X]
  • Knowledge of diagonalization of matrices
  • Concept of isomorphism in algebraic structures
NEXT STEPS
  • Study the structure theorem for finitely generated abelian groups
  • Learn about the properties of polynomial rings and their modules
  • Explore the concept of direct sums in module theory
  • Investigate the implications of diagonalization in linear algebra
USEFUL FOR

Mathematicians, algebraists, and students studying abstract algebra, particularly those focusing on module theory and the structure of abelian groups.

Jim Kata
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So say you have a presentation matrix A for a module, and you diagonalise it and you get something like diag[1,5] well you can interpret that as A can be broken down as the direct sum of 1+Z/5Z. That is the trivial group of just the identity plus cyclic 5. What if your module is defined over the ring of polynomials, and after you diagonalise your presentation matrix you get something like
diag[1,(t-1)^2(t-2)]? How do I interpret this? What is the cyclic group something like
(t-1)^2(t-2) would represent?
 
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I assume you are working with k[X]. Your presentation matrix of M is

[tex]A=\left(\begin{array}{cc} 1 & 0\\ 0 & (X-1)^2(X-2)\\ \end{array}\right)[/tex]

Let the map corresponding to this matrix by [itex]\alpha:k[X]^2\rightarrow k[X]^2[/itex]. Then we know that

[tex]M\cong k[X]^2/\alpha k[X]^2[/tex]

Now, let {e,e'} be a basis corresponding to A, then we know that [itex]\alpha k[X][/itex] is generated by [itex]\{e,(X-1)^2(X-2)e'\}[/itex]

Thus

[tex]M\cong k[X]^2/\alpha k[X]^2=k[X]\times (k[X]/(X-1)^2(X-2))[/tex]
 

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