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Theorem 211 (Structure Theorem for Finitely Generated Modules) Let M be a finitely generated R-module. Then there exists a non-negative integer r, called the (torsion-free) rank of M and non-zero, non-unit elements $d_i$ $∈$ R, known as the invariant factors such that $d_1|d_2|d_3| · · · |d_k$ and such that $M ∼= R^r ⊕\frac{R}{<d_1>}⊕\frac{R}{<d_2>}⊕ · · · ⊕\frac{R}{<d_k>}$ . The rank r is unique and $d_1, . . . , d_k$ unique up to multiplication by units.

I am kind of OK with this. Then the lecture notes give the examples:

(a) $Z_6 ⊕ Z_{12} ⊕ Z_{16} ∼= Z_2 ⊕ Z_3 ⊕ Z_3 ⊕ Z_4 ⊕ Z_{16} ∼= Z_2 ⊕ Z_{12} ⊕ Z_{48}.$

(b) $\frac{Q[x]}{(x^2 − 4)(x^3 − 8)}$ . This is already cyclic (it is generated by $1$) and so is in the required form.

(c) $\frac{Z[i]}{ 2} ⊕ \frac{Z[i]}{ 4} ⊕ \frac{Z[i]}{ 5} ∼= \frac{Z[i]}{ 2} ⊕ \frac{Z[i]}{ 20}$

I am ok with deriving those. The lecture notes then say that all of the above three modules all have zero rank which I honestly can't see why holds. Can anyone talk me though the explanation of this and generally help me understand better the Structure Theorem, Smith Form and all the definitions such as rank, torsion etc. Though I know the statements I don't feel comfortable with them.

asdf
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  • All the examples are torsion modules. The rank is the number of copies of $R$ in the decomposition. For example modifying the first example to $\Bbb{Z}^2 \oplus \Bbb{Z}6 \oplus \Bbb{Z}{12} \oplus \Bbb{Z}_{16}$ is now a rank 2 module. – sharding4 May 27 '17 at 23:00
  • I see, so $r+k$ (using the notation above) always gives the number of generators? – asdf May 27 '17 at 23:11
  • Yes the number of generators for the decomposition in terms of a free module and the torsion by invariant factors. – sharding4 May 27 '17 at 23:19

1 Answers1

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This theorem is valid only for modules over P.I.D.s. $R^r$ is isomorphic to a free submodule of $M$, which has rank $r$.

The background is the following:

Over a P.I.D. $R$, with field of fractions $K$, a finitely generated torsion-free module $M$ is free. Furthermore, the canonical morphism $M\longrightarrow M\otimes_R K$ is injective, and the rank of $M$ is equal to the dimension of the $K$-vector space $M\otimes_R K$.

For a general module $M$, its torsion submodule $T$ (the set of torsion elements in $M$) is the kernel of the canonical map $M\longrightarrow M\otimes_R K$, and it is a direct summand of $M$, i. e. there exists a submodule $E$ of $M$ such that $M\simeq T \oplus E$, and $E\simeq M/T$ is a finitely generated torsion-free $R$-module, hence it is free. $r$ is the rank of this free $R$-module, i.e. the dimension of the $K$-vector space $M\otimes_RK$

Bernard
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