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Let $M$ and $N$ be finitely generated $R$-modules where $R$ principal domain. Show that if $M \oplus M \simeq N \oplus N$ then $M \simeq N.$

Croos
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    You could use the structure theorem, and induction on the number fo summands the structure theorem gives. – Pedro Jun 21 '14 at 18:20
  • theorems you say are the theorems of elementary divisors and invariant factors? – Croos Jun 25 '14 at 02:14
  • Yes. ${}{}{}{}$ – Pedro Jun 25 '14 at 02:15
  • We could decompose $M= L_1 \oplus T_1$ and $N = L_2 \oplus T_2$, where $L_1,L_2$ is free and $T_1, T_2$ modules of torsion, and give $(L_1 \oplus T_1,L_1 \oplus T_1) \simeq ( L_2 \oplus T_2, L_2 \oplus T_2)$. By the way would there? – Croos Jun 26 '14 at 23:11
  • To use said theorem effectively here, notice that it contains a word most mathematicians love to see in such a statement: "unique". – Tobias Kildetoft Jun 27 '14 at 12:30

2 Answers2

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Apply primary decomposition to these modules.

Clearly, the primary decomposition of $M\oplus M$ is just the primary decomposition of $M$ "doubled" in the sense that all factors appear twice as many times in the decomposition of $M\oplus M$ as they do in $M$. We can be certain of this because the "doubled" decomposition of $M$ clearly provides a decomposition for $M\oplus M$, and we are guaranteed uniqueness of types and multiplicities of pieces in the decomposition by the linked theorem.

Suppose $M\ncong N$. Then at least one of two things happens:

  1. $M$ has a indecomposable primary piece that $N$ doesn't have; or
  2. All the indecomposable primary pieces of $M$ and $N$ are the same, they just differ in number.

In case #1, $M\oplus M$ would also have an indecomposable primary piece which $N\oplus N$ lacks, so they would be nonisomorphic. In case #2, we would argue that the multiplicities of the primary piece differing in $N$ and $M$ produce differing multiplicities in $M\oplus M$ and $N\oplus N$, again making them nonisomorphic.

This proves the contrapositive of the statement.

rschwieb
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Another way of putting it is to consider that the isomorphism class of a finitely-generated module $M$ over a PID $R$ is uniquely determined by the sequence of its invariant factors $$ (a_{1}) \supseteq (a_{2}) \supseteq \dots \supseteq (a_{k}), $$ with all $a_{i}$ not units, and $$ M \cong \bigoplus_{i=1}^{k} \frac{R}{(a_{i})}. $$ Clearly the sequence of invariant factors for $M \oplus M$ is just $$ (a_{1}) \supseteq (a_{1}) \supseteq (a_{2}) \supseteq (a_{2}) \supseteq \dots \supseteq (a_{k}) \supseteq (a_{k}). $$ Now do the same for $N$ and $N \oplus N$, and compare.

Andreas Caranti
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  • I would $$N \cong \bigoplus_{i=1}^{l} \frac{R}{(b_{i})}.$$. Thus, $l=k$, but I would have two equal factors in each invariant direct sum, but how can I conclude that they are isomorphic? It's straight? – Croos Jun 29 '14 at 01:29
  • @Croos, as I said, the sequence of invariant factors determines the isomorphism class of the module. Thast is, two modules are isomorphic if and only if they have the same sequence of invariant factors. And if for a module $L$ you have $L = \oplus_{i=1}^{k} R/(a_{i})$, with $ (a_{1}) \supseteq (a_{2}) \supseteq \dots \supseteq (a_{k}), $ then the invariant factors are $ (a_{1}) , (a_{2}) , \dots , (a_{k}). $ – Andreas Caranti Jun 29 '14 at 09:38