$M \oplus M \simeq N \oplus N$ then $M \simeq N.$

Question

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

Answer 1

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.

Answer 2

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.