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I want to prove the following:


Let $A$ be a ring and $n$ a natural number. If the left $A$-module $A^n$ contains a free subset of $n+1$ elements, then $A^n$ already contains an infinite free subset.


Since we can embed $A^{n+1}$ into $A^n$, we can embed $A^{n+2}$ into $A^{n+1}$, and so on. So $A^n$ contains finite free subsets of all sizes, but I really have no clue how to construct an infinite free subset from that.

Can someone help me?

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1 Answer 1

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As you have said, we can embed $A^m$ for any $m$. In particular, there is an embedding of $A^{2n}=A^n\oplus A^n$. To clarify a bit, write the embedding as $A_0\oplus A_0'$, both $A_0$ and $A_0'$ submodules of $A^n$ isomorphic to $A^n$.

Then we can inductively construct a sequence: given $A_j,A_j'$ each isomorphic to $A^n$, we can embed $A^{2n}$ into $A_j'$ as $A_{j+1}\oplus A_{j+1}'$. Then $A_\infty=\bigoplus_j A_j$ is free of rank $\aleph_0$.

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Excellent, thank you! –  Stefan Walter Jul 7 '12 at 18:56

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