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Observe the following theorem for vector spaces.

Let $V$ be a finite-dimensional, nontrivial vector space, and suppose $V=\operatorname{span}(v_1,v_2)$. Then some subset of $\{v_1,v_2\}$ will be a basis for $V$.

I would like to know if there is a corresponding theorem for R-modules (where $R$ is a commutative ring with 1), or if not, at least for free R-modules. Stating such a theorem would be useful for me, but attaching a simple proof to accompany it would be even better.

Edit

Just to give some context, I want to work with the ring $R=\mathbb{Z}[\sqrt{-5}]$. So a corresponding theorem which will work for at least this ring (even if it doesn't work for all rings) should be good enough for me.

Bernard
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Pascal's Wager
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2 Answers2

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No: take $V = \Bbb{Z}$ viewed as a $\Bbb{Z}$-module. $V$ is spanned by any coprime pair of integers $(v_1, v_2)$, e.g., $v_1 = 2$ and $v_2 = 3$, but it only has two bases: $\{1\}$ and $\{-1\}$.

Rob Arthan
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  • If it doesn't work for general R-modules, what about for $\mathbb{Z}[\sqrt{-5}]$-modules? (see my edit) – Pascal's Wager Apr 13 '18 at 18:14
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    @Pascal'sWager $\Bbb Z[\sqrt{-5}]$ as a module over itself is generated by $\sqrt{-5}$ and $4$, but neither of these two generates it alone. – Hagen von Eitzen Apr 13 '18 at 18:24
  • Thank you. I was trying to find a lemma to help me prove this https://math.stackexchange.com/questions/2733318/prove-that-v-is-not-a-free-module problem. If you want some more rep, you might want to check this problem out as well. – Pascal's Wager Apr 13 '18 at 18:30
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The statement is wrong for general $R$-modules, even for free ones: Consider $\mathbb Z$ as a $\mathbb Z$-module, then $\mathbb Z=\langle 2,3\rangle$, but neither $2$ nor $3$ generates $\mathbb Z$, hence $\{2,3\}$ does not contain a basis even though $\mathbb Z$ is free.

asdq
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