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Let $D$ be a principal ideal domain and $M$ a free finitely-generated D-module. Prove or disprove

(1) If a generator of $M$ has the minimum possible elements, then is a basis.

(2) If a set is LI and has the maximum possible elements, then is a basis.

Trying to prove (1): Let $r$ be the rank of $M$ and $G$ a generator with $r$ elements. Define $L=D\cdot G$ the $D-$module of basis $G$. If $\phi:L\to M$ is shown to be an isomorphism, then we are done.

$\displaystyle \frac{\ker \phi}{L}\oplus\ker \phi\simeq L$ because $0\to\ker \phi \to L \to \frac{\ker \phi}{L}$. I want to have $\ker \phi =0$ because that shows $L\simeq M$ and that they have the same rank, making $G$ a basis. Now this doesn't seem possible because the rank of $\frac{\ker \phi}{L}$ will be rank of $\ker \phi$ minus rank of $L$. Because of this, I think it might be false.

Trying to prove (2): If it has the maximum possible elements and it is linearly independent then it is also a generator (unless there is no basis?) which means (2) is true.

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Hint: Consider $F$ the fraction field of $D$, and $M_F$ the $F$-vectorspace associated to $M$, show that $(m_1,..,m_n)$ is a minimal system of generator of $M$ if and only if $(p(m_1),...,p(m_n))$ is a basis of $M_F$ where $p:M\rightarrow M_F$ is the canonical map.

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  • $\begingroup$ So it proves the linear independence of $m_1,\dots,m_n$ and thus is a basis? $\endgroup$ – Cure Nov 22 '17 at 19:22
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(2) Is not true. Consider $R=D=\Bbb Z$, then $\{2\}\subset \Bbb Z$ is a linearly independent set (and there can't be any larger linearly indepedent set), but it is not a generator.

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