the ℤ-module ℚ has no minimal generating set

hbghlyj

In @rapunzel's answer, if I understand correctly, \eqref{1} is from the assumption $\Bbb Q=\left<a,a_1,\cdots,a_n\right>$ where $\left<a,a_1,\cdots,a_n\right>$ is a finite subset of the generating set $S$, and $T=S\setminus\{a\}$

\begin{equation}a=a\cdot k_0+\sum_{i=1}^na_i\cdot k_i,\label1\end{equation}for some $k_i\in\Bbb{Z}$ and $a_i\in T$.
Then $$a(1-k_0)=\sum_{i=1}^na_i\cdot k_i.$$ Letting $b_i=a_i/(1-k_0)$

Why does he assume $k_0\ne1$ in this division? I think in \eqref{1} we could write $a=a\cdot 1+\sum_{i=1}^na_i\cdot 0$

hbghlyj

Letting $b_i=a_i/(1-k_0)$ -- clearly if $a_i$ generate $\Bbb{Q}$ then $b_i$ also generate it. Also we can assume that $a\not=b_i$ for any $i$ -- we get $$\sum_{i=1}^nb_i\cdot k_i=\dfrac{1}{1-k_0}\sum_{i=1}^na_i\cdot k_i=a.$$

He said “$b_i$ generates $\Bbb Q$”. But I think he should write “$S\setminus\{a,a_1,\cdots,a_n\}\cup\{b_1,\cdots,b_n\}$ generates $\Bbb Q$” ?
Even if he proved “$S\setminus\{a,a_1,\cdots,a_n\}\cup\{b_1,\cdots,b_n\}$ generates $\Bbb Q$”, we want to find a proper subset of $S$ that still generates $\Bbb Q$. But here $b_i\notin S$, which cannot to prove the theorem.

Comparing with other answers: @Julian Rosen showed $S\setminus\{a\}$ which is a proper subset of $S$, generates $\Bbb Q$, so the theorem is proved.