I'm attempting to solve the following two problems and I've unfortunately hit a wall.
Q1. Show that if $M$ is a finite module over an infinite commutative ring $A$, then $M$ is a free module $\iff$ $M=\{0\}$.
Q2. Let $M$ be a module over a commutative ring $A$ and let $u_1, u_2, ..., u_n \in M$. Show that $\{u_1, u_1, ..., u_n\}$ is a basis of $M$ if and only if $M= Au_1\bigoplus Au_2\bigoplus ... \bigoplus Au_n$ and none of the $u_i$ are torsion.
Attempt at Q1. $\Leftarrow$ This direction is clear from a convention/definition given previously.
$\Rightarrow$ This is the direction I am struggling with. I know that M is a free module means that M admits a basis and if M is a finite module, then M is generated by a finite number of elements. But how do I combine this information to give M={0}?
(My not so attempt at Q2.) I honestly don't know where to go with this question. My guess is that the left to right implication comes from the definition of a basis, but I'm honestly at a loss (I've been staring at this problem on and off all day).
Any help with these questions would be very much appreciated.
As for Q1: take an element $u$ of $M$ contained in some basis of $M$. Then the elements $ru$, $r\in A$, must all be distinct.
As for Q2: The elements $u_1,\ldots ,u_n$ by definition form a basis of $M$ if and only if every $m\in M$ can be written in the form $m=r_1u_1+\ldots +r_nu_n$ with unique coefficients $r_i$. The implication $\Rightarrow$ is therefore clear, as you already suspected.
Assume now that the direct-sum-decomposition holds. This means that every $m\in M$ can be written as $m=r_1u_1+\ldots +r_nu_n$, where the summands $r_iu_i$ are unique. Assume next that $r_1u_1+\ldots +r_nu_n=s_1u_1+\ldots +s_nu_n$, then $(r_1-s_1)u_1+\ldots +(r_n-s_n)u_n=0$. Since the summands $(r_i-s_i)u_i$ are uniquely determined by the element $m=0$, they must all be $0$: $(r_i-s_i)u_i=0, i=1,\ldots ,n$. By assumption the submodules $Au_i$ are torsion-free, hence one gets $r_i-s_i=0$, $i=1,\ldots ,n$, that is the uniqueness of the coefficients.