Let $A$ be a commutative Noetherian ring with unity with $\mathrm{Spec}(A)$ finite and discrete. For any $A$-module $M$ and any homothety $f_r:M\to M,\ m\mapsto mr,\ r\in A$, if $\ker(f_r)=\{0\}$, then $f_r$ is surjective.
I do not know whether I am on the right track. I am failing to reason it out clearly.
Proof: If $A$ is a commutative Noetherian ring with unity, then Spec($A$) is finite and discrete iff $A$ is Artinian. Hence A is a finite product of commutative Artinian local rings, say $$A\cong A_1\times\ldots\times A_n,n\in\Bbb Z_{>0}.$$ For any $x\in A, x=(x_1,\ldots,x_n)$ with $x_i\in A_i$ and each $x_i$ is either nilpotent or invertible because $A_i$ is Artinian local. If $\ker(f_r)=\{0\}$, then $\ker(f_r)=\{m\in M:f_r(m)=mr=m(r_1,\ldots,r_n)=(0)\}$ implies that $r$ is not a zero divisor on $M$.
Also $\ker(f_r)=\{0\}$ iff $f_r$ is injective on $M$ iff $r\notin \mathcal{P}$ for all primes $\mathcal{P}\in\text{Ass}(M)$ where $\text{Ass}(M)$ are the associated primes of $M$.
Best Answer
It suffices to verify the statement after localizing at each maximal ideal of $A$, so we may assume $A$ is an Artinian local ring. (Or, in terms of your decomposition $A\cong A_1\times\dots\times A_n$, $M$ also decomposes as a product $M_1\times\dots\times M_n$ where each $M_i$ is an $A_i$-module, and it suffices to check that multiplication by $r_i$ is surjective on each $M_i$.) So, $r$ is either nilpotent or invertible. If $r$ is invertible, then $f_r$ is an isomorphism and in particular is surjective. If $r$ is nilpotent, let $n\geq 0$ be minimal such that $Mr^n=0$. If $n>0$, then by minimality of $n$ there is some nonzero $x\in Mr^{n-1}$, but then $xr=0$, contradicting the assumption that $f_r$ is injective. Thus $n=0$, i.e. $M=0$, and $f_r$ is trivially surjective.