Proofs involving minimal generating sets and ideal (submodule) quotients If $M$ is a finitely generated nonzero module over a commutative ring $R$, we can show that $M^{\otimes k}$ is nonzero for all $k$, as follows: 

Select a minimal generating set $x_1, \dotsc, x_d$. Let $N :=
 \mathop{\sum}\limits_{i\neq d} Rx_i$ and $I := \{r \in R\mid rx_d \in
 N \} $. Note $I$ is a proper ideal. Then any $m \in M$ can be written
  as $n + rx_d$ with $n\in N $, and claim that $r$ is well-defined mod $I$. Therefore
  $$ M^{\otimes k} \to R/I \\ (n_1 + r_1x_d) \otimes \dotsb \otimes (n_k
 + r_k x_d) \mapsto r_1 \cdot \dotsc \cdot r_k  $$ is a nonvanishing linear map out of $M^{\otimes k}$. (Since its value on $x_d \otimes \dotsb \otimes x_d$ is one.)

I would summarize the engine of the proof as "the component of a generator $x$ in a minimal generating set is well-defined mod $(\mathop{\sum}\limits_{i\neq d} Rx_i : x_d).$" 
What's really going on in this proof, morally? Can you help solidify it in my mind by relating it to some other results? 
I thought of Nakayama's lemma, but there a minimal generating set was manipulated somewhat differently.
 A: One natural way to (more or less) rephrase your argument if you assume $R$ is Noetherian is the following.  Let $N$ be a maximal proper submodule of $M$ (this is guaranteed to exist by the Noetherian hypothesis).  By maximality of $M$, $M/N$ must be cyclic, so it is isomorphic to $R/I$ for some proper ideal $I$ (in fact, $I$ must be a maximal ideal, since $M/N$ must be a simple module).  So we have a surjection $f:M\to R/I$.  Now it is clear that $(R/I)^{\otimes k}$ is nonzero and in fact isomorphic to $R/I$, by repeatedly using the natural isomorphism $R/I\otimes P\cong P/IP$ for any $R$-module $P$.  But our map $f$ induces a map $f^{\otimes k}:M^{\otimes k}\to (R/I)^{\otimes k}$, which is surjective since a tensor product of surjections is surjective.  Thus $M^{\otimes k}\neq 0$.
Now if $R$ is not Noetherian, it's not obvious a priori that you can get a maximal proper submodule $N$ of $M$.  However, using a minimal generating set of $M$, you can still get $N$ such that $M/N$ is cyclic, by taking $N$ to be generated by all but one of the generators.  The rest of the argument then goes through as above (and in fact, since $M/N\cong R/I$ for some ideal $I$, by picking a maximal ideal containing $I$ you can prove that $M$ actually does have a maximal proper submodule).  Indeed, the map you construct is exactly my map $f^{\otimes k}$ composed with the canonical isomorphism $(R/I)^{\otimes k}\to R/I$.
Yet another way to formulate what's going on is that you could first tensor everything to $R/I$, which is a field since $I$ is a maximal ideal.  Let $M'=M\otimes_R R/I$.  Since we have a surjection $M\to R/I$, $M'\neq 0$ (the surjection $M\to R/I$ gives a surjection $M'\to R/I\otimes R/I=R/I$).  But now we can take the $n$-fold tensor product of $M'$ as an $R/I$ module, which will be nonzero because we are just taking a tensor product of nonzero vector spaces over a field.  But $(M')^{\otimes_{R/I}k}$ is naturally isomorphic to $M^{\otimes_R k}\otimes R/I$, so $M^{\otimes_R k}$ must be nonzero.
