I've managed to prove the following exercise:
(21.) Let $\mathfrak{m}$ be a maximal ideal of a commutative ring $R$. Prove the following: if $A$ is a finitely generated $R$-module, and $x_1, x_2, \ldots, x_n$ is a minimal generating subset of $A$, then $x_1 + \mathfrak{m} A, \ldots, x_n + \mathfrak{m} A$ is a basis of $A/\mathfrak{m}A$ over $R / \mathfrak{m}$.
in Grillet's Abstract Algebra, page 377, only under the hypothesis that $\mathfrak{m}$ is the only maximal ideal, i.e., that $R$ is a local ring.
Weak proof: Firstly, the scalar multiplication is defined as $(r\!+\!\mathfrak{m})(a\!+\!\mathfrak{m}A) \!:=\! ra\!+\!\mathfrak{m}A$. To see this is well defined, notice that if $r\!+\!\mathfrak{m} \!=\! 0$, i.e. $r\!\in\!\mathfrak{m}$, then $ra\!\in\!\mathfrak{m}A$, i.e. $ra\!+\!\mathfrak{m}A\!=\!0$; and also that if $a\!+\!\mathfrak{m}A\!=\!0$, i.e. a\!=\!ma'\!\in\!\mathfrak{m}A, then ra\!=\!rma'\!\in\!\mathfrak{m}A, i.e. $ra\!+\!\mathfrak{m}A\!=\!0$. Actually, the $R/\mathfrak{m}$-module $A/\mathfrak{m}A$ is constructed by first creating the $R$-module $A/\mathfrak{m}A$ (this is possible because $\mathfrak{m}A$ is a submodule of $A$, since $\mathfrak{m}$ is an ideal of $R$) and then turning it into an $R/\mathfrak{m}$-module (this is possible because $\mathrm{Ann}_R(A/\mathfrak{m}A)\supseteq\mathfrak{m}$).
To prove $x_1\!+\!\mathfrak{m}A,\ldots,x_n\!+\!\mathfrak{m}A$ generate $A/\mathfrak{m}A$, we must show that $(R/\mathfrak{m})(x_1\!+\!\mathfrak{m}A)\!+\!\ldots\!+\!(R/\mathfrak{m})(x_n\!+\!\mathfrak{m}A)\!=\!A/\mathfrak{m}A$, which means $Rx_1\!+\!\ldots\!+\!Rx_n\!+\!\mathfrak{m}A\!=\!A/\mathfrak{m}A$, or equivalently, $Rx_1\!+\!\ldots\!+\!Rx_n\!+\!\mathfrak{m}A\!=\!A$, but this is true since $Rx_1\!+\!\ldots\!+\!Rx_n\!=\!A$ by hypothesis. Alternatively, we could argue that since $x_1,\ldots,x_n$ generate the $R$-module $A$, they generate the $R$-module $A/\mathfrak{m}A$, and therefore generate the $R/\mathfrak{m}$-module $A/\mathfrak{m}A$.
If $x_1\!+\!\mathfrak{m}A,\ldots,x_n\!+\!\mathfrak{m}A$ were $R/\mathfrak{m}$-linearly dependent, then WLOG $x_n\!+\!\mathfrak{m}A$ could be expressed as a $R/\mathfrak{m}$-linear combination of the others, so already $x_1\!+\!\mathfrak{m}A,\ldots,x_{n-1}\!+\!\mathfrak{m}A$ would generate $A/\mathfrak{m}A$, which would mean $Rx_1\!+\!\ldots\!+\!Rx_{n-1}\!+\!\mathfrak{m}A\!=\!A$. But since $J(R)\!=\!\mathfrak{m}$, by Nakayama's Lemma this would mean $Rx_1\!+\!\ldots\!+\!Rx_{n-1}\!=\!A$, a contradiction with the hypothesis on minimality of $x_1,\ldots,x_n$. $\blacksquare$
Question: how can I prove the general version of the exercise? I am somewhat skeptical of the claim...