${\rm Hom}_R(M, R/M) =\{0\} \implies R$ is a field. Let $R$ be a local ring with maximal ideal $M$. Suppose $M$ is finitely generated. Prove that if ${\rm Hom}_R(M, R/M) =\{0\}$, then $R$ is a field.
${\rm Hom}_R(M, R/M)$ stand for the group of $R$-module homomorphisms from $M$ to the quotient $R/M$.
 A: $\DeclareMathOperator{\Hom}{\operatorname{Hom}}$By tensor-Hom adjointness, $$\Hom_R(M, R/M) \cong \Hom_R(M, \Hom_{R/M}(R/M, R/M)) \cong \Hom_{R/M}(M \otimes_R R/M, R/M) \cong \Hom_{R/M}(M/M^2, R/M) = (M/M^2)^*$$ where $*$ denotes vector space dual over the field $R/M$. Thus $(M/M^2)^* = 0 \iff M/M^2 = 0 \iff M = 0$ by Nakayama's Lemma (since $M$ is finitely generated), i.e. $R$ is a field.
A: Of course i find zcn's answer very instructive, however i think there is a more low-tech proof:
Suppose $R$ is not a field. Then $M \neq 0$ or equivalently by Nakayama $M \neq M^2$ (since $M$ is finitely generated).
Hence there exists an element $x \in M$ whose class $\bar{x}$ in $M/M^2$ is non-zero. Since $M/M^2$ is an $R/M$ vector space, we can define a nonzero $R/M$ homomorphism $f:M/M^2 \rightarrow R/M$ which is the projection onto the subspace spanned by $\bar{x}$. Now compose $f$ with the natural projection $\pi: M \rightarrow M/M^2$ to get a non-zero $R$-homomorphism $f \circ \pi : M \rightarrow R/M$.
This shows that if $R$ is not a field we must have $\operatorname{Hom}_R(M,R/M) \neq 0$.
