If $I$ injective then $\operatorname{Hom}_R(M,I)=0$ imply $M=0$?

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Let $R=(R, \mathfrak{m},k)$ be a local ring with maximal ideal $\mathfrak{m}$ and residue field $k=R/\mathfrak{m}$. Let $I_k$ the injective hull of $k$, i.e., it's an essential extension of $k$ and an injective object in $\operatorname{R-Mod}$.

Let $M$ an arbitrary $R$-module. Question:

Why $\operatorname{Hom}_R(M,I_k)=0$ imply $M=0$?

First or all is it neccessary to assume that $I_k$ is injective hull or can the claim be weakended to:

If $I$ injective object then for every $M \in \operatorname{R-Mod}$ the condition $\operatorname{Hom}_R(M,I)=0$ imply $M=0$?

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If $M\neq 0$, let $0\neq m\in M$. Then you have a surjection $Rm\to k$ and thus $\mathrm{Hom}_R(Rm, I_k)\neq 0$. Since $I_k$ is injective, such a non-zero homomorphism can be extended to $M$ and thus a contradicts the assumption.

This is not true for other injectives in general. For example, take $R=\mathbb{Z}_{(p)}$ and $I=\mathbb{Q}$, which is injective. But $\mathrm{Hom}_R(R/pR,I)=0$.