Let $R$ be a ring and let $J$ be a left ideal of $R$. Let $s \in R$.
We can make the module $R/J$. Let $\DeclareMathOperator{\ann}{ann}\ann_R(t + J) = {\{r \in R : rt + J = 0}\}$ be the annihilator.
I want to show that there is an injective homomorphism $$\theta : R/\ann_R(t + J) \rightarrow R/J$$.
Now, I defined the map to be $\theta(r + \ann_R(t + J)) = r + J$, and showed it was a homomorphism.
I'm having trouble showing it's injective. I tried to show that the kernel was trivial, like so;
$\ker(\theta) = {\{r + \ann_R(t + J) \in R/\ann_R(t + J) : r + J = 0}\}$. So it is the elements $r + \ann_R(t + J)$ such that $r \in J$. To show this is trivial we could show $J \subseteq \ann_R(t + J)$, as then it would just be elements of the form $0 + \ann_R(t + J)$, which is the zero in the module $R/\ann_R(t + J)$, but I don't see that this is true?
Any help would be appreciated.
Consider the $R$-module morphism $$ \alpha : R \to R / J, \ \ \ \alpha(r) = (rt + J).$$
Notice that the kernel of $\alpha$ is $ {\rm ann}_R(t + J)$; this is true by the definition of ${\rm ann}_R(t + J)$.
Hence $\alpha$ descends to an injective $R$-morphism $\theta$, whose domain is the quotient space $R / {\rm ker}(\alpha)$: $$ \theta : R / {\rm ann}_R(t + J) \to R / J, \ \ \ \theta(r + {\rm ann}_R(t+J)) = (rt + J).$$ I believe this is the $\theta$ that you were looking for. It's slightly different from the $\theta$ that you introduced in your question.