Let $R$ be a ring. Let $z$ be a fixed element of $R$ and $J$ be an ideal of $R$. Define $I = \{r\in R|rz \in J\}$. Prove that $I$ is a left ideal of $R$.
My Attempt $I \neq \emptyset$, since $0_R\cdot z = 0_R \in J \Rightarrow 0_R \in I$
Let, $x,y \in J \Rightarrow x = az$ and $y = bz,$ where $a,b \in R \Rightarrow x - y = az - bz = (a-b)z \in J.$ Thus, $x-y \in I.$
$rx = r(az) = (ra)z \in J.$ Thus, $rx \in I.$
Therefore, $I$ is a left ideal of $R$. Do you think my proof is correct?
You are a little off on the defining characteristic of $I$. If $x\in I$, then $xz\in J$. So, you start with $x,y\in R$ such that $xz,yz\in J$. Then $$ (x-y)z=xz-yz. $$ Since both $xz,yz\in J$ and $J$ is an ideal, $xz-yz\in J$. Thus $x-y\in I$. Do something similar for the last part.