Please consider these two questions:
- Let $R$ be a ring and $z \in R$, which is fixed. Let, $J = \{az \mid a \in R\}$. Prove that $J$ is a left ideal of $R$.
Skipping the subtraction part, this is what I did to show that multiplication holds: Let, $r \in R \Rightarrow r(az) = (ra)z \in J, ra \in R$. Thus, $J$ is a left ideal of $R$.
- Let $R$ be a ring, $z \in R$ (also fixed) and $J$ be a left ideal of R. Define $I = \{a \in R \mid az \in J\}$. Prove that $I$ is a left ideal of R.
Again, skipping the subtraction part, I did the same thing as before for multiplication(Let, $r \in R \Rightarrow r(az) = (ra)z \in J, ra \in R$. Thus, $I$ is a left ideal of $R$) and got wrong. Intuitively looking, these two conditions look the same to me. I mean don't they have the same meanings? If not,then what is the difference between these two conditions $J = \{az \mid a \in R\}$ and $I = \{a \in R \mid az \in J\}$?
There's an obvious difference insofar as $J$ of part 1 is defined in terms of a ring element $z$, whereas $I$ of part 2 is defined in terms of a ring element $z$ and an ideal $J$ (which should not be confused with the ideal $J$ defined in the first part of the problem). Actually if you do take the $J$ as defined in part 1 and also the same $z$, then you automatically obtain $I=R$, which will differe from $J$ (unless $z$ is a unit).
By the way, a correct proof that $I$ is an ideal (here, the multiplicative property) might gol like this: Let $a\in I$ and $r\in R$. Then $az\in J$ and hence (as $J$ is an ideal) $raz\in J$, which shows $ra\in I$.