If $f(x)=ax$ for an idempotent element $a$ in a commutative ring $A$, then $\text{range} f = \langle a \rangle$?

If $f(x)=ax$ for an idempotent element $a$ in a commutative ring $A$, then $\text{range} f = \langle a \rangle$?
This is as a review of the Fundamental Homomorphism Theorem. The problem states to prove that $f$ is a homomorphism, which is not difficult to prove, and that the $\text {ker} f$ is the annahilator of $a$, which follows by definition. However, proving that $\text{range} f = \langle a \rangle$ has turned out to be more difficult, since I could think of an easy counterexample (e.g. $A= \mathbb R$ and $a= 1$, then $\text {ker} f = \{0\}$ and $\text{range} f = \mathbb R \ne \langle a \rangle$ regardless of whether $\langle a \rangle$ is the multiplicative group of $1$ or $\mathbb Z$. The FHT would come in later, when proving that $A/I_a \cong \langle a \rangle$ where $I_a$ is the annahilator of $a$. Is the problem ill stated or am I wrong?

Instead of the subring $\langle a\rangle$ we have to take the ideal $(a)$ generated by $a$, which in a commutative ring $A$ with unit is just $(a)=aA=\{ax\mid x\in A\}$.
• Does this mean the wrong notation is used? I'd think $\langle a\rangle$ means the cyclic group generated by $a$, not the ideal generated by $a$. Or is it a common practice to share the same notation? – hchar Nov 17 '19 at 0:15