I am stuck on the following exercise:
Suppose $\mathfrak a$ is an ideal of a ring $A$ (commutative with unity) and $r(\mathfrak a)=\{x\in A\mid\exists n \in \mathbb{N}: x^n \in \mathfrak a\}$. I need to show that $r(\mathfrak a)=(1)$ iff $\mathfrak a=(1)$.
How do I show this? If $\mathfrak a=(1)$, then take $x \in (1)$. Then $x \in \mathfrak a$ and so $x^1 \in \mathfrak a$ which shows that $x \in r(\mathfrak a)$ and hence $r(a)=(1)$. However, I am unable to proceed in the other direction [$r(\mathfrak a)=(1)$ implies $\mathfrak a=(1)$]. Perhaps I am missing some information here?
$r(\mathfrak a)=1$ implies that $1\in r(\mathfrak a)$ this implies that $1^n=1\in a$.