1. ## Checking answer of ring and ideal

Let I be a subset of R, where R is a commutative ring and let I be an ideal of R.
Show that:
(i) I = R <=> 1 belongs to I <=> R/I = 0

(ii) For a in R, (a) = 0 ={0} <=> a = 0
and (a) = R <=> a in U(R)
Where (a) = Ra = {ra: r in R} is the principal ideal generated by a

This is my attempt:

(i) (=>) Since I = R and 1 is in R, so 1 belongs to I
(=>) 1 in R, then 1 + I = 0+I = R/I = 0
Hence R/I = 0 = { 0+I}
Coversely, (<=) R/I = 0, so 1+ I = 0 + I implies 1 belongs to I and since I in R then I = R
Is that enough for this argument?

Part(ii) I am not sure how to do it, please give some hints

Thank you so much in advanced

2. Since $\displaystyle 0\in I$ and $\displaystyle 0+I=I,$ from $\displaystyle 1+I=0+I$ you can deduce $\displaystyle 1=1+0\in 0+I=I,$ so it is ok.

$\displaystyle (a)=\{0\} \Leftrightarrow a=0$ is simple, just use the definitions. For instance, $\displaystyle (a)=\{0\}$ means for all $\displaystyle r\in R,\ ra=0.$ Then with $\displaystyle r=1\ ...$

$\displaystyle (a)=R\Leftrightarrow a\in U(R)$ can be shown using (i): prove that $\displaystyle 1\in (a)\Leftrightarrow a\in U(R) .$

In principal ideals, every element is a product of the generator with something in the ring: $\displaystyle x$ belongs to $\displaystyle (a)$ iff there is a $\displaystyle b\in R$ such that $\displaystyle ba=x.$