Q:

Help in this proof on DVRs

I'm trying to understand this proof:

Anyone could clarify the converse please?
I really need help.
Thanks a lot

A:

Which part(s) do you want clarified?
If you can fill in the details for the first sentence of the proof, then you can use the result that a commutative ring is local if and only if the set of non-units is an ideal  (in which case, it is the unique maximal ideal).  You can find a proof of this result here.
Now for the statement about ideals.  Given a nonzero ideal $I$ of $R$, let $m = \min \{ n \in \mathbb{Z}_{\geq 0} : t^n \in I\}$.  (You should check that this set is nonempty so that it has a minimum.)  I claim that $I = (t^m)$.
First since $I \neq 0$, then there is a nonzero $x \in I$.  By assumption $x = u t^k$ for some unit $u$ and some $k \in \mathbb{Z}_{\geq 0}$.  Then $t^k = u^{-1} (u t^k) = u^{-1} x \in I$ since $I$ is an ideal.  Thus $\{ n \in \mathbb{Z}_{\geq 0} : t^n \in I\}$ is nonempty, hence has a minimum which we denote by $m$.  Since $t^m \in I$, then $(t^m) \subseteq I$, so it remains to show the reverse inclusion.  Given $r \in I$, then $r = u t^n$ by assumption.  Since $m$ is the minimum, then $n \geq m$, so $r = u t^n = (u t^{n - m}) t^m \in (t^m)$.  Thus $I = (t^m)$, as desired.

