local PID that is not a field is a DVR

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I would be very happy if someone would check my proof of the fact that a local PID that is not a field is a DVR:

Let $A$ be a local PID that is not a field. Since irreducibles generate maximal ideals in a PID, it follows that $A$ has only one (up to multiplication by units) irreducible element, say $t \in A$. Hence (since PIDs are UFDs) every non-zero $a \in A$ can be written uniquely as $ut^n$ for a unit $u \in A$ and some $n \geq 0$ which we denote by $\nu(a)$. Letting $K$ be the field of fractions of $A$, we define a map $\nu:K^*\to \mathbb{Z}$ by $\nu(ab^{-1})=\nu(a)-\nu(b).$ It is clear that $\nu$ defines a discrete valuation on $K$ and, by construction, its valuation ring is precisely $A$. Q.E.D.

Many thanks!

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Yes, that's the idea.

It's clear that this is a valuation and that it's onto $\Bbb Z$.