Proof that a DVR is a Euclidean domain

Question

I am independently studying valuation rings and I am currently trying to prove that a DVR is a Euclidean domain using the definition here http://abstract.pugetsound.edu/aata/section-factorization-domains.html.

Here is what I have:

Let $V = \{a \in F| \nu(a)\geq 0\}$ where $\nu: F\rightarrow \mathbb{Z}\cup \{\infty\}$ is a discrete valuation. Then

1.) For all nonzero $a,b \in V$ $\nu(a)\leq \nu(ab)$ since $ab \in \left <a \right >$ (this is a theorem).

2.) Let $a,b \in V$ with $b\neq 0$. Then if $ab^{-1} \in V$, we can write $a =b(ab^{-1}) +0$. If $ab^{-1} \not\in V$, then $\nu(b) > \nu(a)$. Now we can write $\nu(a) = \nu(b +(a-b)) \geq \min\{\nu(b), \nu(a-b)\}.$

I would like to be able to conclude that $\nu(a) \geq \nu(a-b)$ so that I can write $a = 1(b) + (a-b)$. It seems that this follows from $\nu(b) > \nu(a)$, but I am not completely confident that this is the case.

Thus, my question is: does $\nu(b) > \nu(a) \Rightarrow \nu(a) \geq \nu(a-b)$? If so, why? Or is my approach flawed?

Answer 1

The definition of the link refers to a Euclidean valuation, while a a discrete valuation on $F$ is a surjective function satisfying $$ \nu(ab)=\nu(a)+\nu(b),\;\nu(a+b)\ge \min\{\nu(a),\nu(b)\}. $$ The DVRfor $\nu$ then is the ring $R=\{0\}\cup \{x\in R\mid \nu(r)\ge 0\}$. It satisfies the Euclidean valuation property (is a Euclidean ring), because we have $\nu(a/b)=\nu(a)-\nu(b)$, see here, page $3$.

Answer 2

For a valuation, $\nu(a + b) = \min\{\nu(a),\nu(b)\}$ if $\nu(a) \ne \nu(b)$. Observe that

$$ \nu(b) = \nu(a + b - a) \ge \min\{\nu(a + b), \nu(-a)\}. $$

If we suppose that $\nu(a) < \nu(b)$ then we cannot have $\min\{\nu(a + b), \nu(-a)\} = \nu(a)$ (recall that $\nu(-a) = \nu(a)$). Therefore, in this case, it must be that

$$\min\{\nu(a + b), \nu(-a)\} = \nu(a + b).$$

Thus $\nu(a + b) \le \nu(a)$ and since $\nu(a + b) \ge \min\{\nu(a),\nu(b)\} = \nu(a)$, it follows that $\nu(a + b) = \nu(a)$.

So yes, if $\nu(b) > \nu(a)$ then $\nu(a) = \nu(a - b)$.