Showing that the $\text{ord}(a+b) = \text{min}(\text{ord}(a),\text{ord}(b))$ in a DVR This is Problem 2.29 from Fulton's  Algebraic Curves. First a bit of background because I don't know how standard his terminology is.
For a discrete valuation ring $R$ with maximal ideal $\mathfrak{m} = \langle t \rangle$, any element $z$ of $K := \text{Frac}(R)$ can be written uniquely as $z = u t^n$ where $u \in R^*$ and $n \in \mathbb{Z}$. In this case, we define the order of $z$ to be $\text{ord}(z) =n$.

Let $R$ be a discrete valuation ring with quotient field $K$. If $a,b \in K$ and $\text{ord}(a) < \text{ord}(b)$, show that $\text{ord}(a+b) = \text{ord}(a)$.

So far, I've got the following. Suppose $a = ut^m, \; b = vt^n$, where $m < n$. Then 
$$a+b = (u + v t^{m-n})t^m$$
and since the exercise tells me that the order of $a+b$ should be $m$, I'm inclined to believe that it remains for me to show that $u + v t^{m-n}$ is a unit in $R$. However, I can't seem to find a combination of $u,v,u^{-1},v^{-1}$ and $t$ that is inverse to $u + v t^{m-n}$. Am I attacking this problem incorrectly?
 A: $u + v t^{m-n}\notin\mathfrak m$, so it is invertible.
A: You can prove it by only using the properties of  valuations (discrete or not).
Let $a$, $b$ so that  $\text{ord}(a) < \text{ord}(b)$
By considering the difference we get
\begin{eqnarray*}
\text{ord}(b) \ge \min ( \text{ord}(a+b), \text{ord}(a))
\end{eqnarray*}
But $\text{ord}(b)< \text{ord}(a)$. We conclude that 
\begin{eqnarray*}
\text{ord}(b) \ge \text{ord}(a+b)
\end{eqnarray*}
However, we also have 
\begin{eqnarray*}
\text{ord}(a+b) \ge \min ( \text{ord}(a), \text{ord}(b)) = \text{ord}(b)
\end{eqnarray*}
From the two preceding inequalities we get 
\begin{eqnarray*}
\text{ord}(a+b) = \text{ord}(b)
\end{eqnarray*}
A: First we suppose $a=0$, then
$$ord(b)>ord(0)=\infty\Rightarrow ord(b)=\infty \Rightarrow b=0\Rightarrow$$
$$ord(a+b)=ord(0)=ord(a)$$
Now we suppose que $a\neq 0$. We are going to prove the two inequalities:
$\geq]$ This inequality is trivial:
$$ord(a+b)\geq min\{ord(a),ord(b)\}=ord(a)$$
$\leq]$ We define $d:=ord(a)$, $e:=ord(b)$, $f:=ord(a+b)$, then
$$a=ut^d,\hspace{0.25cm}b=vt^e,\hspace{0.25cm}a+b=wt^{f}$$
where $u,v,w$ are units. We are going to check the inequality
$$a+b=wt^f\Rightarrow a=wt^f-b=wt^f-vt^e=t^{min\{e,f\}}(wt^{f-min\{e,f\}}-vt^{e-min\{e,f\}})\Rightarrow$$
$$ord(a)=ord(t^{min\{e,f\}}(wt^{f-min\{e,f\}}-vt^{e-min\{e,f\}})) = ord(t^{min\{e,f\}}) + ord(wt^{f-min\{e,f\}}-vt^{e-min\{e,f\}})=$$
$$min\{e,f\} + ord(wt^{f-min\{e,f\}}-vt^{e-min\{e,f\}}) \geq min\{e,f\}=min\{ord(b),ord(a+b)\}$$
Then $ord(a)\geq min\{ord(b),ord(a+b)\}$, thus there are two possibilities, one of them is if $min\{ord(b),ord(a+b)\} = ord(b)$ then $ord(a)\geq ord(b)$, but this is impossible because the hypothesis says $ord(a)<ord(b)$, then it has to be the other possibility, so $min\{ord(b),ord(a+b)\}=ord(a+b)$, so we have
$$ord(a)\geq ord(a+b)$$
