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?

$u + v t^{m-n}\notin\mathfrak m$, so it is invertible.
