Let us now assume that $A$ is a discrete valuation ring. Any element $\pi \in A$ for which $v(\pi) = 1$ is called a uniformizer. Such a uniformizer necessarily exists, since $v$ maps $A$ surjectively onto $\mathbb Z_{\ge 0}$.
If we fix a uniformizer $\pi$, why every $x ∈ k^\times$ can be written uniquely as $x = u\pi^n$ where $n = v(x)$ and $u = x/\pi^n ∈ A^\times$ ?
I think this is fundamental fact of DVR, but how to prove formally from definition?
Thank you in advance.
Well, just define $u=x/\pi^n$. Then $v(u)=0$, so $u\in A$. Also, $v(u^{-1})=0$ so $u^{-1}\in A$ as well, so $u$ is a unit of $A$.