What is the missing discrete valuation ring? Let $k$ be an algebraically closed field and $K$ a function field of dimension 1 over it. Then we have that the projective 1-space $P^1$ over $k$ is isomorphic to the abstract variety $C_{K/k}$ (using the terminology and notations as in Hartshornes book) whose underlying space is the set of all discrete valuation rings of $K/k$, endowed with the finite complement topology. Also, we know the affine line lacks the “point at infinity” with respect to $P^1$.
So my question is, what is the corresponding discrete valuation ring that is the stalk at the point at infinity?
I know that if we embed $A^1$ into $P^1$ by mapping $x$ to $(x,1)$, then the missing point is just $(1,0)$ which has stalk isomorphic to $k[x,y]_{(y)}$ but this is isomorphic to the stalk $k[x,y]_{(x)}$ at $(0,1)$ which is weird since it implies the only discrete valuation ring missing is actually isomorphic to another that is not?? 
 A: $\mathbf{A}^1_k = \operatorname{spec}(k[x])$ has the following valuations:


*

*For any $a \in k$, $v_a(f) = $ order of $a$ as a root of $f = $ the largest power of $(x - a)$ that divides $f$.

*$v_\infty(f) = -\deg f$.
As always, this is extended to $k(x)$ by setting $v(f/g) = v(f) - v(g)$.
You can show that $\deg f = -v_0(f(\tfrac1x))$.
The DVR associated to $v_\infty$ is the ring $\left\{\frac fg \in k(x) : \deg(f) \le \deg(g) \right\}$.

Which is weird since it implies the only discrete valuation ring missing is actually isomorphic to another that is not??

This isn't surprising to me because all the points on $\mathbf{P}^1$ "look the same." $\mathbf{A}^1$ is a line and there is an isomorphism ($x \mapsto x + a$) that takes any point on that line to any other point. Likewise, on $\mathbf{P}^1$ there is an isomorphism taking any point to any other point.
For example, $f : x \mapsto \frac1x$ is an isomorphism that switches $0$ and $\infty$. That will give us an isomorphism $f_* : \mathcal{O}_{\mathbf{P}^1,0} \to  \mathcal{O}_{\mathbf{P}^1,\infty}$.
