I'm trying to understand where the valuation defined by a prime divisor on an integral, Noetherian separated scheme regular in codimension 1 comes from.
In particular, I'm looking at this example: $X = \text{Spec}( \mathbb{C}[x,y,z,w]/(xw-yz))$ and $D= V(x,y)$, the prime divisor corresponding to $x=0$ and $y=0$.
There are two descriptions I've seen of this valuation: one is that $D$ has a generic point $\eta$ (which corresponds here to the minimal prime $(x,y)$ here in $D$). Then the valuation $v_D$ is the valuation on $\mathcal{O}_{X,\eta}$, a DVR.
So in this case $\mathcal{O}_{X,\eta}= \left(\frac{\mathbb{C}[x,y,z,w]}{(x,y)}\right)\bigg|_{(x,y)} \cong \mathbb{C}[z,w]|_0 = \mathbb{C}(z,w)$. This is a field then with maximal ideal $0$ - so the valuation of $z$ and $w$ are 0 are units and the valuation of $x$ and $y$ are undefined (as they are 0). Is this right?
However, I saw another description which had us consider $U$ open for which $D \cap U \neq \emptyset$. Then we apparently would have $D \cap U = \text{Spec} (\mathbb{C}[x,y,z,w]/(xw-yz) / P)$ for some prime ideal $P$ and then we have that $\mathbb{C}[x,y,z,w]/(xw-yz) / P$ localised (at 0?) gives us a DVR with a valuation on it. This I find even more confusing.
Could somebody clarify if my calculation above is correct? Could you also explain the second description please?
Your description of $\mathcal{O}_{X,\eta}$ is not correct. It should be $$\mathcal{O}_{X,\eta} = \left(\frac{\mathbb{C}[x,y,z,w]}{(xw-yz)}\right)_{(x,y)}$$
In this ring, $x,y$ clearly have valuation 1 (they are in the maximal ideal $\mathfrak{m} := (x,y)$ but not in $\mathfrak{m}^2$), and $z,w$ have valuation 0 since they are units (they are not in the ideal $(x,y)$).
The second definition basically says that to compute describe divisors and compute valuations (which basically boils down to computing local rings), it suffices to restrict yourself to an open subscheme which intersects your divisor nontrivially. This is perfectly okay simply because local rings (stalks of the structure sheaf) are local objects, and thus if $\eta\in X$, and $U\subset X$ is open and contains $\eta$, then there is an "equality" $\mathcal{O}_{X,\eta} = \mathcal{O}_{U,\eta}$.