I have following question about the red tagged statement in the image below (from Liu's "Algebraic Geometry", p. 260):
Why is a point $x \in X$ with codim $1$ (therefore $codim (\bar{x}) =1$) with $mult_x(D)= mult_{\mathcal{O}_{X,x}}(D_x)\neq 0$ a generic point in $X \backslash U$?
Here the $mult_A$-map is defined above and $D \in Div(X) =\mathcal{K}^*_X /\mathcal{O}^* _X (X)$ is Cartier Divisor (compare page 256).
My problem is following:
If $x$ is generic then (cosidered locally in an affine Noetherian neighborhood $U = Spec(A)$) it corresponds to a minimum prime ideal $p_x$ of $A$. Then in $\mathcal(O)_{X,x}= A_x$ all regular elements are inverible and vice versa because $A$ is localized at $A \backslash p_x$ and zero divisers are exactly the set $Ass(\mathcal{O}_{X,x}= p_x$ (because $A$ is not zero and p_x is the only prime ideal). Therefore for $Frac(\mathcal(O)_{X,x})^* $ consist of fractions $a/b$ with $a, b \in A \backslash p_x =\mathcal{O}_{X,x}^*$.
But in this case I get that $mult_x(D)$ is zero, because for $D_x = a/b$ with $a, b \in \mathcal{O}_{X,x})^*$ as below I get $mult_{\mathcal{O}_{X,x}}(D_x)= mult_{\mathcal{O}_{X,x}}(a/b)=mult_{\mathcal{O}_{X,x}}(a)-mult_{\mathcal{O}_{X,x}}(b) = 0 -0$ , what obviosly contradicts to the statement.
What causes the trouble? Does anybody see a correct way of proving this?
I think you are mixing up $\mathcal{O}_{X,x}$ and $\mathcal{O}_{X\setminus U,x}$.
First, I should say that there is no mention of scheme structure on $X\setminus U$, so the group $\mathcal{O}_{X\setminus U,x}$ does not mean anything. You can put the reduced structure, and in that case, this is true that $\mathcal{O}_{X\setminus U,x}$ is a field. But this is not important.
On the other hand, the group $\mathcal{O}_{X,x}$ is far from being a field. Therefore $mult_{\mathcal{O}_{X,x}}(a)$ might be non zero.
Check examples ! Let $X=\mathbb{A}^2=\operatorname{Spec}k[x,y]$ and $D$ the divisor defined by $(xy^{-2})$. This example shows that one should not put any scheme structure on $X\setminus U$. This closed set has some closed points (corresponding those for which $x=0$ or $y=0$) and two generic points corresponding to the ideals $(x)$ and the ideal $(y)$.
These ideals are exactly those where the multiplicity of $D$ is non zero : $mult_{(x)}(D)=1$ and $mult_{(y)}(D)=-2$.