Normalization or integral closure of ring over $\mathbb Z_p$ Let $p$ be a prime larger than four. And denote the $p$ adic integers by $\mathbb Z_p$.
Consider the ring $A=\mathbb Z_p[x]$ and its field of fractions $K=\mathbb Q_p(x)$. Now let's extend $K$ to a field $L$, which is a function field of a curve, like $L=\mathbb Q_p(x)[y]/(y^2-(x^5-2)(x-p)(x+p))$.
Now what is the normalization or integral closure of $A$ in $L$?
And (as an algebraic question) why do I also have to look at the situation over $\mathbb F_p$ as well?
The geometric answer is: This normalization describes an affine model $\mathcal Y$ of a curve $Y$ with special fiber $\bar Y=\mathcal Y\otimes\mathbb F_p$, which has a singularity at $(0,0)$.
My ideas so far: 


*

*The curve is nonsingular over $\mathbb Z_p$, so nothing to do there.

*Over $\mathbb F_p$ we have a double point at $(0,0)$. Normally one gets rid of such a point by introducing somethin like $t:=y/x$. But how do I do this here?
I also know the easy examples $y^2=x^3$ and $y^2=x^2(x+1)$.
Kind regards
 A: The integral closure of $\mathbb{Z}_p[x]$ in $L$ equals $\mathbb{Z}_p[x,y]$: as you already remarked the ring $\mathbb{Q}_p[x,y]$ is integrally closed, since the plane curve $y^2=(x^5-2)(x^2-p^2)$ has no singular points (Jacobian criterion).
Let $z\in L$ be an element of the integral closure of $\mathbb{Z}_p[x]$. Then $z\in \mathbb{Q}_p[x,y]$ and therefore $z=p_1+p_2y$, where $p_1,p_2\in \mathbb{Q}_p[x]$. Assume that at least one of the coefficients of the $p_i$ does not lie in $\mathbb{Z}_p$. Then there exists some $c$ in the maximal ideal $p\mathbb{Z}_p$ such that $cp_1,cp_2\in\mathbb{Z}_p[x]$ and at least one of their coefficients is a unit in $\mathbb{Z}_p$. Now take the equation $cz=cp_1+cp_2y$ modulo a prime ideal $P$ of the integral closure of $\mathbb{Z}_p[x]$ in $\mathbb{Q}_p(x,y)$ lying over the maximal ideal $p\mathbb{Z}_p$: $0=\overline{cp_1}+\overline{cp_2}\,\overline{y}$, where overlining means taking residue classes. This is contradicting the fact, that $1,\overline{y}$ are linearly independent over $\mathbb{F}_p(\overline{x})$, since the polynomial $Y^2-(\overline{x}^5-2)\overline{x}^2$ is irreducible over $\mathbb{F}_p(\overline{x})$.
Some remarks: in contrast to the situation for curves over fields taking the normalization of a curve over a ring is not the same thing as removing singularities. The normalization only removes the singularities lying on the generic fibre, while singularities on the other fibres may remain, getting only a bit "smoother" if one has good luck.
