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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

Dan
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1 Answers1

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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.

Hagen Knaf
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  • Tanks for your answer.

    I would have proven this wir the help of Gauss' lemma, like here: http://math.stackexchange.com/questions/678419/normalization-of-a-quotient-ring-of-polynomial-rings-reid-exercise-4-6?lq=1 But there we do not need to check whether the equation $y^2-f(x)$ remains irreducible mod $p$. Why ?

     – Dan Feb 09 '15 at 10:09
  • Because in the problem you cite one is working over a field as a base, and all the rings appearing are of dimension $1$. In your situation the base is itself a $1$-dimensional ring and the ring $\mathbb{Z}_p[x,y]$ has dimension $2$. – Hagen Knaf Feb 10 '15 at 00:22
  • Could you explain the zero in $0=\overline{cp_1}+\overline{cp_2},\overline{y}$

    Take for example $z=1/p+1/p^2*y$ Then $c=p^2$, $p^2z=p+y$ but how do you know that $p^2z$ is in $p\mathbb Z_p[x,y]$? We do not know yet that $z\in \mathbb Z_p[x,y]$

     – Dan Feb 12 '15 at 12:57
  • There is indeed a mistake / weak point in my argument - I'll correct it in a moment. – Hagen Knaf Feb 13 '15 at 07:04
  • Do you at least have a source for me that uses those two assumptions: 1. the equation gives a smooth affine curve and 2. the reduction is irreducible – Dan Mar 03 '15 at 12:15