Is the Fermat scheme $x^p+y^p=z^p$ always normal Let $K$ be a number field with ring of integers $O_K$.
Is the closed subscheme $X$ of $\mathbf{P}^2_{O_K}$ given by the homogeneous equation $x^p+y^p=z^p$ normal?
I know that this is true if $K=\mathbf{Q}$. (My method of proof is a bit awkward.)
I expect the answer to be no (unfortunately) in general. What is an equation for the normalization of $X$? In other words, what is an equation for the normalization of $\mathbf{P}^1_{O_K}$ in the function field of $\mathrm{Proj}K[x,y,z]/(x^p+y^p-z^p)$? 
Note that the only difficulty arises modulo $p$. The fibre is then reduced.
 A: This answer amplifies some of QiL's remarks in the comments above:
To investigate such a question, you can use Serre's criterion: normal $= R_1 + S_2$.
In this case your scheme is a projective hypersurface, thus Cohen--Macaulay,
and so in particular $S_2$.  So the only issue is $R_1$. Now the codimension
one points either live in the generic fibre, which is smooth, or are generic points in one of the special fibres.
So you are reduced to computing the local ring around a generic point in positive 
characteristic.  As you note, if this characteristic is different from $p$ then the fibre is smooth.  In particular, it is generically smooth (equivalently, generically reduced, or again equivalently, $R_0$) and all is good.  So you are reduced to the generic points in the char. $p$ fibres.  Here you can hope
to compute explicitly:
You can work in affine coords. where your curve is $x^p + y^p = 1$, and so your problem is to determine whether the ring $O_K[x,y]/(x^p + y^p -1)$ is regular after localizing at the  prime ideals dividing $p$.  Reducing mod $p$, you are asking if the minimal primes of the ring $(O_K/p)[x,y]/(x+y - 1)^p$ are prinicipal.  
Let me assume that there is a unique prime in $O_K$ above $p$, just to ease my typing.
If $O_K/p$ is a field $k$ (i.e. $p$ is unramified), you'll be in good shape, since the minimal prime of $k[x,y]/(x+y-1)^p$ is generated by $x+y-1$.  But if $p$ is ramified,
of degree $e$,
then you get problems: you need two generators: $(x+y - 1)$ and some nilpotent $\pi$ such that $\pi^e = 0$. 
