Singularities of arithmetic surface

Question

I have a problem understanding the discussion of example 8.3.54 in Liu's Algebraic Geometry and Arithmetic Curves. The setting is the following: We have a DVR with uniformiser $t$, characteristic of the residue field not $2$ or $3$ and the arithmetic surface $\operatorname{Spec}(R[x,y]/(y^2-t(x^3+t^3))$. He claims the surface has a unique singular point corresponding to the ideal $(x,y,t)$.

My thoughts on that are that as the surface is flat and of finite presentation over $R$, I can check smoothness on fibres. It is clear that the generic fibre is smooth and modulo $t$ the equation reduces to $y^2=0$ which just gives me a line, which is smooth as well, so I would conclude that the special fibre and therefore the whole surface is smooth, which contradicts Liu's discussion. Could someone point out the mistake in my considerations?

Maybe it has something to do that the equation is $y^2=0$, so I get a non-reduced structure in the special fibre? But if this was true then the surface would not be a curve over the DVR.

This I also posted here: https://math.stackexchange.com/questions/4446892/singularities-of-arithmetic-surfaces but maybe mathoverflow might be more suitable

Answer 1

There seems to be some confusion concerning "regularity" and "smoothness".

First of all, if $X$ is a noetherian integral scheme and $x\in X$, then $X$ is regular at $x$ if $$\mathrm{dim} \ m_x/m_x^2 = \mathrm{dim} X.$$ On the other hand, given a morphism of schemes $\mathcal{X}\to S$, one can also look at its smoothness.

It is not true that, if $\mathcal{X}$ is regular (i.e., nonsingular), then $\mathcal{X}\to S$ is smooth (unless $S$ is the spectrum of an algebraically closed field).

Let's look at your scheme $X = \mathrm{Spec} R[x,y]/(y^2-t(x^3+t^3))$, and determine what its regular points are. (Such points are also called nonsingular points.) We will use that "smoothness over a point implies regularity".

First, $X\to \mathrm{Spec} R$ is smooth over $D(t)$. This is indeed by what you say. The morphism to Spec $R$ is flat, and the special fibre over $k(t)$ is regular, assuming $\mathrm{char}(k(t)) \neq 2,3$.

So, this means that every point of $X$ over $D(t)$ is nonsingular. Thus, the singular points (read: nonregular points) of $X$ lie on the special fibre.

As pointed out in the comments, the special fibre itself is given by setting $t=0$. Denoting $k= R/tR$, we get that this special fibre is given by $y^2=0$ in $k[x,y]$. This is a non-reduced scheme, and highly singular.

Nevertheless, this doesn't prevent the scheme $X$ from being regular! First, let's agree that $\mathrm{dim}(X) = 2$, as it is an arithmetic surface (as you say).

Now, let's look at the maximal ideal $m$ of the point $P=(x,y,t)$ on $X$ lying on the special fibre. Squaring this ideal gives us $m^2=(x^2, xy, y^2, t^2, xt, yt)$.
Clearly $x,y,t$ span the vector space $m/m^2$. Since they are linearly independent, it follows that $\mathrm{dim} \ m/m^2 = 3\neq 2$, so that $P$ is singular (read: nonregular).

What about the other points? Well, for all other points $Q$ on the special fibre, you can show that $\mathrm{dim} \ m_Q / m_Q^2 =2$. So they are regular.