Let's say we have a projective algebraic set $X = \{p_1,…,p_n\}$ that's just a finite set of points. Is X non-singular? My understanding is that a variety/algebraic set is non-singular if the tangent space at each point has the same dimension as the variety/algebraic set itself, but I don't know what the tangent space of a single point would be.
Are finite sets of points in projective space non-singular.
algebraic-geometrycommutative-algebra
Related Solutions
I'll use the more comfortable notations $w,x,y,z$ in place of $x_0,x_1,x_2,x_3$ and assume the base field has characteristic zero..
i) The curve $V$ is isomorphic to the plane curve $V'$ in the $x,y$ plane defined by $y^2=x^3$.
The isomorphism is $p:V\to V':(x,y,z)\mapsto (x,y)$ with inverse $p^{-1}:V'\to V:(x,y)\mapsto (x,y,x^3)$.
The curve $V'$ is irreducible because the polynomial $y^2-x^3$ is irreducible (or because it is the image of $\mathbb A^1\to \mathbb A^2: t\mapsto (t^2,t^3)$). Hence the isomorphic curve $V$ is irreducible too.
The only singularity of $V'$ is $(0,0)$ where the tangent space has dimension $2$ (immediate from the Jacobian criterion you mention) so that $V$ has $(0,0,0)$ as only singularity, with tangent space of dimension $2$.
ii) The surface $Y$ is irreducible because the polynomial $xy^2-z^3\in k[w,x,y,z]$ that defines it is irreducible (notice that it is of degree $1$ in $x$).
The absence of the variable $w$ in the equation indicates that $Y$ is the cone with vertex $[1:0:0:0]$ over the curve $C\subset \mathbb P^2_{x:y:z}=V(w)$ with equation $xy^2-z^3=0$.
The curve $C$ has $[1:0:0]$ as only singularity (cf. Jacobian), hence your surface has the line $V(y,z)\subset \mathbb P^3$ as set of singularities . At every point of that line the tangent space has dimension $3$.
Tricks of the trade
a) You can use the Jacobian also in projective space
b) In an affine or projective space of dimension $n$ a hypersurface has tangent space of dimension $n$ at a singularity and of dimension $n-1$ at all non-singular points.
Edit: Cones
Since Jonathan asks, here is why a homogeneous polynomial $f(x,y,z)$ not involving $w$ defines the cone $C\subset \mathbb P^3$ with vertex $S=[1:0:0:0]$ over the projective curve $V(f)\subset \mathbb P^2_{0:x:y:z}$
A point $R$ on the line joining $S$ to a point $Q=[0:a:b:c]\in V(f)$ has coordinates $[u:va:vb:vc]$ for some $[u:v]\in \mathbb P^1$.
Since $f(R)=f(va,vb,vc)= v^{deg(f)} f(a,b,c)=0$, we see that indeed $R\in C$ and $C$ is the claimed cone.
- variety: integral scheme of finite type over a field. (This may vary a lot in literature: one can add "separated", remove "integral", add "reduced" again...).
- curve: one-dimensional variety.
- projective (over a field): closed subvariety of some projective space (over that field).
- nonsingular: all local rings at closed points are regular local rings.
Hence a projective nonsingular curve is a one-dimensional projective variety all of whose local rings (at closed points) are discrete valuation rings (as DVR means: regular local ring of dimension one).
The equivalence of categories you are working with is provided by the function field functor: $$C\mapsto K(C)=\{\textrm{rational functions on }C\}.$$ It is important to note that for $K(-)$ to be an equivalence, one has to consider the category of curves with arrows the dominant morphisms of curves. In addition, a third category is equivalent to the tho above: that of curves, with dominant rational maps.
Best Answer
When your base field $k$ is characteristic $0$ or algebraically closed, $X$ is always non-singular, and the morphism $X\rightarrow spec(k)$ is smooth(and etale) in these cases.
If you are in the setting of classical algebraic geometry, when $k$ is algebraically closed, the dimension of X is $0$, and the tangent space at each point is again $0$, so it is indeed non-singular.
But in modern setting, when $k$ is an arbitrary field, the answer to your question depends on whether the residue field at each $p_i$ is separable over $k$ or not. As a scheme, X is just the spectrum of a product of finite extension of your base field(assuming each $p_i$ is a closed point, but this is necessary for X to be a closed subset of $P^n$), i.e. $X= spec(\prod k_i)$, where each $k_i$ is a finite extension of $k$, your base field. When all these extensions are separable, the morphism $X\rightarrow spec(k)$ is smooth(actually, etale), and $X$ is a non-singular variety(if you call it a variety). The separable condition is always satisfied when $k$ is char $0$ or algebraically closed, but not always so in char $p$. Just like @KReiser mentioned, the case $F_p(t)/F_p(t^p)$ may happen, when you let your base field $k=F_p(t)$, and let $p$ to be the point in $P^1=Proj(k[x,y])$ corresponds to the polynomial $x^p=ty^p$.
Generally, the tangent space of a variety at a point is defined to be the dual of the stalk of the sheaf of relative differentials, i.e. $T_xX=\Omega_{X/k,x}^*$ . And smoothness is defined to be flat + relative differential locally free of the same relative dimension. When K/k separable, it is known that $Ω_{K/k}=0$.
In any case, $X$ is always a regular scheme.