I'm currently looking at the following two questions:
i) Consider $V = Z(I) \subset \mathbb A_k^3$ where $I$ is generated by $X_1^3 – X_3$ and $X_2^2-X_3$. Find the points at which $V$ is singular and compute the dimensions of the tangent spaces there.
ii) Determine the singular points of the surface $Y$ in $\mathbb P^3$ defined by the polynomial $X_1X_2^2 – X_3^3 \in k[X_0, X_1, X_2, X_3]$ and compute the dimensions of the tangent spaces there.
Now I should state the definitions I have.
For an affine variety $X = Z(f_1, \ldots, f_r) \subset \mathbb A_k^n$ and a point $p \in X$, we define the tangent space of $X$ at $p$ to be $T_pX = \mathrm{Der}(k[X], \mathrm{ev}_p)$, where $\mathrm{Der}(A,\phi)$ denotes the space of derivations $A \to k$ centred at $\phi$.
This can be shown to be equivalent to $T_pX = \mathrm{ker} \begin{pmatrix} \frac{\partial f_1}{\partial x_1}(p) & \ldots & \frac{\partial f_1}{\partial x_n}(p) \\ \vdots & \ldots & \vdots \\ \frac{\partial f_r}{\partial x_1}(p) & \ldots & \frac{\partial f_r}{\partial x_n}(p) \end{pmatrix}: k^n \to k^r$.
For $X$ an arbitrary variety, we define $T_pX = T_pU$ where $U$ is any affine open neighbourhood of $p$ (this can be shown to be well-defined).
If $X$ is an irreducible variety, define $\mathrm{dim}(X) = \mathrm{min} \{ \mathrm{dim} \ T_pX \ | \ p \in X \}$.
If $X$ is irreducible and $k$ is algebraically closed, say $p \in X$ is a smooth point if $\mathrm{dim} \ T_pX = \mathrm{dim} X$, and say $p$ is singular otherwise.
For question i), seeing as my definition of singular only applies to irreducible varieties, I better show $V$ is in fact irreducible. I'm not sure how easy this is – I can't really spot how to do it. With some guesswork I'd say that $k[V] \cong k[x^2, x^3]$ which is clearly in integral domain so $V$ is irreducible, but I'm not sure about the details. Now let's write out the relevant matrix:
$\begin{pmatrix} 3p_1^2 & 0 & -1 \\ 0 & 2p_2 & -1 \end{pmatrix}$ where $p = (p_1, p_2, p_3)$. We have the points $p = (\frac{1}{\sqrt{3}}, \frac{1}{2}, p_3)$ where the rank is $2$ i.e. $\mathrm{dim}T_p X = 1$. The rank can obviously never be $3$, so we have that $\mathrm{dim}X = 1$. So the singular points are those for which the rank is 1. That's true precisely for the points $\{ (0,0,t) \ | \ t \in k \}$.
For question ii), I'm also unsure. Again, why is the surface $Y$ irreducible? Now I'm not sure how to deal with the fact that we're in projective space, so what I'm writing from now onwards may be complete nonsense. We have the standard affine open cover $\mathbb P^3 = U_0 \cup \ldots \cup U_3$, and it's obvious that $Y \cap U_0 \neq \emptyset$. So if the first coordinate of $p$ is non-zero, we can use the definition above to ascertain $T_pY = T_p (Y\cap U_0)$. If the first coordinate is zero, I'll worry about that later. So we're now assuming the first coordinate of $p$ isn't zero, and we're living in (irreducible if $Y$ is) affine space. The "affine" (de-homogenised) polynomial corresponding to $X_1 X_2^2 – X_3^3$ is just $xy^2 – z^3 \in k[x,y,z]$. Now there's a lemma that says if $f \in k[x_1, \ldots , x_n]$ is prime, then $\mathrm{dim} Z(f) = n – 1$. I'll assume our polynomial is prime (I don't know what else to do, so I might as well check later). So the singular points are those for which $\mathrm{dim} T_p (Y \cap U_0) = 3$, which are precisely those whose partial derivatives all vanish. These are the points $\{ (t,0,0) \ | \ t \in k^\times \}$, and the tangent spaces all have dimension 3. Now I think about it, if the first coordinate of $p$ is 0 everything seems to break, so let's leave the answer there.
I greatly appreciate the effort it takes to wade through this potential nonsense. My method seems very long-winded; I don't know if I've missed the point somewhere, or if the definitions I'm using are bad, or if this genuinely is how to go about solving these problems. I'm especially unconfident in my answer to ii).
Thank you.
Best Answer
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.