A point $p$ of a variety $X$ is smooth if and only if it is smooth in $U\subset X$ an open sub variety.

Question

I want to show the following:

Let $X$ be an algebraic variety, let $U$ be an open subvariety of $X$ containing $p$. Then $X$ is smooth at $p$ if and only if $U$ is smooth at $p$.

Here is my definition of being smooth at a point

Definition: Let $X$ be an algebraic variety. We say $X$ is smooth at $p$ if there is an open subvarity $U$ of $X$ containing $p$ and an isomorphism $\varphi: U \rightarrow Z(f_1,..,f_{n-d})\subset \mathbb A^n$ for some $d\leq n$ for some $f_1,..,f_{n-d}$ such that the rank of $$\left(\frac{\partial f_i}{\partial x_j}(\varphi(p)\right)$$ is maximal.

---

Attempt:

($\Leftarrow$) Let $U$ be smooth at $p$, then there is some $\overline{U}$ an open subvariety of $U$ and $\varphi:U\rightarrow Z(f_1,...,f_{n-d})$ satisfying the smoothness criterion. Since $U$ is a subvariety of $X$ so to is $\overline{U}$. Again this $\varphi:U\rightarrow Z(f_1,...,f_{n-d})$ still satisfies the smoothness crieterion so we are done.

($\Rightarrow$) Let $X$ be smooth at $p$, then there is some $\overline{U}$ an open subvariety of $X$ and an isomorphism $\varphi:\overline{U}\rightarrow Z(f_1,..f_{n-d}$ satisfying the smoothness criterion.

We want an open subvariety of $U$, define $V=U\cap\overline{U}$ this certainly is an open subvariety of $U$. And $\varphi\mid_V$ is an isomorphism onto its image. But how do we know its image is of the form $Z(f_1,...,f_{n-d})$ and how do we know that the Jacobian of the restriction still has full rank?

Answer 1

The other answer has some valuable content, but it's long, not an easy read to my eyes, and could be more self-contained. Here I present what I feel is a more direct solution.

---

One direction is easy: if $p\in X$ is smooth as a point of $U\subset X$, then the open neighborhood $U'\subset U$ obtained from the definition of smoothness is also an open neighborhood of $p\in X$ satisfying the same required properties.

For the other direction, suppose $p\in X$ is a smooth point and $U$ is an open neighborhood of $p$. Let $U'$ be the open neighborhood of $p$ guaranteed by the definition of smoothness. If $U'\subset U$, we're done. Else, $U'\cap U$ is an open subset of $U'$, with closed complement $V\subset U'$.

As $V$ is a closed subvariety of the affine variety $U'\cong Z(f_1,\cdots,f_{n-d})\subset \Bbb A^n$ which is disjoint from $p$, we can find a function $f$ on $U'$ vanishing on $V$ and nonvanishing on $p$. Then $D(f)\cap U'$ is an open subset of $U\cap U'$ and also an affine variety, cut out in $\Bbb A^{n+1}$ by the equations $(fx_{n+1}-1,f_1,\cdots,f_{n-d})$.

To see that the Jacobian matrix of this system has rank $n+1-d$, it suffices to check that the entry corresponding to $\frac{\partial (fx_{n+1}-1)}{\partial x_{n+1}}$ is nonzero: every other term of the form $\frac{\partial f_i}{\partial x_{n+1}}$ is zero because $f_i$ doesn't depend on $x_{n+1}$. But $\frac{\partial (fx_{n+1}-1)}{\partial x_{n+1}}=f$, which is nonvanishing on $D(f)\cap U'$. So our Jacobian matrix looks like $$\begin{pmatrix} \frac{\partial(f_1,\cdots,f_{n-d})}{\partial(x_1,\cdots,x_n)} & 0 \\ * & f \end{pmatrix}$$ and this has rank $n+1-d$, so we're done.

Answer 2

If $X:=V(I)\subseteq \mathbb{A}^n_k$ is an affine algebraic variety over a field $k$ of characteristic zero (the ideal $I$ is assumed to be a prime ideal) it follows from Hartshorne's book (HH) and Thm I.3.2 that $\Gamma(X,\mathcal{O}_X)=A$ where $A:=k[x_1,..,x_n]/I$. Again by HH Thm I.3.2 it follows $\mathcal{O}_{X,p} \cong A_{\mathfrak{p}}$ where $\mathfrak{p} \subseteq A$ is the prime ideal corresponding to the point $p\in X$. By HH.II.1 (the section on stalks) it follows the stalk $\mathcal{O}_{X,p}$ is the limit of $\mathcal{O}_X(V)$ over all open subsets $V$ containing $p$. We may restrict the sheaf $\mathcal{O}_X$ to the open set $U$ to get the structure sheaf $\mathcal{O}_U$ on the open sub-variety $U$. It follows from the defnition there is an isomorphism $\mathcal{O}_{X,p} \cong \mathcal{O}_{U,p}$. Hence the sheaves $\mathcal{O}_X$ and $\mathcal{O}_U$ have the same stalk at the point $p$. Again by HH.I.5 we define the point $p$ to be non-singular using the Jacobian matrix of a set of generators $f_i$ of the ideal $I$ defining $X$. This definition is independent of choice og generators $f_i$ of the ideal $I$. By HH.I.5.1 it follows $p$ is non-singular iff the local ring $\mathcal{O}_{X,p}$ is a regular local ring. Since $\mathcal{O}_{X,p}\cong \mathcal{O}_{U,p}$ it follows we may define non-singularity using any open set $U$ containing $p$. Hence we may use the local ring at $p$ to define non-singularity for any variety of finite type over $k$, where $k$ is a field of characteristic zero.

Example 0. If $X$ is a (projective) variety of finite type over $k$ and $p \in X$ is a closed point, we may choose an open affine set $p\in U \subseteq X$. By definition $U:=Spec(A)$ with $A=k[x_1,..,x_n]/I$, where $I:=\{f_1,..,f_l\}$ is a set of polynomials. Then we may use the Jacobian criterion to check if $p$ is a non-singular point of $X$. This does not depend on the choice of open affine subset $U$ containing $p$ and the choice of generators $f_i$ of the ideal $I$.

Let us prove your claim using the Jacobian criterion. Let $U:=Spec(A)$ and $V:=Spec(B)$ be two open affine subvarieties containing $p$ with $A:=k[x_1,..,x_n]/(f_1,..,f_l)$ and $B:=k[y_1,..,y_n]/(g_1,..,g_t)$. Let $U$ be smooth/non-singular at $p$ by the Jacobian criterion. It follows from HH that $\mathcal{O}_{U,p}$ is a regular local ring. There is an isomorphism of local rings $\mathcal{O}_{U,p}\cong \mathcal{O}_{V,p}$ (this is a formal property of limits, and the proof may be found in Atiyah-Macdonalds (AM) book on commutative algebra), hence it follows $\mathcal{O}_{V,p}$ is a regular local ring. Again by HH it follows the Jacobian criterion holds for the generators $g_i$ at the point $p \in V$. The converse is similar and the claim is proved.

Note: This proof uses Chapter I in HH which is "elementary", a formal result on limits in AM - this is also "elementary" - and the definition of the stalk of a sheaf in Chapter II in HH. This is also an "elementary" definition. Hence this proof is in a sense "elementary". To students: Do you think this proof is "elementary enough" or is there a more "elementary" proof "somewhere out there"?

Example 1. There are problems arising in characteristic $p>0$. Let $k$ be a field of characteristic $p$ and let $f(x,y) \in k[x^p,y^p]$ be any polynomial. It follows $\partial f/\partial_x=\partial f/\partial_y=0$, hence the Jacobian matrix of $f$ is identically zero.

If $k$ is any algebraically closed field and $f:X \rightarrow Y$ a morphism of schemes of finite type over $k$, we say "f is smooth of relative dimension $n$" if the following holds:

$f$ is flat. For any irreducible components $X' \subseteq X$ and $Y' \subseteq Y$ with $f(X') \subseteq Y'$ it follows $dim(X')=dim(Y')+n$. For any point $x\in X$ it follows $dim_{\kappa(x)}(\Omega_{X/Y}\otimes \kappa(x))=n$.

Theorem III.10.2 in Hartshorne says that $f$ is smooth of relative dimension $n$ iff $f$ is flat and for any $y \in Y$ it follows the fiber $X_y:=f^{-1}(y)$ is geometrically regular, which means that the fiber product $X_y \times Spec(\overline{\kappa(y)})$ is regular (meaning all local rings are regular local rings). Here $\overline{\kappa(y)}$ is the algebraic closure of $\kappa(y)$. Hence if $Y=Spec(k)$ with $k$ an algebraically closed field it follows the morphism $f:X \rightarrow Spec(k)$ is smooth iff the local ring $\mathcal{O}_{X,p}$ of $p \in X$ is a regular local ring for any point $p$. Hence if the base field $k$ is algebraically closed we may use the module of relative differentials $\Omega^1_{X/Y}$ to define "regularity/non-singularity": We define $X$ (a scheme of finite type over an algebraically closed field $k$) to be "non-singular" iff the canonical map $f:X\rightarrow Spec(k)$ is smooth of relative dimension $dim(X)$.

Example 2: Let $k$ be an algebraically closed field of characteristic $p>0$ and let $f(x_1,..,x_n)\in k[x_1^p,...,x_n^p]$ be a polynomial. It follows there is a polynomial $g(x_1,..,x_n)$ with $g^p=f$, hence $f$ is never irreducible.

The module of Kahler differentials is "functorial" in the following sense: If $f: X \rightarrow Y$ is a morphism of schemes and $i:U \rightarrow X, j:V \rightarrow Y$ are open subschemes with $f(U) \subseteq V$, it follows there is an isomorphism of sheaves $i^*\Omega^1_{X/Y} \cong \Omega^1_{U/V}$. Hence for any point $p \in U$ it follows there are isomorphisms at the stalks $\Omega^1_{U/V,p}\cong \Omega^1_{X/Y,p}$ and at the fiber $\Omega^1_{U/V}\otimes \kappa(p) \cong \Omega^1_{X/Y}\otimes \kappa(p)$. Hence the stalk and fiber of the module of Kahler differentials at $p$ does not depend on the choice of open subscheme $U \subseteq X$ containing $p$. We may choose $U:=Spec(B)$ and $V:=Spec(A)$ and then $i^*\Omega^1_{X/Y}$ is the sheafification of the module $\Omega^1_{B/A}$ of relative differentials.

Example 3: If $f(x_1,..,x_n)\in k[x_1^p,..,x_n^p]$ is any polynomial and $char(k)=p>0$ it follows $\partial f/\partial_{x_i}=0$ for all $i$. It follows the differential $df:=\sum_i \partial f/\partial_{x_i} dx_i=0$. If $X:=Spec(k[x_1,..,x_n]/(f))$ it follows the cotangent bundle $\Omega^1_{X/k}$ is a trivial $\mathcal{O}_X$-module of rank $n$.