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?
Best Answer
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.