In the July3123 version of Algebraic geometry book by Vakil, he defined the notion of smooth of relative dimension in 13.6. I will write down the definition first:
Definition. A morphism $\pi:X\to Y$ is smooth of relative dimension $n$ if there exists open covers $\{U_i\}$ of $X$ and $\{V_i\}$ of $Y$ such that $\pi(U_i)\subseteq V_i$ and the diagram commutes
\begin{array}{}
U_i & \xrightarrow{\cong} & W& \xrightarrow{open} \operatorname{Spec}B[x_1,\ldots,x_{n+r}]/(f_1,\ldots,f_r)\\
\downarrow & &\downarrow_{\rho|_W}&\swarrow_{\rho}\\
V_i&\xrightarrow{\cong}&\operatorname{Spec}B
\end{array}
where $\rho$ comes from the obvious ring map and $W$ is an open subscheme, such that the determinant $det(\frac{\partial{f_j}}{\partial{x_i}})_{i,j\leq r}$ of the Jacobian of the $f_i$'s with respect to the $x_i$'s is an invertible function on $W$.
My question.
- I tried to verify that $\mathbb{A}^n_B\to \operatorname{Spec}B$ is smooth of relative dimension $n$. So I cover $\operatorname{Spec}B$ by itself and $\mathbb{A}^n_B$ by the distinguished opens $D(f)$ where $f\in B[x_1,\ldots,x_n]$. Now I have such commutative diagram
\begin{array}{}
D(f) & \xrightarrow{\cong} &\operatorname{Spec}B[x_1,\ldots,x_n,x_{n+1}=t]/(1-ft)& \xrightarrow{open} \operatorname{Spec}B[x_1,\ldots,x_{n+1}]\\
\downarrow & &\downarrow_{\rho|_W}&\swarrow_{\rho}\\
\operatorname{Spec}B&\xrightarrow{\cong}&\operatorname{Spec}B
\end{array}
but the Jacobian matrix is $0$. So I replace the $\operatorname{Spec}B[x_1,\ldots,x_{n+1}]$ by the distinguished open itself $\operatorname{Spec}B[x_1,\ldots,x_n,x_{n+1}=t]/(1-ft)$. Now the Jacobian is still $0$ since now $r=1$, $f_1=1-ft$ and we only consider the partial derivatives wrt $x_1$. Of course, if we put $t$ in the first position, we would get $-f$, which is indeed a nowhere zero function on $D(f)$. But shouldn't this be independent of our choice? - I am not so sure how he concluded that the locus where $\pi$ is smooth of relative dimension $n$ is open, as such notion is not even defined on points. Am I missing something?
- How do we actually show that any open embedding is etale? According to the definition, we need to locally embed our source into a finite type scheme $\operatorname{Spec}B[x_1,\ldots,x_{r}]/(f_1,\ldots,f_r)$. Is that possible? I tried $\operatorname{Spec}B[x_1,\ldots,x_r]/(x_1,\ldots,x_r)$ but it seems too good to be true..
I apologize for the triviality as it seems to be simply playing around the definition. Any help is appreciated!
Best Answer
Forgive me if I make any mistakes. I am also reading the note by Vakil.
1.I think to have $\mathbb{A}^n_B\to \text{Spec} B$ smooth of relative dimension $n$, we just need to take itself as a cover for both $\mathbb{A}^n_B$ and $\text{Spec} B$, where $\mathbb{A}^n_B=\text{Spec}B[x_1,...,x_{n+1}]/x_1$.
Also, you mention that this should be independent of the choice. I am not sure but I don't see any reason for the commutative diagram to be unique.
I think it's just like the definition of locally free sheaves: we need to make some choice of covering where the sheaf is free on every piece. This doesn't mean it holds for every choice of covering.
2.I don't think he gives the definition of the locus of etaleness. Probably we can mimic the definition of the locus where two morphisms agree. For example, let $f:X\to Y$ be a morphism. The locus of etaleness of $f$ is a locally closed subscheme $i:Z\to X$ where $f\cdot i$ is etale, and for every morphism $g:W\to X$ with $f\cdot g$ etale, $g$ uniquely factors through $i$.
I am not sure how to show that such a locus is open though.
3.Let $U\to X$ be an open embedding. Pick any affine open cover $\{\text{Spec}(B_i)\}_i$ of $X$. The intersection of $U$ with $\text{Spec}(B_i)$ is the union of $\{\text{Spec}(B_{f_{i,j}})\}_j$. Form the new cover $\{\text{Spec}(B_{f_{i,j}})\}_{i,j}$ of $U$, where every piece is an open subscheme of $\text{Spec}(B_i[x]/x)$. This shows that $i$ is etale.