# Understanding proof that smooth point is regular ( Second quetion, Gortz's Algebraic Geometry )

I am reading the Gortz's Algebraic Geometry, proof of Lemma 6.26 and trying to understand some statement.

First, I propose associated question.

Q. Let $$A \subseteq B$$ be a subring with prime ideals $$\mathfrak{p} \subseteq A$$ and $$\mathfrak{q} \subseteq B$$ such that $$\mathfrak{p} = \mathfrak{q} \cap A$$ ( lying over ). Then its residue fields $$\kappa(\mathfrak{p}) := A_{\mathfrak{p}}/\mathfrak{p}A_{\mathfrak{p}}$$ and $$\kappa(\mathfrak{q}):=B_{\mathfrak{q}}/\mathfrak{q}B_{\mathfrak{q}}$$ are isomorphic? If not, when ? I don’t think this statement will work..

This question originates from following proof ( Lemma 6.26 of the Gortz's book ).

Lemma 6.26. Let $$k$$ be a field, and $$X$$ a $$k$$-scheme, locally of finite type. Let $$x\in X$$ be a point such that $$X$$ is smooth at $$x$$ of relative dimension $$d$$ over $$k$$. Then the local ring $$\mathcal{O}_{X,x}$$ is regular of dimension $$\le d$$. If $$x$$ is closed, then $$\dim \mathcal{O}_{X,x}=d$$.

Here the smoothness is defined as

Proof of the Lemma 6.26. Let $$U$$ be an open neighborhood of $$x$$ such that $$U$$ is smooth over $$k$$. As the closed points of $$X$$ are very dense, there exists a closed point $$x'$$ of $$X$$ with $$x'\in U \cap \overline{\{x\}}$$. If we have shown that $$\mathcal{O}_{X,x'}$$ is regular, then $$\mathcal{O}_{X,x}$$, being a localization of $$\mathcal{O}_{X,x'}$$, is also regular by Proposition $$B.77 (1)$$ and $$\dim \mathcal{O}_{X,x} \le \dim \mathcal{O}_{X,x'}$$. Thus we may assume that $$x$$ is a closed point.

We may accept this argument. If needed, I will upload details.

( Continuing proof ) We embed a neighborhood of $$x$$ into an affine space as in the definition of smoothness and denote $$y\in \mathbb{A}^{n}_k$$ the image of $$x$$. Then $$\mathcal{O}_{X,x} \cong \mathcal{O}_{\mathbb{A}^{n}_k ,y} / ( f_1, \dots , f_{n-d})$$, with polynomials $$f_j \in k[T_1,\dots, T_n]$$ such that the matrix $$((\partial f_i / \partial T_j )(y))_{i,j}$$ has rank $$n-d$$. I think that more correct notation is $$\mathcal{O}_{\mathbb{A}^{n}_k ,y}/(\frac{f_1}{1} , \dots , \frac{f_{n-d}}{1})$$, where $$\frac{f_i}{1}$$ is the image of $$f_i$$ in $$k[T_1,\dots T_n]_{\mathfrak{p}_y}$$. By Example 6.5 this means that the images of $$f_1 ,\dots , f_{n-d}$$ in $$(T_y \mathbb{A}^{n}_k )^{*} = \mathfrak{m}_y/\mathfrak{m}_y^2$$ are linearly independent ( where $$\mathfrak{m}_y$$ is the maximal ideal of $$\mathcal{O}_{\mathbb{A}^{n}_k,y}$$). In fact, if $$\kappa(y) = k$$, then this follows immediately. ( $$\because$$ In proof that smooth point is regular ( First quetion, Gortz's Algebraic Geometry ). We may accept this statement. ) The general case is reduced to this case as follows. Consider the base change

$$\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} \bar{X}:= X\times_k \kappa(y) & \ra{p} & X \\ \da{\overline{f}} & & \da{f} \\ \operatorname{Spec}\kappa(y) & \ra{l} & \operatorname{Spec}k \\ \end{array}$$

Then $$\bar{X}=X \otimes_k \kappa(y) \subseteq \mathbb{A}^{n}_{\kappa(y)}$$ ( $$\because$$ I first stuck at this point. If we may assume $$X \subseteq \mathbb{A}^{n}_k$$ ( possible? ), then this is true? ) and the $$p : X \otimes_k \kappa(y) \to X$$ is surjective ( $$\because$$ his book Corollary 5.45 ). And we fix a point $$y' \in \mathbb{A}^{n}_{\kappa(y)}$$ lying over $$y\in \mathbb{A}^{n}_k$$, with corresponding maximal ideal $$\mathfrak{m}_{y'} \subset \kappa(y)[T_1, \dots , T_n]$$. P.s. I think that more correct notation is $$\mathfrak{m}_{y'} \subset \mathcal{O}_{\mathbb{A}^{n}_{\kappa(y)},y'}= \kappa(y)[T_1, \dots , T_n]_{\mathfrak{p}_{y'}}$$ ; i.e., I think that he made typo. Do you agree?

The rank of the Jacobian matrix of the polynomials $$f_j$$ is the same, regardless of whether we consider it over $$k$$ or over $$\kappa(y)$$. I understood this statement as

$$\operatorname{rank} J_{\iota(f_1), \dots, \iota(f_{n-d})}(y') = \operatorname{rank} J_{f_1,\dots f_{n-d}}(y) , \tag{1}$$

where $$\iota : k[T_1,\dots T_n] \hookrightarrow \kappa(y)[T_1, \dots , T_n]$$ is the inclusion. From the $$\bar{f}$$ in the above commutative diagram there exists a field extension $$\kappa(y) \subseteq \kappa(y')$$ and since the rank of matrix is invariant under field extension, the above $$(1)$$ holds. Am I folloiwng well? Anyway he argues continuosly as follows.

Now consider the $$\kappa(y)$$-linear map $$\psi : \mathfrak{m}_y/\mathfrak{m}_y^2 \to \mathfrak{m}_{y'}/\mathfrak{m}_{y'}^2$$. I understood this map as follows : Again consider the inclusion $$\iota : k[T_1, \dots T_n] \hookrightarrow \kappa(y)[T_1, \dots, T_n]$$ and let $$\bar{p} : \mathbb{A}^n_{\kappa(y)} \to \mathbb{A}^{n}_k$$ be the induced morphism between affine schemes. Then by the choice of $$y'$$, $$\bar{p}(y') =y$$ so that $$\iota^{-1}(\mathbb{p}_{y'}) = \mathbb{p}_y$$ ; i.e., $$\mathfrak{p}_{y'} \cap k[T_1, \dots , T_n] = \mathfrak{p}_y$$. In particular, $$\mathfrak{p}_y \subseteq \mathfrak{p}_{y'}$$. Now consider next map

$$\psi_0 : \mathfrak{m}_y:=\mathfrak{p}_y k[T_1, \dots, T_n]_{\mathfrak{p}_y} \to \mathfrak{m}_{y'}:= \mathfrak{p}_{y'} \kappa(y)[T_1, \dots, T_n]_{\mathfrak{p}_{y'}}$$ $$\frac{b}{s} \mapsto \frac{\iota(b)}{\iota(s)}.$$

( Note that from $$s\in k[T_1, \dots , T_n] -\mathfrak{p}_y$$ and $$\mathfrak{p}_y^{c} = \mathfrak{p}_{y'}^{c} \cup (k[T_1, \dots, T_n])^{c}$$, we have that if $$s \notin \mathfrak{p}_y$$, then $$\iota(s) \notin \mathfrak{p}_{y'}$$. So the map $$\psi_0$$ is well-defined. ) Then I think that $$\psi : \mathfrak{m}_y/\mathfrak{m}_y^2 \to \mathfrak{m}_{y'}/\mathfrak{m}_{y'}^2$$ is induced from the $$\psi_0$$. Note that $$\psi(\frac{f_i}{1}+\mathfrak{m}_y^2 ) = \frac{\iota(f_i)}{1} + \mathfrak{m}_{y'}^2$$.

By the previous case, the images of the $$\frac{f_j}{1} + \mathfrak{m}_y^2$$ in $$\mathfrak{m}_{y'}/\mathfrak{m}_{y'}^2$$ form a linearly independent system, and it follows that the same holds in $$\mathfrak{m}_y/\mathfrak{m}_y^2$$, as desired. As $$\mathcal{O}_{\mathbb{A}^{n}_k ,y}$$ is regular, Proposition $$B.77 (3)$$ implies that $$\mathcal{O}_{X,x}$$ is regular of dimension $$d$$. QED.

Q. I don't understand the bold statement at all. As noted in the proof, we showed the linear independence if $$\kappa(y)=k$$. In the above commutative diagram, we have a morphism $$\bar{f} : \bar{X}:= X \times_k \kappa(y) \to \operatorname{Spec}\kappa(y)$$ with $$y' \in \bar{X}$$. I think that if $$\kappa(y') = \kappa(y)$$, then we may apply the reduced case to show that $$\{ \psi(\frac{f_i}{1}+\mathfrak{m}_y^2 ) = \frac{\iota(f_i)}{1} + \mathfrak{m}_{y'}^2 \}_{i}$$ is linearly independent. We showed that $$\mathfrak{p}_{y'} \cap k[T_1, \dots , T_n] = \mathfrak{p}_y$$. So if my first question at the starting point is true, then $$\kappa(y') = \kappa(y)$$. But is it really true? Or can we show $$\kappa(y') = \kappa(y)$$ by other approach? Or is there any other route to show the linear independence or to reduce to the case that $$k=\kappa(y)$$?

Yes. The proof is not really accurate. For any $$y' \in \mathbb{A}^{n}_{\kappa(y)}$$ lying over $$y\in \mathbb{A}^{n}_k$$, it is not in general $$\kappa(y') =\kappa(y)$$, so we cannot apply the reduced case directly. But we can choose suitable $$y'$$ such that $$\kappa(y') =\kappa(y)$$. This is as follows.
Again let $$f: X \to \operatorname{Spec} k$$ , $$l : \operatorname{Spec} \kappa(y) \to \operatorname{Spec}k$$, and $$i_y : \operatorname{Spec}\kappa(y) \to X$$ be the canonical morphism ( Gortz's book p.71 ). Since $$l \circ id_{\operatorname{Spec} \kappa(y)} =l = \operatorname{Spec}( k \hookrightarrow \kappa(y)) := \operatorname{Spec}(\Gamma(f \circ i_y)) = f\circ i_y$$, by universal property, there is an unique morphism $$g:=\operatorname{Spec}\kappa(y) \to \overline{X}:=X \otimes_k \kappa(y)$$ such that $$p\circ g = i_y$$ and $$\overline{f} \circ g = id_{\operatorname{Spec}\kappa(y)}$$. ($$p$$ and $$\overline{f}$$ are as in the commutative diagram in the question. ) And let $$y'$$ be its image point. Then $$y'$$ is lying over $$y$$. And since $$g: \operatorname{Spec}\kappa(y) \to \overline{X}:=X \otimes_k \kappa(y)$$ is a morphism of locally ringed spaces, it induces a local homomorphism $$\mathcal{O}_{\overline{X},y'}\to \kappa(y)=\mathcal{O}_{\operatorname{Spec}\kappa(y),p}$$, and hence a homomorphism $$\iota :\kappa(y') \to \kappa(y)$$. And also, we have field homomorphism $$\eta : =\Gamma(\overline{f} \circ i_{y'}) : \kappa(y) \to \kappa(y')$$, where $$\overline{f} : \overline{X} \to \operatorname{Spec}\kappa(y)$$ is the base change map in the question. Note that $$\iota \circ \eta = id_{\kappa(y)}$$ (?) so that $$\iota$$ is surjective hence an isomorphism ; i.e., $$\kappa(y') \cong \kappa(y)$$.