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

enter image description here

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 $k=\kappa(y)$?


1 Answer 1


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., $\kappa(y') \cong \kappa(y)$.

So we may apply the reduced case so we can understand the bold statement in the question.

P.s. I would like to thank Gortz for his advice through E-mail.


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