# Hartshorne Proposition 9.5

The Proposition is that : Let $f:X\longrightarrow Y$ be a flat morphism of schemes of finite type over a field $k$. For any point $x\in X$, let $y=f(x)$. Then $\dim_x(X_y)=\dim_x(X)-\dim_y(Y)$. Here for any scheme $X$ and any point $x\in X$, by $\dim_x(X)$, we mean the dimension of the local ring $\mathcal{O}_{x,X}$.

They begin the proof as follows :

First we make a base change $Y'\longrightarrow Y$ where $Y'=\textrm{Spec } \mathcal{O}_{y,Y}$ and consider the morphism $f':X'\longrightarrow Y'$ where $X'=X\times_{Y} Y'$. Then $f'$ is also flat, $x$ lifts to $X'$ and the three numbers are the same.

What is meant by : $x$ lifts to $X'$. It is not the inverse image, because the inverse image could contain more than one element. What does it mean?

• I do not have the book in front of me, but in general lifts need not be unique. One can still choose a lift. Nov 5, 2014 at 16:13
• Think about $x$ as being a morphism from Spec of some field to $X$. To say it lifts to $X'$ means that there is a morphism to $X'$ such that the obvious triangle commutes.
– user64687
Nov 5, 2014 at 16:30
• How do we think of $x$ as a morphism? Nov 5, 2014 at 16:32
• @poorna Consider a scheme X. For a scheme T the set of scheme morphisms Hom(T,X) is named the set of points of X with values in T (EGA I,3.5). This concept generalizes the intuitive notion of a point of a scheme: In case T=Spec(A) with a local ring A the morphisms Spec(A)⟶X correspond bijectively to the local morphisms $\mathscr O_{X,x}⟶A$ with x∈X. Notably for A=K a field we obtain Hom(Spec K,X)={k(x)⟶K:x∈X}, i.e. the points of X with values in Spec K, the K-valued points of X, are the points x∈X - taken in the topological sense - with K an extension field of the residue field k(x). Nov 6, 2014 at 9:42

@poorna The answer to your question follows from the universal property of the fibre product:

We have the fibre product

$$\begin{array} X\ \ \ X' & \stackrel{}{\longrightarrow} & X \\ \ \ \downarrow{f'} & & \downarrow{f} \\ Spec(\mathscr O_{Y,y}) & \stackrel{j_y}{\longrightarrow} & Y \end{array}$$

Consider the local ring $A = Spec(\mathscr O_{X,x})$. Due to the universal property of the fibre product the commutative square defining in a topological sense $x$ and $y=f(x)$

$$\begin{array} XSpec \ A & \stackrel{j_X}{\longrightarrow} & X \\ \ \ \downarrow{} & & \downarrow{f} \\ Spec(\mathscr O_{Y,y}) & \stackrel{j_y}{\longrightarrow} & Y \end{array}$$

maps into the fibre product via a unique morphism

$$Spec \ A \stackrel{j_{x'}}{\longrightarrow} X'$$

Referring to my previous comment, the latter morphism is a point of $X'$. It lifts the 'point' $x$, i.e. $Spec \ A \stackrel{j_X}{\longrightarrow} X$, to $X'$.

• Thank you! I am still not clear actually. We get the unique morphism $j_{x'}:Spec\ A\longrightarrow X'$ which factors through $X$. Do they refer to this morphism as the lift of $x$. Or is it the image of the $\mathcal{m}_x$, the unique closed point of $Spec\ A$ in $X'$ Nov 6, 2014 at 12:55
• If you consider points the elements of the topological space$|X|$ underlying a scheme $X$ then you have the point $j_x(\mathcal m_A) \in |X|$ and its lift $j_{x'}(\mathcal m_A) \in |X'|$. I assume you think along these lines. When refering to the more general concept from EGA the morphism $j_x: Spec A \longrightarrow X$ is a point of $X$ with value in $Spec A$ and $j_{x'}: Spec A \longrightarrow X'$, the point of $X'$ with value in $Spec A$, is a lift of $j_x$. Nov 7, 2014 at 5:12
• I tried working along these lines, but I am still unable to understand, why the three numbers are the same. In particular, if $x'\in X'$ is the lift of $x$, why is $dim_{x'}X'=dim_x X$? Nov 12, 2014 at 18:59
• @gradstudent I am also having serious difficulty understanding the proof. There is a much easier proof in Algebraic Geometry I -Ulrich Gortz, Torsten Wedhorn, Corollary 14.95, page-464 (with a different hypothesis) Apr 25, 2016 at 21:09