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I have a question about an argument in the proof of Thm 8.2.5 (Dimension formula) from Liu's "Algebraic Geometry" (page 334):

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Firstly the problem is reduced to the affine case with $Y:=Spec \text{ }\mathcal{O}_{Y,y}$ and $X$ closed subscheme of $Z:= \mathbb{A}^1_Y=Spec\text{ }\mathcal{O}_{Y,y}[T]$. Therefore $X= Spec\text{ }\mathcal{O}_{Y,y}[T]/I$.

Let $\zeta $ be the (unique) generic point of $Y$ and let $X_{\zeta}$ (resp. $Z_{\zeta}$) the generic fiber of $\zeta$ under morphism $f$ (resp. $Z \to Y$).

Now the question: Why the (obvious) fact that $codim(X_{\zeta},Z_{\zeta}) \le 1$ implies already $codim(X,Z) \le 1$ ?

Indeed $Z_{\zeta}= O_{Y,y}[T] \otimes k(\zeta)=k(\zeta)[T]$ where $k(\zeta)= O_{Y, \zeta}$ is the localisation of $O_{Y,y}$ at generic point $\zeta \in Y$. Therefore $Z_{\zeta}$ has Krull dimension $1$ and therefore $X_{\zeta}$ can also only have codimension $1$ or $0$ wrt $Z_{\zeta}$.

The question is why does it suffice to deduce that $codim(X,Z) \le 1$?

My considerations:

By definition if $X$ is an irreducible (because integral) subset of $Z$ then codimension $codim(X,Z)$ is defined as follows:

Let $\nu \in X$ the unique generic point of $X$. Then

$$codim(X,Z)= min_{x \in X} \dim \mathcal{O}_{Z,x}= \dim \mathcal{O}_{Z,\nu}$$

and respectively

$$codim(X_{\zeta},Z_{\zeta})= min_{x \in X_{\zeta}} \dim \mathcal{O}_{Z_{\zeta},x}$$

Remark: $X_{\zeta}$ might be possibly not more irreducible and therefore could have no more a unique generic point as $X$ have.

So the question why $min_{x \in X_{\zeta}} \dim \mathcal{O}_{Z_{\zeta},x}=\dim \mathcal{O}_{Z,\nu}$?

Can these considerations lead to an argument solving the problem? Or does there exist another standard method / tool / kind of a "dimension formula" to relate codimensions of integral subschemes and their corresponding generic fibers.

Remark: I have already asked this question in MSE without finding a satisfying solution: https://math.stackexchange.com/questions/3297166/dimension-formula-from-lius-ag-ac?noredirect=1&lq=1

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  • $\begingroup$ Morphism being dominant in this case means there is a point in X sent to the generic point of Y so one could pick an irreducible component of X containing such point and show that the codimension of the irreducible component is <=1. So we are reduced to the case when X is irreducible. If codim of X is bigger than one then there is a chain of closed irreducibles of length 2 in between X and Z but then this when restricted to the generic fiber gives contradiction (everything is closed irreducible so generic fiber can recover the whole thing by taking closure). $\endgroup$
    – GTA
    Commented Sep 17, 2019 at 7:02
  • $\begingroup$ @GTA: ah yes I think I see (partially) the point. So the essential observation is that if we have such chain $X:= X_0 \subset X_1 \subset ... \subset X_n =:Z$ of irreducibles with unique generic points $\zeta_i \in X_i$ for each $i$ then the fact that $X \to Y$ is dominant implies that all induced maps $\phi_i:X_i \to Y$ are dominant and therefore again for every $i$ the generic fiber $(X_i)_{\zeta}$ always contain the generic point $\zeta_i$ of $X_i$ but not $\zeta_{i+1}$. $\endgroup$
    – user267839
    Commented Sep 17, 2019 at 12:26
  • $\begingroup$ The remaining point that stays unclear is that following your argument that "...there is a chain of closed irreducibles of length 2 in between X and Z but then this when restricted to the generic fiber gives contradiction (everything is closed irreducible so generic fiber can recover the whole thing by taking closure)..." when you restict the chain $X:= X_0 \subset X_1 \subset ... \subset X_n =:Z$ to generic fibers then the resulting chain $X_{\zeta} \subset (X_1)_{\zeta} \subset ... \subset (X_n)_{\zeta} =:Z_{\zeta}$ not more consists of irreducible $\endgroup$
    – user267839
    Commented Sep 17, 2019 at 12:34
  • $\begingroup$ members; the $(X_i)_{\zeta}$ retain only the property beeing closed in $Z_{\zeta}$ (since $X_i$ closed in $Z$ and using base change argument which preserves closed immesions) and dense in $X_i$ (since it contains the generic). So the chain need to be modified,right? I think that at this point it requires more carefull analysis of it? Like construct a chain of $K_i$s where $K_i$ is defined as closure of $\zeta_i$ in $(X_i)_{\zeta}$ then check that $K_i$ is closed contained in $K_{i+1}$ and blabla... $\endgroup$
    – user267839
    Commented Sep 17, 2019 at 12:35
  • $\begingroup$ Or do I make the life here too awkward for me and the fact that $(X_i)_{\zeta}$ are dense in $(X_i)$ provide immediately the desired result (maybe by a pure topological lemma which I don't know)? $\endgroup$
    – user267839
    Commented Sep 17, 2019 at 12:40

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