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Let $f: X \to Y$ a dominant morphism of varieties over a field $k$. My 'varieties' are by definition irreducible and let denote as $g_X \in X$ and $g_Y \in Y$ their generic points. Dominant means that $f(g_X)=g_Y$. Let $n= \dim(X), m= \dim(Y)$. Then it is known that the transcendence degree of function fields of varieties coinside with their dimensions. Therefore $\operatorname{trdeg}_k \ K(X) =\dim(X)=n$ and $\operatorname{trdeg}_k \ K(Y) =\dim(Y)$.

I want to check that the dimension of the generic fiber $X_{g_Y} = f^{-1}(g_Y)$ is $e= n-m$.

My 'proof' has some gaps I would like to plug. Assume that both varieties are affine: $X= \operatorname{Spec}(A), Y=\operatorname{Spec}(R)$. The map $f$ is dominant, therefore $R \to A$ is injective. The affine ring of generic fiber $X_{g_Y}$ is $A \otimes_R K(Y)$. Since $g_X \in X_{g_Y} $ there exist a map $A \otimes_R K(Y) \to K(X)= \operatorname{Frac}(A)$ und therefore, since $K(X)$ is a field, we get the induced inclusion of fields $K(A \otimes_R K(Y)) \subset K(X)$. $K(A \otimes_R K(Y))$ is the function field of generic fiber $X_{g_Y}$.

Question: what do we know about the field extension $K(A \otimes_R K(Y)) \subset K(X)$? which transcendence degree does it have?

My idea is obvious: if I can prove that $\operatorname{trdeg}_{K(A \otimes_R K(Y)} K(X)= m$ then I can use additivivity formula for transcendence degrees of field extensions to show that $\operatorname{trdeg}_k K(A \otimes_R K(Y))=n-m$ because of $\operatorname{trdeg}_k \ K(X) =n$ and additivity for tower of extensions $k \subset K(A \otimes_R K(Y) \subset K(X)$. But I don't know how I can figure out that $\operatorname{trdeg}_{K(A \otimes_R K(Y)} K(X)= m$.

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  • $\begingroup$ why the downvote? $\endgroup$ – Isak the XI Mar 24 at 23:02
  • $\begingroup$ I believe those two fields are equal. Maybe you can try an example such as $A^1 \to spec k$ induced by $k \to k[x]$. $\endgroup$ – Youngsu Mar 25 at 7:15
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As Youngsu pointed out, the fields $K(A\otimes_RK(Y))$ and $K(X)$ are indeed equal. To see this, note that $A\otimes_RK(Y)$ is the localisation of the domain $A$ at its multiplicative subset $R\setminus\{0\}\subseteq A$, and $K(A\otimes_RK(Y))$ is obtained from $A\otimes_RK(Y)$ by localising at the remaining non-zero elements. By transitivity of localisations, this shows that $K(A\otimes_RK(Y))$ can also be written as the localisation of $A$ at all its non-zero elements, which is $K(X)$.

So what goes wrong here? The problem is that the generic fibre $X_{g_Y}$ is usually not a variety over $k$! Suppose, for example, that $X=Y$ and $f\colon X\rightarrow X$ the identity. Then the generic fibre is $\operatorname{Spec} K(X)$, and $K(X)$ is usually not of finite type as a $k$-algebra, unless it is a finite field extension of $k$ (by Hilbert's Nullstellensatz), i.e., unless $X$ is $0$-dimensional.

However, $X_{g_Y}$ is a variety* over $K(Y)$ (can you show that?), and it is indeed true that $X_{g_Y}$ is $(n-m)$-dimensional as a $K(Y)$-variety! So instead of $\operatorname{trdeg}_kK(A\otimes_RK(Y))=n-m$, you need to show that $$\operatorname{trdeg}_{K(Y)}K(A\otimes_RK(Y))=n-m\,.$$ Now your idea of using transitivity of transcendence degrees works: We've seen above that $K(A\otimes_RK(Y))=K(X)$, and $\operatorname{trdeg}_{K(Y)}K(X)=\operatorname{trdeg}_kK(X)-\operatorname{trdeg}_kK(Y)=n-m$.


* Usually, $K(Y)$ won't be algebraically closed, but I think you're working with the scheme-theoretic definition of varieties anyway.

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  • $\begingroup$ Thank you for your detailed answer. on your question why $X_{g_Y}$ is variety over $K(Y)$ i think the reason is that we need just to such that $A\otimes_RK(Y)$ has finite type over $K(Y)$, but that's clear because $A$ is finitely generated $R$-algebra and base change preserves this property, that's it? $\endgroup$ – Isak the XI Mar 26 at 1:10
  • $\begingroup$ About your argument on the observation that $A\otimes_RK(Y)$ is the localization of $A$ by certain mult set. Here you implicitely use that $A\otimes_R K(Y)$ is a domain (not contains zero divisors except $0$). Otherwise, if a ring $S$ isn't adomain, the function field of $S$ cannot simply constructed by only inverting all elements except zero. How the proceed in that case? $\endgroup$ – Isak the XI Mar 26 at 1:11
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    $\begingroup$ If $\operatorname{nil}(A)$ is the nilradical of $A$, we can replace $A$ by $A/\operatorname{nil}(A)$ and similarly $R$ by $R/\operatorname{nil}(R)$. This doesn't change the function fields $K(X)$ and $K(Y)$; or more precisely, $K(A)=K(A/\operatorname{nil}(R))$ and likewise for $R$. Moreover, the fibre $A\otimes_R K(Y)$ gets replaced by $A/\operatorname{nil}(A)\otimes_RK(Y)$, which is again a quotient by a nilpotent ideal. Hence also $K(A\otimes_RK(Y))=K(A/\operatorname{nil}(A)\otimes_RK(Y))$. Thus, everything can be reduced to the situation where $A$ and $R$ are domains. $\endgroup$ – Florian Adler Mar 26 at 10:45
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    $\begingroup$ Oops, I made a typo in my previous comment: It should be $K(A)=K(A/\operatorname{nil}(A))$, of course. In general, if $A$ is a ring such that $\operatorname{Spec} A$ is irreducible, then $A$ has a unique minimal prime ideal $\mathfrak q$ (which is then necessarily given by $\mathfrak q=\operatorname{nil}(A)$, since the nilradical is the intersection of all prime ideals), corresponding to the generic point. I would define $K(A)$ as the residue field at $\mathfrak q$. That is, $K(A)=k(\mathfrak q)=A_{\mathfrak q}/\mathfrak q A_{\mathfrak q}$. This coincides with the above. $\endgroup$ – Florian Adler Mar 26 at 18:16
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    $\begingroup$ by the way the irreducibility of $X_{g_y}$ can be also showed pure topologically: If $Y \subset X$ and $X$ is irreducible then $Y$ is irred iff the closure $\overline{Y}$ is irred. But the generic fiber contains the generic point of $X$ so it's closure is $X$ $\endgroup$ – Isak the XI Mar 26 at 18:54

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