# Dimension of generic fiber of a morphism between varieties

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 $$\operatorname{trdeg}_{K(A \otimes_R K(Y)} K(X)= m$$.

• why the downvote? – Isak the XI Mar 24 at 23:02
• I believe those two fields are equal. Maybe you can try an example such as $A^1 \to spec k$ induced by $k \to k[x]$. – Youngsu Mar 25 at 7:15

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.

• 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? – Isak the XI Mar 26 at 1:10
• 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? – Isak the XI Mar 26 at 1:11
• 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. – Florian Adler Mar 26 at 10:45
• 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. – Florian Adler Mar 26 at 18:16
• 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 $X$ – Isak the XI Mar 26 at 18:54