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In Joe Harris's book "Algebraic Geometry: A First Course", we find the following proposition:

Let $f:X\longrightarrow Y$ be a dominant rational map, for $X$ and $Y$ be two varieties.

Proposition 7.16. The general fiber of the map $f$ is finite if and only if the inclusion $f^*$ expresses the field $K(X)$ as a finite extension of the field $K(Y)$. In this case, if the characteristic of $K$ is $0$, the number of points in a general fiber of $f$ is equal to the degree of the extension.

My question is very objective: what does "general fiber" mean? In other words, what characterizes $f^{-1}(y)$, for $y \in Y$, as a general fiber?

Thanks.

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  • $\begingroup$ Possibly it means the fiber of all the points except those on a subvariety. $\endgroup$
    – Fan Zheng
    Commented Jan 14, 2018 at 20:36
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    $\begingroup$ Have you read the book? The notion of "general object" is explained quite precisely in Lecture 5. $\endgroup$
    – abx
    Commented Jan 14, 2018 at 20:53
  • $\begingroup$ Yes, I read. But it is not clear to me. I am thinking that it means the following: there is a Zariski open $U \subset Y$ such that $f\mid V:V\longrightarrow U$ is an unbranched, where $V=f^{-1}(U)\subset X$. Is this correct? Does the set U need to be dense? $\endgroup$
    – Manoel
    Commented Jan 14, 2018 at 22:14
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    $\begingroup$ The book is a wonderful source of examples and intuition, but not techniques of proof. So don't sweat too much trying to prove various things asserted in that book, because often some relevant concepts (whether from commutative algebra or elsewhere) or techniques are not discussed there; just keep those puzzles in mind to figure out for yourself later when you learn from a book that spends more time on systematic development of the modern theory. (That Prop. 7.16 is not rocket science in the end, but the proof involves a "spreading out" perspective that is not discussed in that book.) $\endgroup$
    – nfdc23
    Commented Jan 14, 2018 at 22:28
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    $\begingroup$ @Manoel Your understanding is correct. Whether the open set is dense depends on whether a variety is irreducible by definition. $\endgroup$
    – Fan Zheng
    Commented Jan 15, 2018 at 3:42

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You cannot characterize the subset $Z \subseteq Y$ of points over which the fibre is not finite by looking at the field extension, since by the function fields $K(Y)$ and $K(Y-Z)$ are the same. In fact, the function field is a birational invariant, not a biregular one.

If you want to figure out what "general" means in this context, you can argue as follows.

Let us assume that $X$ and $Y$ are irreducible. If we assume that $K(X)$ is a finite extension of $K(Y)$, then necessarily $\dim X = \dim Y$. Supposing that you have resolved the singularities of the map so that $f \colon X \to Y$ is defined everywhere, you can look at the differential map $$df_x \colon T_xX \longrightarrow T_{f(x)}Y.$$ Then the subset $Z \subseteq Y$ of points where $f$ is finite is contained in the subset $Z'$ of points where $f$ is unramified, namely the complement the image of the set of critical points of $df$ (a point $x \in X$ is critical when $df_x$ is not of maximal rank).

So being unramified (in characteristic $0$) is a general property, since the locus of critical values has zero measure in $Y$ (for instance, by Sard lemma).

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