I have two question abouts some steps in the proof of Proposition 4.4.16 from Liu's "Algebraic Geometry and Arithmetic Curves" (page 155):
Denote $f:X \to S$ the morphism from above. Let $X_s=f^{-1}(s) \neq \varnothing$ the fiber of certain point $s \in S$ of Dedekind scheme $S$.
Then we take an irreducible component $F$ of $X_s$.
Here my first question/understanding problem:
Liu makes following strange step: He replaces $X$ by open affine subscheme $U \subset X$ with the property that $U$ meets $F$ (therefore $U \cap F \neq \varnothing$) but not other components of $X_s$.
The question is why the assumption that $X$ is irreducible implies that such $U$ with desired properties exist and the proof can be reduced to the case $X=U$?
Why such reduction step would fail /be not allowed in the proof if $X$ would be not irreducible?
My considerations: By definition irreducible implies that any open set of $X$ is dense and so contain the (unique) generic point of $X$ and therefore any open is also irreducible. So naively replacing $X$ by $U$ doesn't change the initial conditions of the proposition. Is this exactly the main reason for this reduction step?
Futhermore the fiber $X_s$ has subspace topology induced by $X$ (since $X_s \subset X$) and by general property of induced subspace topology any open subset $V$ of $X_s$ arises from an open $U \subset X$ via $V=U \cap X_s $ (schematically $U \cap X_s \cong U \times_S X_s$).
So naively one can always "avoid" other irreducible components of $X_s$ by open sets from $X$ pure topologically. Or am I wrong? So where does here play role the assumption on $X$ that it is irreducible?
Second Question: Why $K(X_{\eta})=K(X)$?
We know that "birational" isn't stable under base change so I don't see an argument for the equality above?
Best Answer
Assume that $X$ is not irreducible. Say that $X=X_1 \cup X_2$, where each $X_i$ is irreducible, and $\dim(X_1)=\dim(X_2)$. Say that $X_1$ dominates $S$, while the irreducible variety $X_2$ is glued to $X_1$ along a closed point in $X_{1,s}$. In particular, $X_2$ is an irreducible component of the fiber $X_s$. By construction, this variety does not satisfy the conclusion of the theorem. So, let's try to see where the argument breaks down.
Say you want to do the cut down as in the proof to isolate $F=X_2$, avoiding the other irreducible components of $X_s$. Call the open affine $U$ (i.e., what replaces $X$ in the proof above). Then, as you want to avoid all the other irreducible components of $X_s$, it means that you throw away all of $X_{1,s}$. Then, as $U$ contains both the generic points of $X_1$ and $X_2$, it follows that $U$ is disconnected. In particular, $U= \mathrm{Spec}(A)$, where $A$ is not an integral domain. Then, the proof references other statements that you did not include, but my guess is that you need an integral domain to apply them.
As for the fields of fractions, it follows from a property of localization. Say that $R$ is an integral domain. Then, $K(R)$ is obtained by localizing $R$ at $R \setminus \lbrace 0 \rbrace$. In particular, this is how you obtain $K(X)$, once you pick an open affine $\mathrm{Spec}(R)$. On the other hand, say that $T$ is a multiplicative set in $R$. If you first create $T^{-1}R$ (that is the local ring of $X_\eta$ by inverting the functions that come from $S$), and then take the field of fractions of $T^{-1}R$ (that is, the field of fractions of $X_\eta$), you get again $K(R)$. See Proposition 10.9.10 here https://stacks.math.columbia.edu/tag/00CM.