Questions about Milne book I'm reading Milne's book about algebraic geometry and I have following questions?


*

*Suppose I have an irreducible algebraic variety $V$, i.e. an algebraic prevariety with the separation condition, then why is $U \cap U'$ an open affine if $U$ and $U'$ are$?$

*Now, since $U$ and $U'$ are nonempty open subsets then they are irreducible and so $K[U],K[U']$ and $K[U \cap U']$ are domains, then why $K[U \cap U'] \subset K(U)?$

*Why $K(V) = K(U)$ for every open affine $U \subset V$?
Everything it's on page 85. Thanks for your help.
 A: Let $k$ be the ground field.


*

*Consider the diagonal morphism $\Delta:V\to V\times_kV$. Then $$U\cap U'=\Delta^{-1}(U\times_kU').$$ Since $\Delta$ is a closed immersion (by definition of separatedness), the preimage of an affine open is affine, and our $U\times_kU'$ is certainly affine.


The answer to 2 and 3 (once $U\cap U'$ is assumed affine) does not depend on separatedness but only on irreducibility of $V$. 
There are three crucial facts: (i) an irreducible variety $V$ has a unique generic point (call it $\eta$), (ii) for every $x\in V$, the stalk $\mathcal O_{x,V}$ is equal to $\mathcal O_{x,U}$ if $U$ is open in $V$ (and contains $x$!), (iii) the generic point lies in every open subset of $V$ and it is the generic point in each of them.
Let $U=\textrm{Spec }A\subset V$ be open. The function field $K(V)$ is by definition the stalk $\mathcal O_{\eta,V}$ at the generic point $\eta$ of $V$, and $$K(U)\overset{(iii)}{=}\mathcal O_{\eta,U}\overset{(ii)}{=}\mathcal O_{\eta,V}=:K(V).$$
In other words, all the domains $A=\bigcap_{P\in U}A_P\subset \mathcal O_{\eta,V}$ share the same function field! you can think it this way as well: it is clear that $K(U)\subset K(V)$ because $A\subset \mathcal O_{\eta,V}=K(V)$. The reverse inclusion is given by $K(V)=\mathcal O_{\eta,V}\to\mathcal O_{\eta,U}=K(U)$, corresponding to $U\hookrightarrow V$ (sending $\eta\mapsto \eta$). 
The fact that every open affine has the same function field of the whole variety answers question 2 as well.
Edit. Avoiding generic points and stalks. Claim: if $U,U'$ are affine open subsets of an irreducible variety $V$, then $\textrm{Frac }\mathcal O_V(U)=\textrm{Frac }\mathcal O_V(U')$.
Let us take an affine open subset $P\subset U\cap U'\subset U$, principal in $U$, and an affine open subset $P'\subset P\subset U'$, principal in $U'$. Then, say, $\mathcal O_V(P)=\mathcal O_V(U)_f$ and 
$$
\textrm{Frac }\mathcal O_V(U)=\textrm{Frac }\mathcal O_V(U)_f=\textrm{Frac }\mathcal O_V(P).
$$
(An integral domain shares its field of fractions with all of its localizations.)
On the other hand,
$$
\textrm{Frac }\mathcal O_V(U')\subseteq \textrm{Frac }\mathcal O_V(P)\subseteq \textrm{Frac }\mathcal O_V(P')=\textrm{Frac }\mathcal O_V(U'),
$$
hence $\textrm{Frac }\mathcal O_V(U')=\textrm{Frac }\mathcal O_V(P)=\textrm{Frac }\mathcal O_V(U)$.
So one can define the function field of $V$ to be $\textrm{Frac }\mathcal O_V(U)$, where  $U$ is any affine open in $V$. 
