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

  1. 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$?$

  2. 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 $K[U \cap U'] \subset K(U)?$

  3. Why $K(V) = K(U)$ for every open affine $U \subset V$?

Everything it's on page 85. Thanks for your help.

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Let $k$ be the ground field.

  1. 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 $V$.

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  • $\begingroup$ Thanks a lot, but can you explain me the same omitting 'generic points', 'schemes' and 'stalks'? $\endgroup$ – Diego Silvera Feb 17 '13 at 23:54
  • $\begingroup$ I edited my answer. Please, let me know if it is still unclear. $\endgroup$ – Brenin Feb 18 '13 at 1:04
  • $\begingroup$ It's all right now. Another thing, can you recommend me another books or notes in the spirit of Milne, but mooore explicit with its arguments? Because I have had serious headaches with this book witch my teacher decided to follow. Thanks again. $\endgroup$ – Diego Silvera Feb 18 '13 at 14:07
  • $\begingroup$ I have to say that I never used Milne's book on AG. I feel very comfortable with Ravi Vakil's notes Foundations of Algebraic Geometry where all the concepts are neatly explained. You find the current version of these notes on his webpage. $\endgroup$ – Brenin Feb 18 '13 at 14:28

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