0
$\begingroup$

Let $X$ be a prevariety. Define a equivalnce class of $(U,f)$ where $\emptyset \neq U \subset X $ open and $f \in \Gamma (U)$ by setting $(U,f) \sim (V,g)$ if there is $\emptyset \neq W\subset U\cap V$ open such that $f|_W =g|_W$

Define the structure of a field extension of $k$ (algebraic closed field) on the set of equivalence classes, and show that this field extension is isomorphic to the function field of $X$

This is exercise 1.17 from Ulrich Gortz and Torsten wedhorn algebraic geometry 1

Any ideas?

Edit: In this book, a prevariety is a connected space with functions $(X,\mathcal{O}_X)$ with the property that there is a finite covering $X=\cup_{i=1}^n U_i$ such that the space with functions $(U_i,\mathcal{O}_{X|U_i})$ is an affine variety for all $i=1...n$

  • 0
    Since not anybody has access to the book, you should probably say what a prevariety is in this book. – 2017-02-21
  • 0
    added definition @MooS – 2017-02-21

1 Answers 1

2

Here is an outline, you should be able to fill in the details:

Denote the set of equivalence classes by $K$ and the function field by $K(X)$.

  • Define $(U,f) \cdot (V,g) := (U \cap V, fg)$ and $(U,f) + (V,g) := (U \cap V, f+g)$ (Note that I actually mean the respective restrictions of $f,g$ on the right hand side.)
  • Check that this gives you a field $K$. For the multiplicative inverse, note that you can invert any non-zero function on the (open!) locus of its non-zeros.
  • Obviously $k$ is a subfield (constant functions).
  • For any affine open $U$, we clearly have an injection $\Gamma(U) \hookrightarrow K$. Since $K$ is a field, it factors through the quotient field, i.e. we get an injection $K(X) := K(U) \hookrightarrow K$.
  • This is surjective because any element in $K$ can be represented as $(U,f)$ with $U$ affine and then $(U,f)$ is contained in the image of $K(X) := K(U) \hookrightarrow K$.