# Function Field of Variety and Scheme

Let $V\subseteq \mathbb{A}^n_k$ be a closed irreducible algebraic set ("affine variety") over a closed field $k$.

Construct the topological space $X$ consisting of all closed irreducible subsets of $V$ (it is given the Zariski topology). Note that we have a continous map $f:V\to X$ given by $f(x) = \overline{\{x\}}$.

Let $\mathcal{O}_V$ be the sheaf of regular functions over $k$ on $V$. Push the sheaf, the ringed space $(X,f_*\mathcal{O}_V)$ is an affine scheme.

Is the following true? Given $V'\in X$, $k(V')\simeq K(V')$. Where $k(V') =$ residue field of the scheme $X$ at $V'$, and $K(V') =$ the field of rational functions of $V'$.

I think exactly how to answer your question may depend on the definitions you are using.

To set definitions:

On the side of varieties: $K(V') = frac(k[X]/I(V'))$, where $k[X]$ is the coordinate ring of $X$. (And $V'$ is serving double duty, here as an algebraic set.)

The residue field of the scheme $X$ at $V'$ is the residue field of the ring $k[X]$ localized at the prime $V'$.

So the commutative algebra side of your question is:

Given a ring $R$ and a prime ideal $P$, is $frac(R/P) \cong R_P / PR_P$. The answer is yes: localization commutes with quotients ("localization is exact"), and localizing $R/P$ at $P$ is the same as constructing its field of fractions, because we are inverting all nonzero elements.

• What is the coordinate ring of a scheme? I never seen this notation before. Also, my definition of $k(V')$ is the quotient of the stalk of $X$ at $V'$ modulo its maximal ideal. – Nicolas Bourbaki Jul 24 '15 at 7:12
• Your definition of residue field at a point in a scheme agrees with mine, all that is going on here is that localization is exact. I thought you were asking about the field of rational functions of the corresponding variety - if not then I am somewhat confused about what your question is. The notation k[X]? It denotes the coordinate ring of a variety X over a field k. Anyway I think I am being a little sloppy with notation, but probably the commutative algebra fact I mentioned answers your question. – Lorenzo Najt Jul 24 '15 at 9:38
• But X is a scheme over k. What do you mean by k[X]? I understand what k[V] means, since V is a variety over k. – Nicolas Bourbaki Jul 24 '15 at 18:04
• @NicolasBourbaki Yes, I mean $k[V]$, sorry. But it is also probably reasonable to interpret $k[X]$ as the ring of global sections of $X$, which is isomorphic. – Lorenzo Najt Jul 24 '15 at 22:20