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.

  • $\begingroup$ 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. $\endgroup$ – Nicolas Bourbaki Jul 24 '15 at 7:12
  • $\begingroup$ 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. $\endgroup$ – Lorenzo Najt Jul 24 '15 at 9:38
  • 1
    $\begingroup$ 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. $\endgroup$ – Nicolas Bourbaki Jul 24 '15 at 18:04
  • 1
    $\begingroup$ @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. $\endgroup$ – Lorenzo Najt Jul 24 '15 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.