7

Consider the context of this question: Local Ring of a Subvariety (problem 1.3.13 in Hartshorne).

In trying to prove that $\dim \mathcal{O}_{Y,X} = \dim X + \dim Y$, I proved that $\mathcal{O}_{P,X} / m_{Y,X} \cong \mathcal{O}_{P,Y}$, where $P$ is a point in $Y$. If the local rings $\mathcal{O}_{P,X}$ and $\mathcal{O}_{P,Y}$ are localizations of an affine ring (i.e. a finitely generated $k$-algebra), then we have that $\operatorname{ht} m_{Y,X} + \dim \mathcal{O}_{P,Y} = \dim \mathcal{O}_{P,X}$ by the Cohen-Macaulay property, and the result follows.

Question 1: Why can we make the above assumption?

Hartshorne's problem statement also mentions that $K(Y)$ is NOT algebraically closed if $Y$ is not a point.

Question 2 (edited): How can we see that? What if $Y$ is not affine variety?

PS: I am looking for an instructive answer showing how we can reduce to the affine case (I assume this is how these two questions can be answered), as I am still working on getting used to reduction to affine case type of arguments.

Community
  • 1
Manos
  • 25,833

1 Answers1

7

Here are some points you might find helpful.

(Note that I will use "affine" to denote either affine or quasi-affine. This is in accordance with the general terminology of schemes which you'll encounter in the next chapter)

In general whenever you're trying to prove a 'local' statement, you can always reduce to the affine case, since schemes/varieties are locally affine. For example, if $X$ is a projective variety inside $\mathbb{P}^n$, then the cover of $\mathbb{P}^n$ by $n+1$ affine opens naturally gives you an open covering of $X$ by affine opens.

You should try to think of points as just being particular (ie, maximal codimension) subvarieties, as opposed to being something particularly unique. In particular, many of the things you can do with points you can also do with arbitrary subvarieties. For example, if you take the affine plane with ring $k[x,y]$, the local ring at the origin is just $k[x,y]_{(x,y)}$. This is a case of "localization at a prime", so in particular, you can consider other localizations like $k[x,y]_{(x)}$. This is also just a localization at the prime ideal $(x)$, which is now not the local ring of a point as you are used to. However, drawing upon the philosophy that ideals correspond to their vanishing, you should think of this ring as the local ring of the subvariety $\{x = 0\}$ of the affine plane.

In fact, it's true that every subvariety of any affine variety $U$ is the vanishing of a prime ideal of the ring of regular functions on $U$. From the algebraic properties of localization you should be able to prove that if $Y$ is a subvariety of an affine variety $U$ which is the vanishing of the prime ideal $p$, then $\mathcal{O}_{Y,U} = (\mathcal{O}_U)_p$.

For your second question, since the function field of any variety (projective, quasi-proj, affine, quasi-affine) is the same as the function field of any open subset, we may assume $Y$ is affine. Then, if $Y$ is a dimension $n > 0$ affine variety over $K$, then you know that if $A$ is the ring of $Y$, then $A$ is a finitely generated $K$-algebra, and $K(Y)$ is the fraction field of $A$. Since $A$ is a finitely generated $K$-algebra, it looks like $K[x_1,\ldots,x_m]/(f_1,\ldots,f_r)$. You know $K(Y)$ has trdeg $n$ over $K$, so $K(Y)$ is an algebraic extension of $K(t_1,\ldots,t_n)$ where the $t_i$ are algebraically independent elements of $K(Y)$. You also know that $K(t_1,\ldots,t_n)(x_1,\ldots,x_m) = K(Y)$, where you view the $x_j$ as algebraic elements over $K(t_1,\ldots,t_n)$, so $K(Y)$ is a finitely generated field extension of $K(t_1,\ldots,t_n)$. However, the algebraic closure of $K(t_1,\ldots,t_n)$ is necessarily infinitely generated, since adjoining any finite collection of algebraic elements gives you a finite extension, and the algebraic closure of $K(t_1,\ldots,t_n)$ has infinite degree over $K(t_1,\ldots,t_n)$, since for example for any $n$, $\sqrt[n]{t_1}$ has degree $n$ over $K(t_1,\ldots,t_n)$.

This is a rigorous proof, but sometimes it's easier to understand math just by doing examples. Take for example some finitely generated $K$ algebra like $K[x,y]/(y^2 - xy + x^3)$. Then its function field is just the extension of $K(x)$ obtained by adjoining a root of of the polynomial $T^2 - xT + x^3$ over $K(x)$. This is obviously a finite extension of $K(x)$, so it can't be algebraically closed, since the algebraic closure of any field of the form $K(x_1,\ldots,x_n)$ is infinite over $K(x_1,\ldots,x_n)$.

oxeimon
  • 12,279
  • That's correct. If $X$ is affine with ring $A$, and $Y\subset X$ is quasi-affine, the the ring of functions of $Y$ is just $A_f$, where $f\in A$ (this is the localization at the powers of $f$, ie we're adding in $f^{-1}$). If $P\in Y$, then $P$ corresponds to a prime ideal $\mathfrak{p}$ of $A_f$, which are precisely the primes of $A$ which do not contain $f$. It's easy to see then that localizing at $\mathfrak{p}$ means you're adding in $f^{-1}$ anyway, so $A_{\mathfrak{p}} = (A_f)_\mathfrak{p}$. – oxeimon Jun 02 '14 at 19:18
  • I've added an answer to your second question. – oxeimon Jun 02 '14 at 20:04
  • Question 1: In your comment above, what exactly is $f$? Question 2: According to Hartshorne's definition of a subvariety (Problem 3.10), a subvariety is locally closed and irreducible, but need not be closed. Now if $X$ is an affine variety with ring $A$, then prime ideals of $A$ correspond to closed subvarieties of $X$. Do you assume in your 5th paragraph that the subvariety is closed? – Manos Jun 03 '14 at 15:34
  • Question 3: Again regarding your 5th paragraph, I am trying to prove $\mathcal{O}{Y,U} = (\mathcal{O}_U)_p$ assuming $Y$ is a closed subvariety of $U$. Now, I have shown that there exists an injective ring homomorphism $(\mathcal{O}_X)_p \rightarrow \mathcal{O}{Y,U}$ and I am stuck in proving surjectivity. In particular, let $\langle V, f \rangle \in \mathcal{O}_{Y,U}$, with $V$ open in $U$, $f$ regular on $V$ and $V \cap Y \neq \emptyset$. I want to show that $f$ extends to a regular function on $U$ so that $\langle V, f \rangle$ is the preimage of $f/1$, but then $f$ may not extend... – Manos Jun 03 '14 at 16:40
  • To your first question, yes I did assume closed subvariety, but a locally closed subvariety is just a closed subvariety of an open subset. The open subsets associated to localizations of the form $A_f$ are all affine, and its a fact that they form a basis for the Zariski topology, so you can always find a small enough open affine inside which your subvariety is closed. – oxeimon Jun 04 '14 at 00:03
  • For your second question, by your definition of $\mathcal{O}_{Y,U}$, $f$ doesn't have to extend to all of $Y$. It just has to be regular on a nontrivial open subset. – oxeimon Jun 04 '14 at 00:12