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Let $H_i$ and $H_j$ be the hyperplanes in $\Bbb{P}^n$ defined by $x_i = 0$ and $x_j = 0$ with $i \neq j$. I want to show that any regular function on $\Bbb{P}^n - (H_i \cap H_j)$ is constant. Now I think I have the proof in my hands which is the following.

My proof: We have $$\Bbb{P}^n - (H_i \cap H_j) = (\Bbb{P}^n - H_i) \cup (\Bbb{P}^n - H_j).$$ The ring on regular functions on the left hand side should be equal to $\mathcal{O}(\Bbb{P}^n - H_i) \cap \mathcal{O}(\Bbb{P}^n - H_j)$. We know that $\mathcal{O}(\Bbb{P}^n - H_i) \cong (k[x_0,\ldots,x_n]_{x_i})_0$ and similarly for $x_j$. Thus for any regular function $h $ on $\Bbb{P}^n - (H_i \cap H_j)$, $$h = \frac{f}{x_i^k} = \frac{g}{x_j^m}$$ for some $f,g$ homogeneous polynomials respectively of degrees $k$ and $m$. Then by unique factorization and assuming that the fractions are in the lowest terms I conclude $k = m = 0$ so that $h \in k$.

My problem: The thing is, taking the intersection of two rings of regular functions requires knowing that they are both embedded into something big. What is this thing? I guess it would be the fraction field of $k[x_q,\ldots,x_n]$. Also, I am worried that my answer in some sense is "not canonical" because it requires me to make a choice in embedding the ring of regular functions inside of something.

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    Any rationnal (hence any regular) function on an open subset of $X$ is an element of the field $K(X)$ of rationnal functions on $X$. Here $K(\mathbb{P}^n)$ is (maybe not canonically) isomoprhic to $K(\mathbb{A}^n) = K(x_1,x_2,\dots,x_n)$ (cf thm 3.4 of Hartshorne). – user10676 Apr 30 '13 at 08:38
  • What I mean is that you can use Thm 3.4 to compute $K(\mathbb{P}^n)$. Note that it is the fraction field with $n$ indeterminates (not $n+1$ !). – user10676 Apr 30 '13 at 08:51
  • @user10676 I am looking at Theorem 3.4 now and we have $K(\Bbb{P}^n) \cong S(Y)_{((0))}$. I agree that the latter is the polynomial ring in $n$ - variables not $n+1$ (Sorry I forgot to take the degree $0$ - component) but how does that help me here? –  Apr 30 '13 at 08:54
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    It exactly answers your question. If $X$ is an integral scheme (if you wish, irreducible variety), then $\mathcal{O}X$ is a subsheaf of the constant sheaf corresponding to $K(X)$, and in fact we have $\mathcal{O}_X(U) = \cap{x \in U} \mathcal{O}_{X,x}$, and $\mathcal{O}_X(U \cup V) = \mathcal{O}_X(U) \cap \mathcal{O}_X(V)$, taken inside $K(X)$. – Martin Brandenburg Apr 30 '13 at 09:01
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    If your question is What is this thing?, then the answer is $K({\mathbb P}^n)$, an the embedding ${\mathcal O}(\mathbb{P}^n-H_i) \subset K({\mathbb P}^n)$ is canonical. If not, then I misunderstand you. – user10676 Apr 30 '13 at 09:02
  • Dear Martin, your last sentence of your comment is precisely what I want to know, thanks. So you're saying that for every open subset $U$ of $X$, we always have that $\mathcal{O}_X(U) \hookrightarrow K(X)$ canonically? (At least when $X$ is an irreducible variety) –  Apr 30 '13 at 09:07
  • @user10676 It seems to me that to know that any regular fucntion on an open subset is an element of $K(X)$ requires knowing that $\mathcal{O}X(U) = \cap{x\in U} \mathcal{O}_{X,x}$. Thought this is not proven in Hartshorne section 1.3, I think I can prove it in the case that $X = \Bbb{P}^n$ here. So if we identify the local rings as subrings of $K(\Bbb{P}^n)$ then my question is answered. –  Apr 30 '13 at 09:14
  • This equation is trivial when using the definition of regular functions. – Martin Brandenburg Apr 30 '13 at 09:24
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    @MartinBrandenburg For a beginner like me, why is it trivial? For one it does not come immediately from the definition given in Hartshorne. I have looked at 1.3 again, and he does not even define what is the ring of regular functions on an open subset. –  Apr 30 '13 at 09:44
  • Dear Benja, Martin's characterization is indeed not trivial. It is given as Hartshorne's Theorem 3.2 (a), and its proof (see bottom of page 17) requires the two intermediate results (b) and (c) . – Georges Elencwajg May 01 '13 at 13:24
  • @GeorgesElencwajg Dear Georges, I am looking at the proof now, but the proposition is for affine varieties. I believe Martin's result $\mathcal{O}U(X) = \bigcap{x \in U} \mathcal{O}_{X,x}$ is for any open subset $U$ of any irreducible variety, which is not the same as the result in Hartshorne. Regards. –  May 01 '13 at 13:32
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    Dear Benja, you are absolutely right. The general case however follows from the affine case and you can find it as Proposition 4.18 (c) page 65 in Qing Liu's book, so now you have at least a reference . The reduction from the general to the affine case is not technically difficult but it requires some familiarity with the notion of sheaf. I suggest that you don't spend too much time on this if it causes you trouble, but come back to the question when you feel more familiar with sheaves, an indispensable tool that I am sure you will soon master. Good luck! – Georges Elencwajg May 01 '13 at 14:05
  • How do you know that $\mathcal{O}(\mathbf{P}^{n}-H_{i})\cong(k[x_{0},...,x_{n}]{x{i}})_{0}$? Where does this isomorphism come from? – JDZ Aug 27 '19 at 03:40

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I will try to make clear what is a regular and rational function, which is a good exercise for me because I am also learning this stuff.

Let $X$ be an irreducible affine, quasi-affine, projective or quasi-projective variety. We know what a regular function on $X$ is (cf. Hartshorne p.15) and I won't recall it here. Denote by $\mathcal{O}(X)$ the set of regular functions on $X$.

Now a rational function on $X$ is : the data of an open $U\subset X$ and a regular function $\phi : U \rightarrow K$ modulo the obvious equivalent relation (cf. Hatrshorne p.16). Denote by $K(X)$ this set.

Now there is an obvious morphism of $K$-algebras $\mathcal{O}(X) \rightarrow K(X)$ which sends $f$ to the pair $(X,f)$, and it is injective (Note: (I think) Hartshorne doesn't mention/prove explicitly this point, but it comes from the facts that open sets are dense and $K$ is a separated variety). There are some useful theorems (quite hard to prove) which allow us to compute $\mathcal{O}(X)$ and $K(X)$ (namely Thm 3.2 and Thm 3.4).

If $V$ is an open subset of $X$, then there is an embedding $K(U) \rightarrow K(X)$ (sending a pair $(U,\phi)$ to $(U,\phi)$).

I hope this will help make your mind clear about regular and rational functions (note that I didn't mention the sheaf of functions but you don't need this for this exercise).

Note : I didn't tell you how solve your exercise. You proved that there exist $f,g \in K[T_0,\dots,T_n]$ such that for all $x=(x_i)_{i=0,...,n} \in K^{n+1}$ with $x_0 \neq 0$ and $x_1 \neq 0$ : $$\frac{f(x)}{x_i^k} = \frac{g(x)}{x_j^k}.$$ This can be seen as an equality in $K(\mathbb{P}^n)$, but the 'symbols' $f(x)$, $x_i$, $g(x)$, $x_j$ are not quite elements of $K(\mathbb{P}^n)$. You may want to rewrite this expression and for this I suggest you to find a transcendence basis of $K(\mathbb{P}^n)$ over $K$.

JDZ
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user10676
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  • Dear user10676, sorry for the late reply but why is my solution wrong? Do I really need to go to transcendence bases? Maybe I indexed my $x_i's$ wrongly but I think my solution is correct. –  May 02 '13 at 13:16
  • @BenjaLim: if you want to conclude with the fact that $K[T_0,\dots,T_n]$ is UFD, then you will have to prove that $f(T_0,\dots,T_n)/T_i^k=g(T_0,\dots,T_n)/T_j^m$ AS polynomial (which can be done). But beware that it is senseless to talk about UFD if you work in $K(\mathbb{P}^n)$ because $f(x)/x_i^k$ is NOT the 'quotient' of the 'function' $f(x)$ by the 'function' $x_i^k$. – user10676 May 02 '13 at 14:30
  • Dear user 10676, it is proven in Hartshorne that $K(\Bbb{P}^n) \cong S(\Bbb{P}^n)_{((0))} =k(x_0,\ldots,x_n)_0$. Why can't I identify $f(x)/x_i^k$ say as a quotient of two polynomials? –  May 03 '13 at 04:03
  • Ok, you are right. $f(x)$ and $x_i$ should be seen as elements of the field $K(\mathbb{A}^{n+1})$ which contains $K(\mathbb{P}^{n})$. Now that's fine. – user10676 May 03 '13 at 09:11
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Although I find the comments to the question and user10676's answer quite satisfying, I'll add a new perspective at Benja's explicit request to me in a comment to another question here.

If $X$ is a normal variety (for example a smooth variety), and $Y$ is a closed subvariety of codimension $\geq 2$, then the restriction morphism $\mathcal O(X)\to \mathcal O(X\setminus Y)$ is bijective.
In particular if $X$ is complete we obtain $\mathcal O(X\setminus Y)=\mathcal O(X)=k$ and this applies to your case, where you may take $X=\mathbb P^n_k$ and $Y=H_i\cap H_j$.

Unfortunately I could not locate this theorem in Hartshorne, which seems to be your reference book, but it follows from Chapter 4, Theorem 1.14, page 118 of Liu's fine book .

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Georges Elencwajg
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  • Dear Georges, thanks for your answer. I really appreciate it. I think I sent you an email a few days ago to the email address you have on your math.se page. Regards, –  May 01 '13 at 13:14
  • Dear Benja, you did well to warn me about your e-mail. There is so much spam in my mailbox that I tend not to open it for long intervals and I had not seen your letter. I am very sorry for my negligence and will answer you now. – Georges Elencwajg May 01 '13 at 13:35