Hartshorne Ex. 1.3.8 - Where do I take intersections here? 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. 
 A: 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 $K$.
A: 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 .
