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I have some questions about exercise I.3.20 of Robin Hartshorne's Algebraic Geometry:

Let $Y$ be a variety of dimensions $\geq 2$, and $P\in Y$ be a normal point. Let $f$ be a regular function on $Y − P$. Show that $f$ extends to a regular function on $Y$.

$P$ is a normal point means the local ring $O_P$ at $P$ is integrally closed.

There are some posts about this question here on MSE, here on MO, and again on MSE.

The proofs of the above posts are similar, but I have a common question about the proof. Take the proof in the first post for example:

Firstly, we can reduce this question to the affine case, and use a theorem in commutative ring theory to see that :

$$(\mathcal{O}(Y))_{\mathfrak{m} P}=\bigcap_{\text {height } \mathrm{q}=1, \mathrm{q} \in \operatorname{Spec}\mathcal{O}(Y), \mathfrak{q} \subset \mathfrak{m}_P} \mathcal{O}(Y)_\mathfrak{q}.$$

Now it suffices to prove that the regular function is included on the right-hand side.

My question is, a regular function $f$ on $Y-P$ is locally a rational function of polynomials, which doesn't mean it can be expressed as a rational function globally. So how do we know that $f\in \mathcal{O}(Y)_\mathfrak{q}\simeq A(Y)_\mathfrak{q} \subset \operatorname{Frac}(A(Y))?$ since it means $f$ can be expressed globally.

Eric
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  • Your proposed objection is just wrong - $f\in \operatorname{Frac}(A(Y))$ does not mean it can be expressed globally. (I've also improved your formatting a bit - please take a look at the edit.) – KReiser Apr 18 '22 at 03:19
  • @KReiser Thank you. Could you please make it clear to me? In the above proofs, they show that $f$ can be expressed locally, which implies that $f$ is included in RHS. I cannot understand why it holds. – Eric Apr 18 '22 at 03:34
  • What part? To be honest, I'm not sure where you're getting tripped up - this should all be pretty direct from the definition of the local ring of a subvariety. Are you familiar with that concept? – KReiser Apr 18 '22 at 03:43
  • @KReiser In the proof of the first post, they show that for arbitrary $Q \neq P$, we can find an open neighborhood containing $Q$, on which $f=\frac{g}{h}$, where $g,h\in A(Y)$ and $h \notin \mathfrak{q}$. But I can not understand why this claim shows that $f\in A(Y)\mathfrak{q}$? since I think $f\in A(Y)\mathfrak{q}$ implies that $f$ can be expressed globally on Y but they only show that $f=\frac{g}{h}$ locally. – Eric Apr 18 '22 at 03:55

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You have some strange misconceptions about things being "expressed globally". Let's review the definition of the local ring of a subvariety: for $Z\subset Y$ a subvariety, the local ring of the subvariety $\mathcal{O}_{Y,Z}$ is the set of equivalence classes $(U,f)$ where $U\subset Y$ is open, $U\cap Z\neq\emptyset$, and $f$ is a regular function on $U$. We say two elements $(U_1,f_1)$ and $(U_2,f_2)$ are equal iff $f_1$ and $f_2$ agree as regular functions on $U_1\cap U_2$. (This is why showing $f=g/h$ locally is enough to show $f\in\mathcal{O}_{Y,Z}$.) It is easy to show that if $Z$ is determined by the prime ideal $\mathfrak{q}\subset A(Y)$ that $\mathcal{O}_{Y,Z}\cong A(Y)_\mathfrak{q}$. There is no requirement that anything mentioned here be defined globally.

KReiser
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