Hartshorne exercise I.3.20, is the point required to be normal?

Question

Problem I.3.20 in Hartshorne: Show that if $Y$ is a variety such that $\dim Y \ge 2$ and $Y$ is normal at a point $P$, then any regular function on $Y-P$ extends to a regular function on $Y$

It seems like most solutions use a result from commutative algebra that depends on $\mathcal{O}_P$ being integrally closed. For example Hartshorne Problem I.3.20, or Regular functions extension to normal points of varieties

I believe I have a solution that does not depend on $\mathcal{O}_P$ being integrally closed, but would like to make sure I am not making a mistake.

edit: this is not true

Let $M$ be a set of maximal ideals in an integral domain $A$. Let $S = \{ f \in A \mid f \notin m, \forall m \in M\}$ be a multiplicative set.

We have $A_S = \bigcap\limits_{m \in M} A_m $

The proof of this is a modified version of what is found here:

An integral domain $A$ is exactly the intersection of the localisations of $A$ at each maximal ideal, and Localization at an element is intersection of localizations at primes not containing the element

We can use this fact and basically copy the proof of I.3.2 from Hartshorne but apply it to quasi-affine varieties. Suppose $X$ is a quasi-affine variety open in $\bar{X}$. We have a natural map $A(\bar{X}) \rightarrow \mathcal{O}(X)$, the kernel is just $0$ since $\bar{X}$ is irreducible. Let $S = \{ f \in A(\bar{X}) \mid f \notin m_q, \forall q \in X\}$, we also have an injective map $A(\bar{X})_S \rightarrow \mathcal{O}(X)$ since if $h \in S$, $h(q) \neq 0 \space \forall q \in X$, so $g/h$ is in $\mathcal{O}(X)$ for all $g$. We can also construct an isomorphism $K(\bar{X}) \rightarrow K(X)$ by intersecting open sets with $X$ and use I.3.2 to show an isomorphism $\operatorname{Quot}(A(\bar{X})) \rightarrow K(X)$.

We know $\mathcal{O}(X) \subseteq \bigcap\limits_{q \in X} \mathcal{O}_{q,X}$ and we have isomorphisms $\bigcap\limits_{q \in X} \mathcal{O}_{q,X} \rightarrow \bigcap\limits_{q \in X} A_{m_q} \rightarrow A(\bar{X})_S$. So we have $A(\bar{X})_S \cong \mathcal{O}(X)$.

Now we show that the set $S$ doesn't change when we remove a point from $X$. If $\dim(\bar{X}) \ge 2$, by exercise I.1.8 we have every irreducible component of $\bar{X} \cap Z(f)$ has dimension $\ge 1$ if $\bar{X} \cap Z(f) \neq \emptyset$. If $X \cap Z(f) = \{p\}$ for a single point $p$, we have $\bar{X} \cap Z(f) = ((\bar{X} \backslash X) \cap Z(f)) \cup \{p\}$ so we have an irreducible component with dimension 0. So if $X \cap Z(f)$ has one point, it has more than one point. So if $f \in m_p$ then $f \in m_q$ for some $q \in X \backslash \{p\}$. So if $f \notin m_q, \forall q \in X\backslash \{p\}$ then $f \notin m_q, \forall q \in X$

So the map $\mathcal{O}(X) \rightarrow \mathcal{O}(X\backslash\{p\})$ induced by the inclusion must be surjective, and we have that there must be an extension of the regular function in the exercise. For the projective case we reduce to the affine cover.

So is this proof accurate? My thought is that regular functions on quasi-affine varieties should just be of the form $g/h$ where h could be zero only on $\bar{X}\backslash X$. Is there a counterexample when $\mathcal{O}_P$ is not integrally closed? I'm not sure why Hartshorne put that requirement in.

edit: The commutative algebra assumption I was using is just not true, so everything that follows doesn't work.

Answer 1

The point must be normal. Here is a counterexample: let $X$ be $\Bbb A^1$ with $(0,0)$ glued to $(1,1)$. The coordinate algebra $k[X]$ of $X$ is the subalgebra of $k[x,y]$ generated by $$\{x(x-1),x(y-1),y(x-1),y(y-1),x^2(x-1),x^2(y-1),y^2(x-1),y^2(y-1)\}$$ (ref), which is not integrally closed as $x\in k(X)$ is a solution of $T^2-T-x(x-1)$ but not in the coordinate algebra $k[X]$. Since $X$ without the glued point is isomorphic to $\Bbb A^2\setminus\{(0,0),(1,1)\}$, a normal variety, we must have that this glued point is the non-normal point. This same rational function $x$ is regular on the complement of the glued point, but does not extend to the whole variety $X$, since it would have to take the values $0$ and $1$ at the glued point.