# Regular functions extension to normal points of varieties

I am doing the exercise 3.20 in Robin Hartshorne's Algebraic Geometry, Chapter 1.

Let $Y$ be a variety of dimension $\geq2$, and let $P\in Y$ 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 $\mathcal{O}_P$ at $P$ is integrally closed. I believe we should show that $f$ is integral over $\mathcal{O}_P$.

I have found a solution here using the theory of schemes: Extension of regular function.

While the book does not present scheme theory up to this section, I think there should be another proof.

Any help or hints will be appreciated!

Here is an outline. I'll just assume $Y$ is affine to make the problem easier.

We'll have to use the second part of the lemma here. This says:

$$A(Y)_{\mathfrak{m}_P} = \bigcap_{\mathfrak q \text{ height 1 prime of A(Y)_{\mathfrak m_P}}} (A(Y)_{\mathfrak{m}_P})_{\mathfrak{q}} = \bigcap_{\mathfrak q \text{ height 1 prime of A(Y) contained in \mathfrak m_P}} A(Y)_{\mathfrak q}$$ Note all these calculations are occuring in $K(Y)$, the field of rational functions. Our given $f$ is a rational function so must be of the form $g/h$, where $g$ and $h$ lie in $A(Y)$ and $h \neq 0$. It remains to show that $h$ is not in any height 1 prime $\mathfrak q \subset \mathfrak m_P$, because then $f$ will lie in $A(Y)_{\mathfrak m_P}$ so $f$ will be regular at $P$. If $h$ were in such a prime $\mathfrak q$, then $h$ would vanish on $Z(\mathfrak q)$. We can find a point $Q \neq P$ in $Z(\mathfrak q)$ because $\dim Y \geq 2$ (basically $\mathfrak q \subsetneq \mathfrak m_P$ by Hartshorne Thm 3.2c). But $f$ is regular at $Q$ so $h$ cannot vanish at $Q$. So $h \notin \mathfrak q$, as desired.

This is a response to Mingfeng's question below.

We have a ring isomorphism from the field of fractions of $A(Y)$ to the function field $K(Y)$ (note by Hartshorne's definition, this function field is really equivalence classes of pairs of regular functions defined on some nonempty open set). This map sends a fraction $f/g$ to the class of the pair ($f/g$, $\{ g \neq 0 \}$). This map is injective because this is a nonzero map out of a field. To see surjectivity, take a rational function, and take a representative, which is a regular function $h$ defined on an open set $U$. By definition, regular functions locally look like $f/g$. So shrink $U$ so we have the representation $f/g$. Then map the fraction $f/g$ to the pair ($f/g$, $\{ g \neq 0 \}$). This pair is equivalent to $(h, U)$. Hence this map is an isomorphism.

Notice that $A(Y)_{m_p} \subset frac (A(Y))$, and $\mathcal O_p \subset K(Y)$. I am thinking of $\mathcal O_p$ as on page 16 in Hartshorne. Then, the above map restricts to an isomorphism $A(Y)_{m_p} \to \mathcal O_p$.

Now back to our problem. $f$ is defined on $Y-0$, so I can think of $f$ as a rational function. Under the isomorphism, suppose $f$ corresponds to $g/h$. My goal is to show $g/h$ actually lies in $A(Y)_{m_p}$, because then $f$ will actually lie in $\mathcal O_p$, so $f$ is actually regular at $p$.

We can see $h(Q) \neq 0$ as follows. Since $f$ is regular at $Q$, $f \in \mathcal O_Q$. This means $g/h$ lies in $A(Y)_{m_Q}$. So $h \notin m_Q$, so $h(Q) \neq 0$.

• How do you conclude that $h(Q)\neq0$? Since we only know $f=\frac{g}{h}$ locally. @hwong557 – m-agag2016 Nov 24 '16 at 6:37
• @MingfengZhao I believe $f=g/h$ not just locally, but wherever $g$ and $h$ are not both zero. $f$ is a rational function, and rational functions can be identified with the fraction field of $A(Y)$, just by sending a fraction to the rational function it defines. – hwong557 Nov 24 '16 at 14:33
• I still do not understand why the representation is global. Since $f$ is regular only on $Y\backslash\{0\}$, so we can say $f=\frac{g}{h}$ on $Y\backslash(Z(h)\bigcup \{0\})$. Do you have another proof of the above problem if assume that $f=\frac{g}{h}$ but $h=0$ on $W=Z(\mathfrak{q})$, and then get some contradiction? Thanks a lot. – m-agag2016 Nov 25 '16 at 11:24
• @MingfengZhao You are correct that $g/h$ is not a global representation of $f$. I apologize for the confusion. Nevertheless, I believe the proof is still okay. I added a new section to the answer that answers your original question $h(Q) \neq 0$. Hopefully we are both clear now. – hwong557 Nov 25 '16 at 15:29
• Since $f$ is regular at $Q$, should we say that one representative of $g/h$ lies in $A(Y)_{m_Q}$? I mean, $g_1/h_1=g/h$ (that is, $g_1h=gh_1$ on $Y$), and $h_1(Q)\neq0$. – m-agag2016 Dec 3 '16 at 2:34