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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!

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Chan Ki Fung
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1 Answers1

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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 $h(Q) \neq 0$.

hwong557
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  • How do you conclude that $h(Q)\neq0$? Since we only know $f=\frac{g}{h}$ locally. @hwong557 – m-agag2016 Nov 24 '16 at 06:37
  • @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 03 '16 at 02:34
  • @m-agag2016 Sure if you want. I am identifying $A(Y)_{m_Q}$ with the subring of $K(Y)$ whose denominators are not in $m_Q$. – hwong557 Dec 03 '16 at 02:37
  • Do you mean that if we identify $A(Y){m_Q}$ inside $K(Y)$, since $f\in A(Y){m_Q}$, then every representative of $f$ has a unique global representation in the form $g/h$? – m-agag2016 Dec 03 '16 at 02:40
  • @m-agag2016 The representation is neither unique nor do we have a global representation. I explained that in the bijection above. – hwong557 Dec 03 '16 at 02:43
  • There is still a problem with the proof. If you do it in this way, by starting with a choice of a representation as g/h, there is no way to guarantee that h is non-zero at Q. For example, take f to be identically 1 outside the origin in A^2, and y/y as its representative outside the y-axis in $A^2$. The problem is easy to solve, of course. You just have to give yourself freedom to pick the representation according to Q, and it doesn't really matter as in the end everything should be equivalent to f itself in K(Y). (Again, recall Hartshorne's definitions!) – Artur Araujo Mar 23 '18 at 00:06
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    I think the proof you gave is wrong as written. You cannot conclude $h \neq 0$ at $Q$. What you can conclude from regularity of $f$ is that $f \in A(Y){\mathfrak{m}_Q}$, which is a subset of $A(Y){\mathfrak{q}}$, since $\mathfrak{q} \subset \mathfrak{m}_Q$. Thus $f$ belongs to each set being intersected and $f \in \mathcal{O}_P$. I think it is not necessary to write $f$ as a specific quotient of elements of $A(Y)$— such a representation is not unique, and phrasing the proof in terms of it is confusing. – Vik78 Apr 14 '22 at 17:15