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In Hartshorne, Ex II.5.14, he defines that a closed subscheme $X \in \mathbb{P}^r_A$ is called projectively normal if the homogeneous coordinate ring $S(X)$ is integrally closed. This is also given by many notes and in wikipedia. But the denominators of elements of the fraction field of $S(X)$ may not be homogeneous, which makes things funny and the problem difficult to solve.

My question is whether this definition is the right one. If yes, how does it make sense? If no, what is the common way to fix the definition?

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  • $\begingroup$ You are correct that the denominators of elements of the fraction field need not be homogeneous, but this is not an issue when solving exercise II.5.14 - the strategy is similar to the proof of theorem II.5.19, and the issue you've identified is completely bypassed there. You may also note that part (d) of this problem gives an definition which avoids discussion of the field of fractions of the coordinate ring. $\endgroup$
    – KReiser
    Nov 18, 2020 at 21:48
  • $\begingroup$ @KReiser May you explain in detail? $\endgroup$
    – XT Chen
    Nov 19, 2020 at 2:48
  • $\begingroup$ Explain what? Have you read/understood the proof of theorem II.5.19 and part (d) of exercise II.5.14? $\endgroup$
    – KReiser
    Nov 19, 2020 at 3:30
  • $\begingroup$ @KReiser I have read the proof and maybe I don't understand. My confusion is how can I bypass the degree problem as in 5.14. What we want to show is that $\mathscr{S}_p$ is integrally closed where $\mathscr{S} = \oplus \mathcal{O}(n)$. But for an element in $\mathrm{Frac} (\mathscr{S}_p)$, its denominator and numerator are not necessary homogeneous. What we only know is that $S_{(p)}$, whose elements' denominators and numerators are homogeneous of the same degree, is integrally closed and $S'$ is integral over $S$. $\endgroup$
    – XT Chen
    Nov 19, 2020 at 8:03
  • $\begingroup$ You're right, I didn't see this gap and I guess none of my students have either. I've written an answer below. $\endgroup$
    – KReiser
    Nov 19, 2020 at 9:33

1 Answer 1

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This definition is fine and part (d) gives you a definition which is equivalent and doesn't encounter this issue of your. Your concern can be fixed with a little ingenuity.

Suppose $R=\bigoplus_d R_d$ is a graded domain, and let $K$ be its quotient field. Assume there exists a nonzero element $y\in R_1$. Let $K_d$ denote the subgroup of elements which can be represented as $\frac{r_1}{r_2}$ for $r_1,r_2$ homogeneous and $\deg r_1-\deg r_2=d$. Then $K'=\bigoplus_{d\in\Bbb Z} K_d$ is a subring of $K$, it's isomorphic to $K_0[y,1/y]$, and $K=K_0(y)=\operatorname{Frac} K'$. We observe that $K_0[y,1/y]$ is integrally closed in its field of fractions, being the localization of a polynomial ring over a field, so the integral closure of $R$ must be a subring of $K'$, and in particular consist of sums of homogeneous fractions. So we've eliminated your difficulty.

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  • $\begingroup$ Choosing $y\in S_{>0}\backslash p$ and by this answer, the problem is totally set down. $\endgroup$
    – XT Chen
    Nov 19, 2020 at 10:28

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