1
$\begingroup$

I'm trying to solve Exercise I.3.18 (a) in Hartshorne's Algebraic Geometry. The exercise is stated as follows.

Defn: A variety $Y$ is normal at a point $P \in Y$ if the local ring $\mathcal{O}_{P,Y}$ is an integrally closed ring. $Y$ is normal if it is normal at every point.

Defn: A projective variety $Y \subset \mathbb{P}^n$ is projectively normal (with respect to the given embedding) if its homogeneous coordinate ring $S(Y)$ is integrally closed.

Exercise: If a projective variety $Y \subset \mathbb{P}^n$ is projectively normal, then $Y$ is normal.

It seems like an easy exercise. My attempt is to calculate the local ring of $Y$ at an arbitary point $P \in Y$. By theorem I.3.4 in Hartshorne, we know that $\mathcal{O}_{P,Y} = S(Y)_{(M_P)}$, where $M_P$ is the ideal generated by the set of homogeneous $f \in S(Y)$ such that $f(P)=0$.

Since the localization of an integrally closed ring is again integrally closed, we see that $S(Y)_{M_P}$ is integrally closed. Yet my question is, when we move on, how can we show that the subring of degree zero in the graded ring $S(Y)_{M_P}$ is integrally closed?

I feel really frustrating when facing graded rings for the first time. I'm not sure whether my first steps on the above attempts are correct or not. Say $P=[a_0, \ldots, a_n] \in \mathbb{P}^n$, is the corresponding maximal ideal $M_P$ stil be $(X_0 - a_0, \ldots, X_n - a_n)$ as in the affine case? (Here I omitted the 'bars' representing the coorsponding equivalent classes in $S(Y)$.) Does the localization $S(Y)_{M_P}$ differ from the "ordinary" localization defined in [Atiyah & MacDonald]? I used to think that there's no distinction, but after reading serveral posts on this topic in MSE, I'm getting more puzzled.

$\endgroup$

1 Answer 1

3
$\begingroup$

Consider the integrally closed ring $R=S(Y)_{M_P}$, let $R_0$ be the set of its elements of degree $0$. Let $f=a/b \in Frac(R_0)$ be integral over $R_0$, thus $f$ is integral over $R$ so is in $R$. Thus, we have $a=fb$, $a,b \in R_0$, $f \in R$. As $R$ is integrally closed, $R$ is an integral domain, so $f$ cannot have any component of degree nonzero, thus $f \in R_0$.

$\endgroup$
1
  • $\begingroup$ Thank you a lot! :) $\endgroup$
    – Hetong Xu
    Jul 25, 2020 at 7:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .