# Localization of graded rings and the normality of its subring of degree zero

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.

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 $$f \in R_0$$.