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This is a question about notation in Hartshorne's Algebraic Geometry. According to my understanding $k[x_0,\cdots,x_n]_{(x_i)}$, (see e.g. page 18), consists of the elements of degree zero in the localization of $k[x_0,\cdots,x_n]$ by the prime ideal $(x_i)$. That is, denominators must belong to the complement of $(x_i)$. So why is $g/x_i^N$ an element of $k[x_0,\cdots,x_n]_{(x_i)}$, for $g \in k[x_0,\cdots,x_n]$ (see e.g. bottom of page 18)?

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The notation is quite confusing, but at least we can usually figure it out from context.

The notation $k[x_1,\ldots,x_n]_{(x_i)}$ here is equivalent to $k[x_1,\ldots,x_n][x_i^{-1}]_0$, meaning the subring of degree $0$ elements of the localization at the element $x_i.$ It does not simply denote the localization $k[x_1,\ldots,x_n]_{\frak p},$ where $\frak p$$=\langle x_i\rangle $ is a prime ideal, nor is it simply the localization $k[x_1,\ldots,x_n][x_i^{-1}]$ at the multiplicative set generated by $x_i.$ That is why any $f\in S(Y)_{(x_i)}$ can be written $f=g_i/x_i^N$ with $g_i$ having degree $N,$ where $S(Y)$ is the homogeneous coordinate ring mentioned on that page; $g_i$ and $x_i^N$ both have degree $N,$ meaning $f$ has degree zero.

If you have read a book like Shafarevich before this, you may have seen this localization as "dehomogenization" of a homogeneous polynomial, and it will help to keep this in mind. For example, to dehomogenize $f(x,y,z)=y^2z+xyz$ with respect to $x,$ we divide by $x^3$ since it has degree $3,$ getting $$\dfrac{f(x,y,z)}{x^3}=\dfrac{y^2z+xyz}{x^3}=(\dfrac{y}{x})^2(\dfrac{z}{x})+(\dfrac{y}{x})(\dfrac{z}{x}).$$

We consider $\dfrac{y}{x},\dfrac{z}{x}$ as coordinates on $\Bbb A_x^2$, where $\Bbb A_x^2$ is a standard open subset of $\Bbb P^2,$ where the original variety lived, and then this dehomogenization of $f$ with respect to $x$ cuts out locally the projective variety in this open chart. (Note that the dehomogenization need not be homogeneous. We have an affine variety here!)

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Regarding your comment about $f = g_i / x_i^N$ i can see why $g_i \in S(Y)$ must have degree $N$. But why must it be homogeneous? –  Manos Oct 9 '12 at 15:33
    
Dear @Manos, if $g_i$ were not homogeneous, then $g_i/x_i^N$ would also not be homogeneous, and in particular, would not be homogeneous of degree $0,$ as was assumed. –  Andrew Oct 9 '12 at 19:20

As far as I can tell, the notation $k[x_0,\ldots,x_n]_{(x_i)}$ means the degree zero part of the localization of the polynomial ring at (the multiplicative set generated by) $x_i$, whereas Hartshorne would write $k[x_0,\ldots,x_n]_{((x_i))}$ to mean elements of degree zero in the localization by the ideal $(x_1)$. See how he writes $k[x_0,\ldots,x_n]_{((0))}$, for example.

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