Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question
up vote 4 down vote accepted

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!)

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.