1
$\begingroup$

Let $S$ be a graded ring. Let $f\in S$. Then we can form a multiplicative set $Q=\{f,f^2,f^3,\dots\}$ and then take the localization $Q^{-1}S=S_{f}$. If I talk about the degree $0$ elements of this localization, then do I mean that $a/b\in S_{(f)}$ and $\text{deg}(a)-\text{deg}(b)=0$. Furthermore, by the degree, do we mean homogeneous degree?

I imagine that I do require the homogeneous condition, otherwise I run into problems. I.e. say that I do not require this to be the homogeneous degree and instead only require the numerator and denominator to have the same degree of the maximal degree monomial. Then consider $\Bbb{C}[x_0,x_1]_{(x_0)}.$ This would force me to permit things like $\frac{x_0^2+x_1+c}{x_0^2}=1+\frac{x_1}{x_2^2}+ \frac{c}{x_0^2}$. Since this doesn't look like what I expect degree $0$ to look like, this makes me unhappy. $$$$ But if I consider only homogeneous elements in the numerator: $$\frac{ax_0^3+bx_0^2x_1+cx_0x_1^2+dx_1^3}{x_0^3}=a+b\frac{x_1}{x_0}+c\frac{x_1^2}{x_0^2} + d\frac{x_1^3}{x_0^3},$$and the degrees all make me happy. (This of course would have held for degree $n$ on numerator)

Am I right in thinking the latter is indeed what we desire?

$\endgroup$
1
  • $\begingroup$ That's correct, f,a,b should be homogeneous. $\endgroup$
    – Youngsu
    Oct 3, 2017 at 18:07

1 Answer 1

0
$\begingroup$

Yes, by definition $S_{(f)}$ consists of fractions $\frac{a}{b}$ where $a \in S$ and $b \in Q$ are homogeneous of equal degree. See, for example, the parenthetical remark following the first displayed equation in Proj as a scheme. Or, better yet, any modern text on algebraic geometry or scheme theory like Hartshorne, or maybe Daniel Murfets notes here.

$\endgroup$
2
  • $\begingroup$ I think there is a problem. Suppose that the graded ring $S$ is not a domain, and $fy=0$ for some non-zero element $y$. Then $\frac{x+y}{f^n}=\frac{x}{f^n}\in S_f$ for each $x\in S$. When $\text{deg}(x)=n\text{deg}(f)\ne \text{deg}(y)$, is the above element included in $S_{(f)}$? $\endgroup$
    – user150248
    Jul 7, 2018 at 4:25
  • 1
    $\begingroup$ If $\deg x = \deg f^n$ then yes, $\frac{x}{f^n}$ is in $S_{(f)}$. You've picked up on the following subtlety: For an element of the localization $S_f$ to be in $S_{(f)}$ it needs to be represented as a quotient of homogeneous elements, but that doesn't need to be its only representation. $\endgroup$
    – Jim
    Jul 7, 2018 at 12:38

You must log in to answer this question.

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