In Proposition II.2.5b of Algebraic Geometry, Hartshorne claims that for any graded ring $S$, and for any homogeneous element $f\in S_+$, we have an isomorphism of locally ringed spaces $$\varphi:\left(D_+(f),\mathcal{O}_{\mathrm{Proj\,}S}\vert_{D_+(f)}\right)\to\mathrm{Spec\,}S_{(f)}.$$

My question is about his claim that $\varphi$ acts on topological spaces as a bijection. Note that he defines, for $\mathfrak{p}\in D_+(f)\subset\mathrm{Proj\,}S$, $$\varphi(\mathfrak{p}) = S_f\mathfrak{p}\cap S_{(f)}$$ which is prime by basic commutative algebra, and it is relatively easy to check that it is injective as well. But how do we show surjectivity? This is my work so far:

For $\mathfrak{q}\in\mathrm{Spec\,}S_{(f)}$, we let $\mathfrak{p} = \sqrt{S_f\mathfrak{q}}$, and provided $\mathfrak{p}$ is a homogeneous prime ideal such that $\mathfrak{p}\cap S_{(f)} = \mathfrak{q}$, then this gives us the necessary element of $D_+(f)$.

It is easy to see that $\mathfrak{p}$ must be homogeneous and that $\mathfrak{p}\cap S_{(f)} = \mathfrak{q}$ so that $\mathfrak{p}\ne S_f$. However, I'm struggling to show that $\mathfrak{p}$ is prime. What is the correct approach here?

I'm not sure if StackOverflow or StackExchange is the better place to ask this question, but it seems to be relatively technical for StackExchange, so I'm asking it here.

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    $\begingroup$ See stacks.math.columbia.edu/tag/00JO $\endgroup$ Jan 24, 2017 at 5:30
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    $\begingroup$ I think that questions about understanding stuff in the early chapters of Hartshorne might belong at MSE. $\endgroup$
    – Kevin Buzzard
    Jan 24, 2017 at 8:13

2 Answers 2


It seems this proposition causes quite some trouble for a number of people. Also I think in the literature this statement is all in all treated too concisely. Let me try to give a precise and hopefully complete proof of the surjectivity.

Define $\Psi(\mathfrak{q})$ as the nilradical of the ideal $I$ generated by $X:= \cup_{d\in\mathbb{N}}\{s\in S_{d}\;\vert\;\exists n\in\mathbb{N}:\frac{s}{f^{n}}\in\mathfrak{q}\}$.

Clearly, I is homogeneous, as it is generated by homogeneous elements. Furthermore, the nilradical of a homogeneous ideal is homogeneous. Let $J\subseteq S$ any homogeneous ideal, then we need to show that $\sqrt{J}=\cup_{d\in\mathbb{N}}(S_{d}\cap \sqrt{J})$. If will only show that the left hand side is contained in the right hand side. Pick an element $s\in \sqrt{J}$ and write it as a sum $s_{1}+\dots + s_{n}$ of homogeneous elements in $S$ of strictily ascending degrees. Since there is a $r\in\mathbb{N}$ such that $s^{r}\in J$ we can uniquely write $s^{r}$ as a sum of homogeneous elements of $J$ (here the homogeneity of $J$ comes in) of strictly ascending degrees, and the lowest summand has the form $s_{1}^{r}\in J$, so in particular, $s_{1}\in\sqrt{J}$. Now subtract $s_{1}$ from $s$ and go in inductively so see that each of the summand lie in $\sqrt{J}$.

To show that $\Psi(\mathfrak{q})$ is prime it is enough to show that for all homogeneous elements $s,t\in S$ the relation $st\in \Psi(\mathfrak{q})$ entails $s\in\Psi(\mathfrak{q})$ or $t\in \Psi(\mathfrak{q})$.

So let $st\in \Psi(\mathfrak{q})$. By definiton there is some $r\in\mathbb{N}_{\geq 1}$ such that $s^{r}t^{r}=\sum_{i=0}^{m}a_{i}b_{i}$ so that for each of the $b_{i}$ there exists some $n_{i}$ such that $\frac{b_{i}}{f^{n_{i}}}$ lies in $\mathfrak{q}$. We can choose some $p$ high enough so that $(st)^{p}$ lies in $X$. Namely, if at least one of the $n_{i}$ is larger that zero, set $p:=\deg(f)\cdot r\cdot \sum_{i=1}^{m}n_{i}$ and $n=r\cdot \sum_{i=1}^{m}n_{i}\cdot(deg(s)+deg(t))$. Then $\frac{s^{p}t^{p}}{f^{n}} = \frac{\left(\sum_{i=1}^{m}a_{i}b_{i}\right)^{p/r}}{f^{n}}$ lies in $\mathfrak{q}$.

If all of the $n_{i}$ are equal to zero, then set $p:=\deg(f)\cdot r$ and $n:=(deg(s)+deg(t))r$. Then $\frac{s^{p}t^{p}}{f^{n}}$ lies in $\mathfrak{q}$.

In both cases we can write $\frac{(st)^{p}}{f^{n}}=\frac{s^{p}}{f^{\overline{n_{1}}}}\cdot\frac{s^{p}}{f^{\overline{n_{2}}}}$ as an equation in $S_{(f)}$, so one of the factors of the product must lie in $\mathfrak{q}$, so it follows by definition that $s$ or $t$ must lie in $\Psi(\mathfrak{q})$.

This shows that $\Psi$ is well defined.

To see that $\Phi\Psi=id_{Spec S_{(f)}}$ I pick an element in $\Phi(\Psi(\mathfrak{q}))$, which is of the form $\sum_{i=1}^{q}d_{i}e_{i}$ for suitable $d_{i}\in S_{f}$ and $e_{i}\in\Psi(\mathfrak{q})$ for all $1\leq i\leq q$. Similar to a previous argument, I can take a power of this sum which must lie in $\mathfrak{q}$ (namely $p:=\sum_{i=1}^{q}p_{i}$ where each $e_{i}$ satisfies $\frac{e_{i}^{p_{i}}}{f^{n_{i}}}\in\mathfrak{q}$ for some $n_{i}$). So it lies in $\mathfrak{q}$ itself. On The other hand, of al element $s$ lies in $\mathfrak{q}$, then picking an element $\overline{s}$ and some $n$ such that $\frac{\overline{s}}{f^{n}}=s$ shows directly that $s\in\Phi(\Psi(\mathfrak{q}))$.


Given $\mathfrak{q}$ define

$$\mathfrak{p}=\{g\in S|\exists n \frac{g}{f^n}\in \mathfrak{q}\}$$ note that since the elements of $S_{(f)}$ have degree zero the value of $n$ above is determined by $g$. To show that $\mathfrak{p}$ is prime assume that $gh\in \mathfrak{p}$ then $$\frac{gh}{f^{n+m}}\in \mathfrak{q}$$ and thus $$\frac{g}{f^{n}} \frac{h}{f^{m}} \in \mathfrak{q}$$ and since $\mathfrak{q}$ is prime we have say $\frac{g}{f^{n}} \in \mathfrak{q}$ and thus $g\in \mathfrak{p}$.

Finally we must show $$\mathfrak{q}=\mathfrak{p}S_f\cap S_{(f)}$$ if say $\frac{g}{f^n}\in \mathfrak{q}$ then

$$g\frac{1}{f^n}\in \mathfrak{p}S_f\cap S_{(f)}.$$ On the other hand if $g\in \mathfrak{p}$ and $\frac{h}{f^{n+m}}\in S_f$ is such that $$g\frac{h}{f^{n+m}}\in S_{(f)}$$

then $\frac{h}{f^m} \frac{g}{f^n}\in \mathfrak{q}$ since $\frac{g}{f^n}\in \mathfrak{q}$

  • $\begingroup$ Why is $\mathfrak{p}$ an ideal? $\endgroup$
    – user150248
    Jul 8, 2018 at 3:20
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    $\begingroup$ You seem to implicitely use that if $\frac{gh}{f^{N}}\in\mathfrak{q}\subseteq S_{(f)}$ for some $N\in\mathbb{N}$, then there exist natural numbers $n,m$ such that $\frac{g}{f^{n}}$ and $\frac{h}{f^{m}}\in S_{(f)}$. This sounds incorrect to me as it implies that if the sum of two numbers $a$ and $b$ (the degrees of $g$ and $h$) are divided by a third number $d>0$ (the degree of $f$), then both of $a$ and $b$ are divided by $d$. $\endgroup$
    – user520682
    Jul 8, 2021 at 14:04
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    $\begingroup$ More concretely: If $S=\mathbb{C}[x,y], f=xy,g=h=x$, then $\frac{gh}{f}$ is a non-invertible element in $S_{(f)}$ so it lies in some prime ideal $\mathfrak{q}$ of $S_{(f)}$, but there exist no $n,m\in\mathbb{N}$ so that $\frac{g}{f^{n}}$ or $\frac{h}{f^{m}}$ even lie in $S_{(f)}$. $\endgroup$
    – user520682
    Jul 8, 2021 at 14:10

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