The Stacks project

Proof. First we show that the map is a bijection by constructing an inverse. Let $f \in S_ d$, $d > 0$ be invertible in $S$. If $\mathfrak p_0$ is a prime of $S_0$, then $\mathfrak p_0S$ is a $\mathbf{Z}$-graded ideal of $S$ such that $\mathfrak p_0S \cap S_0 = \mathfrak p_0$. And if $ab \in \mathfrak p_0S$ with $a$, $b$ homogeneous, then $a^ db^ d/f^{\deg (a) + \deg (b)} \in \mathfrak p_0$. Thus either $a^ d/f^{\deg (a)} \in \mathfrak p_0$ or $b^ d/f^{\deg (b)} \in \mathfrak p_0$, in other words either $a^ d \in \mathfrak p_0S$ or $b^ d \in \mathfrak p_0S$. It follows that $\sqrt{\mathfrak p_0S}$ is a $\mathbf{Z}$-graded prime ideal of $S$ whose intersection with $S_0$ is $\mathfrak p_0$.

To show that the map is a homeomorphism we show that the image of $G \cap D(g)$ is open. If $g = \sum g_ i$ with $g_ i \in S_ i$, then by the above $G \cap D(g)$ maps onto the set $\bigcup D(g_ i^ d/f^ i)$ which is open. $\square$

Proof. Since $D_{+}(f) = \text{Proj}(S) \cap D(f)$, these sets are open. This proves (1). Also (2) follows as $D(ff') = D(f) \cap D(f')$. Similarly the sets $V_{+}(I) = \text{Proj}(S) \cap V(I)$ are closed. This proves (8).

Suppose that $T \subset \text{Proj}(S)$ is closed. Then we can write $T = \text{Proj}(S) \cap V(J)$ for some ideal $J \subset S$. By definition of a homogeneous ideal if $g \in J$, $g = g_0 + \ldots + g_ m$ with $g_ d \in S_ d$ then $g_ d \in \mathfrak p$ for all $\mathfrak p \in T$. Thus, letting $I \subset S$ be the ideal generated by the homogeneous parts of the elements of $J$ we have $T = V_{+}(I)$. This proves (9).

The formula for $\text{Proj}(S) \cap D(g)$, with $g \in S$ is direct from the definitions. This proves (3). Consider the formula for $\text{Proj}(S) \cap D(g_0)$. The inclusion of the right hand side in the left hand side is obvious. For the other inclusion, suppose $g_0 \not\in \mathfrak p$ with $\mathfrak p \in \text{Proj}(S)$. If all $g_0f \in \mathfrak p$ for all homogeneous $f$ of positive degree, then we see that $S_{+} \subset \mathfrak p$ which is a contradiction. This gives the other inclusion. This proves (4).

The collection of opens $D(g) \cap \text{Proj}(S)$ forms a basis for the topology since the standard opens $D(g) \subset \mathop{\mathrm{Spec}}(S)$ form a basis for the topology on $\mathop{\mathrm{Spec}}(S)$. By the formulas above we can express $D(g) \cap \text{Proj}(S)$ as a union of opens $D_{+}(f)$. Hence the collection of opens $D_{+}(f)$ forms a basis for the topology also. This proves (5).

Proof of (6). First we note that $D_{+}(f)$ may be identified with a subset (with induced topology) of $D(f) = \mathop{\mathrm{Spec}}(S_ f)$ via Lemma 10.17.6. Note that the ring $S_ f$ has a $\mathbf{Z}$-grading. The homogeneous elements are of the form $r/f^ n$ with $r \in S$ homogeneous and have degree $\deg (r/f^ n) = \deg (r) - n\deg (f)$. The subset $D_{+}(f)$ corresponds exactly to those prime ideals $\mathfrak p \subset S_ f$ which are $\mathbf{Z}$-graded ideals (i.e., generated by homogeneous elements). Hence we have to show that the set of $\mathbf{Z}$-graded prime ideals of $S_ f$ maps homeomorphically to $\mathop{\mathrm{Spec}}(S_{(f)})$. This follows from Lemma 10.57.2.

Let $S = \mathbf{Z}[X_1, X_2, X_3, \ldots ]$ with grading such that each $X_ i$ has degree $1$. Then it is easy to see that

does not have a finite refinement. This proves (7).

Let $I \subset S$ be a graded ideal. If $\sqrt{I} \supset S_{+}$ then $V_{+}(I) = \emptyset$ since every prime $\mathfrak p \in \text{Proj}(S)$ does not contain $S_{+}$ by definition. Conversely, suppose that $S_{+} \not\subset \sqrt{I}$. Then we can find an element $f \in S_{+}$ such that $f$ is not nilpotent modulo $I$. Clearly this means that one of the homogeneous parts of $f$ is not nilpotent modulo $I$, in other words we may (and do) assume that $f$ is homogeneous. This implies that $I S_ f \not= S_ f$, in other words that $(S/I)_ f$ is not zero. Hence $(S/I)_{(f)} \not= 0$ since it is a ring which maps into $(S/I)_ f$. Pick a prime $\mathfrak q \subset (S/I)_{(f)}$. This corresponds to a graded prime of $S/I$, not containing the irrelevant ideal $(S/I)_{+}$. And this in turn corresponds to a graded prime ideal $\mathfrak p$ of $S$, containing $I$ but not containing $S_{+}$ as desired. This proves (10) and finishes the proof. $\square$

Proof. We define a map $\psi : M_{(\mathfrak p)} \to (M_{(f)})_{\mathfrak p'}$. Let $x/g \in M_{(\mathfrak p)}$. We set

This makes sense since $\deg (x) = \deg (g)$ and since $g^{\deg (f)}/f^{\deg (g)} \not\in \mathfrak p'$. We omit the verification that $\psi$ is well defined, a module map and an isomorphism. Hint: the inverse sends $(x/f^ n)/(g/f^ m)$ to $(xf^ m)/(g f^ n)$. $\square$

Proof. We may assume there are no inclusions among the $\mathfrak p_ i$. The result is true for $r = 1$. Suppose the result holds for $r - 1$. Pick $x \in I$ homogeneous of positive degree such that $x \not\in \mathfrak p_ i$ for all $i = 1, \ldots , r - 1$. If $x \not\in \mathfrak p_ r$ we are done. So assume $x \in \mathfrak p_ r$. If $I \mathfrak p_1 \ldots \mathfrak p_{r-1} \subset \mathfrak p_ r$ then $I \subset \mathfrak p_ r$ a contradiction. Pick $y \in I\mathfrak p_1 \ldots \mathfrak p_{r-1}$ homogeneous and $y \not\in \mathfrak p_ r$. Then $x^{\deg (y)} + y^{\deg (x)}$ works. $\square$

Proof. Suppose $f, g \in S$ are such that $fg \in \mathfrak q$. Let $f_ d$ (resp. $g_ e$) be the homogeneous part of $f$ (resp. $g$) of degree $d$ (resp. $e$). Assume $d, e$ are maxima such that $f_ d \not= 0$ and $g_ e \not= 0$. By assumption we can write $fg = \sum a_ i f_ i$ with $f_ i \in \mathfrak p$ homogeneous. Say $\deg (f_ i) = d_ i$. Then $f_ d g_ e = \sum a_ i' f_ i$ with $a_ i'$ to homogeneous par of degree $d + e - d_ i$ of $a_ i$ (or $0$ if $d + e -d_ i < 0$). Hence $f_ d \in \mathfrak p$ or $g_ e \in \mathfrak p$. Hence $f_ d \in \mathfrak q$ or $g_ e \in \mathfrak q$. In the first case replace $f$ by $f - f_ d$, in the second case replace $g$ by $g - g_ e$. Then still $fg \in \mathfrak q$ but the discrete invariant $d + e$ has been decreased. Thus we may continue in this fashion until either $f$ or $g$ is zero. This clearly shows that $fg \in \mathfrak q$ implies either $f \in \mathfrak q$ or $g \in \mathfrak q$ as desired. $\square$

Proof. The first assertion holds because the prime $\mathfrak q$ constructed in Lemma 10.57.7 satisfies $\mathfrak q \subset \mathfrak p$. The second because we may consider $S/I$ and apply the first part. $\square$

Proof. Choose $f_1, \ldots , f_ n \in S$ which generate $S$ as an $R$-algebra. We may assume that each $f_ i$ is homogeneous (by decomposing each $f_ i$ into its homogeneous components). An element of $S_{(f)}$ is a sum of the form

with $\lambda _{e_1 \ldots e_ n} \in R$. Thus $S_{(f)}$ is generated as an $R$-algebra by the $f_1^{e_1} \ldots f_ n^{e_ n} /f^ e$ with the property that $e\deg (f) = \sum e_ i\deg (f_ i)$. If $e_ i \geq \deg (f)$ then we can write this as

Thus we only need the elements $f_ i^{\deg (f)}/f^{\deg (f_ i)}$ as well as the elements $f_1^{e_1} \ldots f_ n^{e_ n} /f^ e$ with $e \deg (f) = \sum e_ i \deg (f_ i)$ and $e_ i < \deg (f)$. This is a finite list and we see that (1) is true.

To see (2) suppose that $M$ is generated by homogeneous elements $x_1, \ldots , x_ m$. Then arguing as above we find that $M_{(f)}$ is generated as an $S_{(f)}$-module by the finite list of elements of the form $f_1^{e_1} \ldots f_ n^{e_ n} x_ j /f^ e$ with $e \deg (f) = \sum e_ i \deg (f_ i) + \deg (x_ j)$ and $e_ i < \deg (f)$. $\square$

Proof. We may write $R' = R[x_1, \ldots , x_ n]/I$ for some ideal $I$. For an element $g \in R[x_1, \ldots , x_ n]$ denote $\tilde g \in R[X_0, \ldots , X_ n]$ the element homogeneous of minimal degree such that $g = \tilde g(1, x_1, \ldots , x_ n)$. Let $\tilde I \subset R[X_0, \ldots , X_ n]$ generated by all elements $\tilde g$, $g \in I$. Set $S = R[X_0, \ldots , X_ n]/\tilde I$ and denote $f$ the image of $X_0$ in $S$. By construction we have an isomorphism

To do the same thing with the module $M$ we choose a presentation

with $k_ j = (k_{1j}, \ldots , k_{rj})$. Let $d_{ij} = \deg (\tilde k_{ij})$. Set $d_ j = \max \{ d_{ij}\}$. Set $K_{ij} = X_0^{d_ j - d_{ij}}\tilde k_{ij}$ which is homogeneous of degree $d_ j$. With this notation we set

which works. Some details omitted. $\square$