Let \(S\) be a graded ring. Show that \(\Proj S = \varnothing\) if and only if every element of \(S_{+}\) is nilpotent.
Let \(\phi: S \to T\) be a graded homomorphism of graded rings, preserving degrees. Let \(U = \ts{\mfp \in \Proj T \st \mfp \not\supseteq \phi(S_+)}\). Show that \(U\) is an open subset of \(\Proj T\), and show that \(\phi\) determines a natural morphism \(f: U \to \Proj S\).
The morphism \(f\) can be an isomorphism even when \(\phi\) is not. For example, suppose that \(\phi_d: S_d \to T_d\) is an isomorphism for all \(d \geq d_0\), where \(d_0\) is an integer. Then show that \(U = \Proj T\) and the morphism \(f: \Proj T \to \Proj S\) is an isomorphism.
Let \(V\) be a projective variety with homogeneous coordinate ring \(S\). Show that \(t(V) \cong \Proj S\).
Solution
Write \[ S_+=\bigoplus_{d>0}S_d \] for the irrelevant ideal.
1. If every element of \(S_+\) is nilpotent, then\[ \Proj S=\varnothing. \]
Every prime ideal contains every nilpotent element. Hence every homogeneous prime ideal contains all of \(S_+\). But by definition \[ \Proj S =\{\mathfrak p:\mathfrak p\text{ homogeneous prime and }\mathfrak p\not\supseteq S_+\}. \] Thus no prime belongs to \(\Proj S\).
2. Conversely, if\[ \Proj S=\varnothing, \]then every homogeneous element of positive degree is nilpotent.
Let \[ f\in S_d, \qquad d>0, \] be homogeneous. Suppose \(f\) is not nilpotent.
Consider the set of homogeneous ideals \(I\subseteq S\) such that \[ I\cap\{1,f,f^2,\ldots\}=\varnothing. \] It is nonempty because \((0)\) belongs to it, and the union of any chain is again a homogeneous ideal disjoint from the powers of \(f\). By Zorn’s lemma there is a maximal such homogeneous ideal \(\mathfrak p\).
We claim \(\mathfrak p\) is prime. Let \(a,b\) be homogeneous elements not in \(\mathfrak p\). By maximality, the homogeneous ideals \[ \mathfrak p+(a) \qquad\text{and}\qquad \mathfrak p+(b) \] each meet the powers of \(f\). Thus for some \(m,n\), \[ f^m=p_1+ra, \qquad f^n=p_2+sb \] with \(p_1,p_2\in\mathfrak p\). Multiplying gives \[ f^{m+n} \in \mathfrak p+(ab). \] If \(ab\in\mathfrak p\), this would put \(f^{m+n}\) in \(\mathfrak p\), contradiction. Hence \[ ab\notin\mathfrak p. \] So the complement of \(\mathfrak p\) is multiplicatively closed on homogeneous elements, which is the homogeneous-prime criterion.
Since \[ f\notin\mathfrak p \] and \(f\in S_+\), the prime \(\mathfrak p\) does not contain \(S_+\). Hence \[ \mathfrak p\in\Proj S, \] contradiction. Therefore every homogeneous positive-degree element is nilpotent.
3. If every homogeneous element of \(S_+\) is nilpotent, then every element of \(S_+\) is nilpotent.
Every element \[ s\in S_+ \] is a finite sum of homogeneous positive-degree components: \[ s=s_{d_1}+\cdots+s_{d_r}. \] Each \(s_{d_i}\) is nilpotent by hypothesis. The nilpotent elements of a commutative ring form the nilradical, which is an ideal, so their finite sum is nilpotent. Hence \(s\) is nilpotent.
4. Therefore\[ \boxed{ \Proj S=\varnothing \iff \text{every element of }S_+\text{ is nilpotent}. } \]
The forward implication is <1>2–<1>3 and the reverse implication is 1.
5. Let\[ \phi:S\longrightarrow T \]be a degree-preserving graded homomorphism and put\[ U = \{\mathfrak p\in\Proj T:\mathfrak p\not\supseteq\phi(S_+)\}. \]Then\[ \boxed{ U = \bigcup_{\substack{s\in S_+\\s\text{ homogeneous}}} D_+(\phi(s)), } \]so \(U\) is open in \(\Proj T\).
A homogeneous prime \(\mathfrak p\in\Proj T\) fails to contain the ideal \(\phi(S_+)\) exactly when there is some homogeneous positive-degree element \[ s\in S_+ \] with \[ \phi(s)\notin\mathfrak p. \] That is exactly the condition \[ \mathfrak p\in D_+(\phi(s)). \] Taking the union over all such \(s\) gives the formula.
6. For \(\mathfrak p\in U\), define\[ f(\mathfrak p)=\phi^{-1}(\mathfrak p). \]Then\[ f(\mathfrak p)\in\Proj S. \]
The inverse image of a prime ideal under a ring homomorphism is prime, and because \(\phi\) preserves degrees, the inverse image of a homogeneous ideal is homogeneous. Thus \[ \phi^{-1}(\mathfrak p) \] is a homogeneous prime ideal of \(S\).
Since \(\mathfrak p\in U\), some \[ s\in S_+ \] satisfies \[ \phi(s)\notin\mathfrak p. \] Hence \[ s\notin\phi^{-1}(\mathfrak p), \] so the inverse-image prime does not contain \(S_+\). Therefore it belongs to \(\Proj S\).
7. The map of points\[ f:U\longrightarrow\Proj S \]is continuous, and for homogeneous \(s\in S_+\),\[ \boxed{ f^{-1}(D_+(s)) =D_+(\phi(s)). } \]
For \(\mathfrak p\in U\), \[ \begin{aligned} f(\mathfrak p)\in D_+(s) &\iff s\notin\phi^{-1}(\mathfrak p)\\ &\iff \phi(s)\notin\mathfrak p\\ &\iff \mathfrak p\in D_+(\phi(s)). \end{aligned} \] The distinguished opens form a basis of \(\Proj S\), so this identity proves continuity.
8. On the distinguished opens from 7, the homomorphism \(\phi\) induces degree-zero localization maps\[ \boxed{ S_{(s)} \longrightarrow T_{(\phi(s))}. } \]
Because \(\phi\) is graded and \[ \phi(s) \] is inverted in \(T_{\phi(s)}\), the universal property of localization gives a graded homomorphism \[ S_s\longrightarrow T_{\phi(s)}. \] It preserves degrees, so it restricts to the degree-zero parts \[ (S_s)_0=S_{(s)} \longrightarrow (T_{\phi(s)})_0=T_{(\phi(s))}. \]
9. The affine morphisms\[ D_+(\phi(s)) =\Spec T_{(\phi(s))} \longrightarrow \Spec S_{(s)} =D_+(s) \]from 8 agree on overlaps and glue to a morphism of schemes\[ \boxed{ f:U\longrightarrow\Proj S. } \]
8. On points it is contraction of primes, hence agrees with the set map in
6.
If \(s,t\in S_+\) are homogeneous, then on the overlap \[ D_+(\phi(s))\cap D_+(\phi(t)) =D_+(\phi(st)) \] both local morphisms are obtained from the same graded homomorphism \(\phi\) after localizing further by \(st\). Hence they agree. Since the opens \(D_+(\phi(s))\) cover \(U\) by 5, they glue uniquely to the desired scheme morphism.
10. Suppose now that there is an integer \(d_0\) such that\[ \phi_d:S_d\xrightarrow{\sim}T_d \]for every \(d\ge d_0\).Then\[ \boxed{U=\Proj T.} \]
Suppose \[ \mathfrak p\in\Proj T \] contained \(\phi(S_+)\). Then for every \[ d\ge\max(d_0,1), \] the isomorphism \[ \phi_d:S_d\to T_d \] would give \[ T_d=\phi(S_d)\subseteq\mathfrak p. \]
Let \[ t\in T_n, \qquad n>0, \] be homogeneous. Choose \(m\) large enough that \[ mn\ge d_0. \] Then \[ t^m\in T_{mn}\subseteq\mathfrak p. \] Since \(\mathfrak p\) is prime, \[ t\in\mathfrak p. \] Thus \[ T_+\subseteq\mathfrak p, \] contradicting \[ \mathfrak p\in\Proj T. \] Therefore no point of \(\Proj T\) is excluded from \(U\).
11. Let \(s\in S\) be homogeneous of positive degree and assume\[ \deg s\ge d_0. \]Then the map\[ \boxed{ S_{(s)} \xrightarrow{\sim} T_{(\phi(s))} } \]from 8 is an isomorphism.
Put \[ r=\deg s. \]
For surjectivity, take a homogeneous degree-zero fraction \[ \frac{t}{\phi(s)^n} \in T_{(\phi(s))}, \qquad t\in T_{nr}. \] Choose \(N\ge0\) such that \[ (n+N)r\ge d_0. \] Then \[ t\phi(s)^N \in T_{(n+N)r}. \] Since \(\phi_{(n+N)r}\) is surjective, there exists \[ a\in S_{(n+N)r} \] with \[ \phi(a)=t\phi(s)^N. \] Hence \[ \frac{t}{\phi(s)^n} = \frac{\phi(a)}{\phi(s)^{n+N}}, \] the image of \[ \frac a{s^{n+N}}\in S_{(s)}. \]
For injectivity, suppose \[ \frac a{s^n}\in S_{(s)} \] maps to zero. Then for some \(m\ge0\), \[ \phi(s)^m\phi(a)=0 \] in \(T\). Choose \(N\) so large that \[ \deg(s^{m+N}a)\ge d_0. \] Then \[ \phi(s^{m+N}a)=0. \] Since \(\phi\) is injective in that degree, \[ s^{m+N}a=0. \] Thus \[ \frac a{s^n}=0 \] in the localization. Hence the map is injective.
12. The opens\[ D_+(s), \qquad s\in S_+\text{ homogeneous}, \quad \deg s\ge d_0, \]cover \(\Proj S\).
Let \[ \mathfrak q\in\Proj S. \] Choose a homogeneous positive-degree element \[ a\in S_+ \] with \[ a\notin\mathfrak q. \] Some power \[ s=a^N \] has degree at least \(d_0\). Since \(\mathfrak q\) is prime and \(a\notin\mathfrak q\), \[ s\notin\mathfrak q. \] Thus \[ \mathfrak q\in D_+(s). \]
13. Under the eventual degreewise-isomorphism hypothesis, the morphism\[ f:\Proj T\longrightarrow\Proj S \]is an isomorphism.
By 10 the domain \(U\) is all of \(\Proj T\).
7. On every such pair, 11 shows that the defining ring map\[ S_{(s)}\to T_{(\phi(s))} \]is an isomorphism.Hence\[ D_+(\phi(s)) \xrightarrow{\sim} D_+(s) \]is an isomorphism of affine schemes.
Thus \(f\) is an isomorphism on an open cover of the target and the corresponding inverse-image cover of the source. The local inverses agree on overlaps, so they glue to a global inverse. Therefore \(f\) is an isomorphism.
14. Let \(V\subseteq\mathbb P^n_k\) be a projective variety with homogeneous coordinate ring\[ S=k[x_0,\ldots,x_n]/I(V). \]The standard open subsets\[ V_i=V\cap\{x_i\ne0\} \]have affine coordinate rings\[ \boxed{ \Gamma(V_i,\mathcal O_V) \cong S_{(x_i)}. } \]
Dehomogenizing with respect to \(x_i\) identifies the affine variety \(V_i\) with the zero set of the dehomogenized equations of \(I(V)\) in the affine chart \[ \{x_i\ne0\}\cong\mathbb A^n. \] Its coordinate ring is precisely the degree-zero part of the localization \[ S_{x_i}, \] namely \(S_{(x_i)}\).
15. The standard affine opens of \(\Proj S\) are\[ D_+(x_i) \cong \Spec S_{(x_i)}. \]Under the affine identification of 14, they are exactly the schemes associated to the affine varieties \(V_i\).
The standard theorem on \(\Proj\) gives \[ D_+(x_i)=\Spec S_{(x_i)}. \] Step 14 gives the same affine coordinate ring for \(V_i\). Hence the scheme associated to \(V_i\) is naturally isomorphic to this standard open of \(\Proj S\).
16. These local isomorphisms agree on overlaps, and therefore\[ \boxed{t(V)\cong\Proj S.} \]
On \[ V_i\cap V_j, \] the transition from one affine chart to the other is obtained by further localizing the homogeneous coordinate ring by \(x_ix_j\) and taking degree zero. The same localization is the overlap \[ D_+(x_i)\cap D_+(x_j) =D_+(x_ix_j) \] inside \(\Proj S\).
Thus the affine identifications from 15 commute with the overlap identifications. The associated scheme \(t(V)\) is obtained by gluing the affine schemes of the \(V_i\) by these transition maps, while \(\Proj S\) is obtained by gluing the same affine schemes by the same maps. The local isomorphisms therefore glue to the displayed global isomorphism.
17. Q.E.D.
Steps <1>1–<1>4 prove part (a), <1>5–<1>9 prove part (b), <1>10–<1>13 prove part (c), and <1>14–<1>16 prove part (d).