Some sanity checks:

  1. If $S_*$ and $T_*$ are two $A$-algebras, then is $Proj(S_* \otimes_A T_*) \cong Proj(S_*) \otimes_A Proj(T_*)$? Here the tensor product means $(S_* \otimes T_*) = \oplus_{n \geq 0} \oplus_{i + j = k} S_i \otimes_A S_j$. (I know this result for the Segre product of the two graded $A$-algebras: $\oplus_{n \geq 0} S_n \otimes_A T_n$. The sketch of both proofs would be: A map from a scheme $X$ to $Proj(S_* \otimes_A T_*)$ is given by a bunch of compatible $X_{s \otimes t} \to Spec ((S_* \otimes_A T_*)_{s \otimes t})_0$, which is the same as $((S_* \otimes_A T_*)_{s \otimes t})_0 \to \Gamma(O_X, X_{s \otimes t})$. From $((S_* \otimes_A T_*)_{s \otimes t})_0 = (S_*[s^{-1}])_0 \otimes_A (T_*[t^{-1}])_0$ this last map is turned into compatible (over $A$ and also in terms of gluing) maps $(S_*[s^{-1}])_0 \to \Gamma(O_X, X_{s \otimes t})$ and $(T_*[t^{-1}])_0 \to \Gamma(O_X, X_{s \otimes t})$, which turns into maps on the corresponding specs, and glues together to give maps $X \to Proj(S_*)$ and $X \to Proj(T_*)$.
  2. If $L$ is a line bundle on an affine scheme $X = \operatorname{Spec} A$, and $c(L) = \oplus_{n \geq 0} L^{\otimes n}$ (does this ring have a name?) is $Proj_A(c(L)) = Spec(A)$, with structure map the identity? (The sketch of this proof would be: over each trivialization $\operatorname{spec} A_f$, with nowhere vanishing global section $l$, $(c(L)_l)_0 = A_f$ (literally equals I think, or at any rate "natural", and the structure map is the identity. Now everything glues.)

(Similarly, global proj of the $c(O_X)$ on a scheme $X$ should be $X$ again, with structure map the identity - "every scheme is projective over itself". Also, by using global spec with $O_X$, every scheme is affine over itself - this is easier to see, because of course the identity map is an affine map, using the criterion that it pulls back open affines to open affines. But I don't know a similar characterization for projective maps, which right now I am thinking of as maps of the form $Proj B_* \to X$, for $B_*$ a finite type quasicoherent sheaf of $\mathbb{Z}_{\geq 0}$ graded $A$ algebras on $X$. )

I think these are basically trivial to prove if true, and have sketched arguments for both, but it is easy to make mistakes because there are so many details, especially when invoking the phrase "this is natural so it glues"...

I am hoping someone can glance at these statements and tell me if they are true or not - or preferably recommend some reference were I can see them. (They are just technical lemmas that would make some other exercise easier to prove / understand.)


(1) No, this isn't true at all. For instance, if $S=A[s]$ and $T=A[t]$ with $|s|=|t|=1$, then $\operatorname{Proj}(S)=\operatorname{Proj}(T)=\mathbb{P}^0_A=\operatorname{Spec}(A)$ so their product over $\operatorname{Spec}(A)$ is again $\operatorname{Spec}(A)$, but $\operatorname{Proj}(S\otimes T)=\operatorname{Proj}(A[s,t])=\mathbb{P}^1_A$. In terms of your argument, the problem is that (for instance) $((S\otimes T)_{s\otimes t})_0$ contains the element $s^{-1}\otimes t$, which comes not from an element of $(S[s^{-1}])_0\otimes (T[t^{-1}])_0$ but from an element of $(S[s^{-1}])_{-1}\otimes (T[t^{-1}])_1$. So in general, your equation $((S \otimes T)_{s \otimes t})_0 = (S[s^{-1}])_0 \otimes (T[t^{-1}])_0$ is not true; instead, you have the much more complicated $((S \otimes T)_{s \otimes t})_0=\bigoplus_{n\in\mathbb{Z}}(S[s^{-1}])_n \otimes (T[t^{-1}])_{-n}$. For a geometric explanation of why you shouldn't expect this to work, you may be interested in the comments and my answer to this question on MO.

(2) Yes, this is true, and your proof sketch is correct. Geometrically, you should think of $\operatorname{Proj}(c(L))\to\operatorname{Spec}(A)$ as a bundle of projective spaces on $\operatorname{Spec}(A)$, whose fiber over a point is the projectivization of the fiber of the sheaf $L$. So if $L$ is a line bundle, each fiber is $1$-dimensional, so the projectivization should be a point, and so it is reasonable to expect $\operatorname{Proj}(c(L))\to\operatorname{Spec}(A)$ to be an isomorphism.

  • $\begingroup$ Thanks! So, if $\otimes^s$ denotes the Segre product, we do have $((S_* \otimes^s_A T_*)_{s \otimes t})_0 = (S_*[s^{-1}])_0 \otimes_A (T_*[t^{-1}])_0$? $\endgroup$ – Lorenzo Najt Jan 18 '16 at 4:14
  • $\begingroup$ Yes, that's correct. In that case we must have $|s|=|t|$, and any element of $(S\otimes^s T)_{s\otimes t}$ is a sum of terms of the form $(s\otimes t)^{-n} s'\otimes t'$ where $|s'|=|t'|$, so if the degree of such a term is $0$, we have $|s'|=n|s|=n|t|=|t'|$ and it can be written as $s^{-n}s'\otimes t^{-n}t'$, where $s^{-n}s'\in (S[s^{-1}])_0$ and $t^{-n}t'\in (T[t^{-1}])_0$. $\endgroup$ – Eric Wofsey Jan 18 '16 at 4:19

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