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Let $S$ be a finitely generated graded $A$ algebra, where $A$ is a commutative ring with unity. The exercise says to describe a natural structure morphism from Proj $S$ to Spec $A$. I would appreciate some assistance! Thanks!

user211392
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    I could answer, but I would like to ask you this : what does mean being a finitely generated graded $A$-algebra ? I mean, you could for sure translate this into the existence of some surjective graded morphism $A[T_0,\ldots,T_n]\to S$, isn't it ? Then, knowing the definition of $\textrm{Proj}(S)$, couldn't you come up with a morphism $\textrm{Proj}(S)\to \textrm{Spec}(A)$ ? How would you, to an homogenous prime ideal of $S$ not containing the irrelevant ideal, associate a prime ideal of $A$, knowing that $S$ is a quotient of an homogeneous polynomial ring over $A$ ?... – Olórin Mar 22 '15 at 00:43
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    To define a morphism from $X:=Proj S$ to $Spec A,$ it is enough to define a ring homomorphism from $A$ to $\Gamma (X, \mathcal O_X)$ (R.Hartshorne, Algebraic Geometry, Ch. II, Ex. 2.4). Now define $\phi: A \to \Gamma (X, \mathcal O_X)$ by $\phi (a)(\mathfrak p) = \frac{a}{1} \in S_{(\mathfrak p)},$ the homogeneous localization of $S$ at the prime ideal $\mathfrak p.$ Then $\phi$ is a ring homomorphism and consequesntly we have a morphism $f: X \to Spec A.$ – Krish Mar 22 '15 at 03:49
  • @Krish There's an explicit formula. – Olórin Mar 22 '15 at 11:04
  • @Krish Could you possibly explain a little more on how $\phi$ is defined? I am little bit confused on how it is defined... – user211392 Mar 23 '15 at 19:36
  • @user10000100_u or could you possibly tell me what the explicit formula is? – user211392 Mar 23 '15 at 19:37
  • Here is a link: http://therisingsea.org/notes/TheProjConstruction.pdf – Krish Mar 24 '15 at 06:17

1 Answers1

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If $B$ is an $A$-algebra, then there is a canonical morphism Spec $B \to$ Spec $A$.

Now if $S$ is a graded $A$-algebra, then Proj $S$ is a union of various Spec $B$'s, as $B$ ranges over certain $A$-algebras, constructed from localizations of $S$. Each Spec $B$ has its canonical morphism to Spec $A$, and these are compatible on overlaps, defining the required morphism Proj $S \to$ Spec $A$.

user224221
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