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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!

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    $\begingroup$ 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 $A$ ?... $\endgroup$ – Olórin Mar 22 '15 at 0:43
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    $\begingroup$ 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 $f: X \to Spec A.$ $\endgroup$ – Krish Mar 22 '15 at 3:49
  • $\begingroup$ @Krish There's an explicit formula. $\endgroup$ – Olórin Mar 22 '15 at 11:04
  • $\begingroup$ @Krish Could you possibly explain a little more on how $\phi$ is defined? I am little bit confused on how it is defined... $\endgroup$ – user211392 Mar 23 '15 at 19:36
  • $\begingroup$ @user10000100_u or could you possibly tell me what the explicit formula is? $\endgroup$ – user211392 Mar 23 '15 at 19:37
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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 $A$.

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  • $\begingroup$ What is the ring morphism from $A$ to $B$ that corresponds to the morphism Spec $B$ to Spec $A$? $\endgroup$ – user211392 Mar 22 '15 at 1:02
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    $\begingroup$ Also, could you possibly explain me why the maps are compatible on the overlaps? Thanks! $\endgroup$ – user211392 Mar 22 '15 at 1:08

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