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If $B$ is a graded ring, then for me is clear that $\operatorname{Proj}B$ with affine covering given by $D_+{(f)}\cong\operatorname{Spec} B_{(f)}$ is a scheme. The problem arises when $B$ is a graded $A$-algebra, because in this case $\operatorname{Proj}B$ should be an $A$-scheme:

clearly we have a morphism of rings $A\longrightarrow B_{(f)}$ that induces a morphism of affine schemes ${\operatorname{Spec}{B}}_{(f)}\longrightarrow\operatorname{Spec} A$. Now, to get the structure of $A$-scheme, we must define a morphism from $\operatorname{Proj}B$ to $\operatorname{Spec} A$, and the simplest thing to do is glueing the above morphisms of affine schemes. But nobody ensures that this glueing is possible; to be precise my question is the following: why do the affine morphisms match on the intersections?

Many thanks in advance.

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Let $f,g \in B$ be homogeneous of degree $n$ resp. $m$. We have to check that the diagram of morphisms of affine schemes $$\begin{array}{c} \mathrm{Spec}(B_{(f)}) \cap \mathrm{Spec}(B_{(g)}) & \rightarrow & \mathrm{Spec}(B_{(g)}) \\ \downarrow && \downarrow \\ \mathrm{Spec}(B_{(f)}) & \rightarrow & \mathrm{Spec}(A)\end{array}$$ commutes. This corresponds to the diagram of commutative rings $$\begin{array}{c}(B_{(f)})_{\frac{g^n}{f^m}} \cong (B_{(g)})_{\frac{f^m}{g^n}}& \leftarrow & B_{(g)} \\ \uparrow && \uparrow \\ B_{(f)} & \leftarrow & A. \end{array}$$ The isomorphism here comes from the gluing construction of $\mathrm{Proj}(B)$, it maps $$\dfrac{b}{f^k} \mapsto \dfrac{b f^{mk-k}}{g^{nk}} \cdot (\dfrac{g^n}{f^m})^k$$ Now if $a \in A$, we also denote by $a$ the image in $B$, which is homogeneous of degree $0$. The image in $B_{(f)}$ is therefore $\frac{a}{1}$, the image in the localization $(B_{(f)})_{\frac{g^n}{f^m}}$ is denoted the same, and the image in $(B_{(g)})_{\frac{f^m}{g^n}}$ is then (take $k=0$ above) also $\frac{a}{1}$. When we map through $B_{(g)}$ we get the same.

Here is a more conceptual answer: Given a gluing data of schemes $(X_i,X_{ij} \hookrightarrow X_i,\psi_{ij} : X_{ij} \cong X_{ji})$ with gluing $X$. Assume that $S$ is a scheme, $X_i$ is an $S$-scheme and the $\psi_{ij}$ are $S$-isomorphisms. Then also $X$ is an $S$-scheme in such a way that each $X_i \hookrightarrow X$ is an $S$-morphism. This can be applied to $X=\mathrm{Proj}(B)$, $X_f = \mathrm{Spec}(B_{(f)})$, $S=\mathrm{Spec}(A)$.

An even more conceptual answer involves the functor of points of $\mathrm{Proj}(B)$. Then you don't have to go into the messy gluing construction at all.

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