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.


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.