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let $B=\oplus_{i\geq0}B_i$ be a graded ring with $B_d=B_1^d$ for every $d\geq1$. Suppose $B_1$ is a finitely generated $A$-module for some ring $A$. Then, is $B$ an $A$-algebra in some canonical way?

In Qing Liu's Algebraic Geometry and Arithmetic Curves, Ex.2.3.11(b), he said in the above settings the scheme $Proj(B)$ was a projective $A$-scheme, I wondered if this is a printing error, should he replace $A$ by $B_0$?

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    I don't understand what you mean when you say that $B_1$ is a finite type $A$-algebra. _A priori_ $B_1$ is just a $B_0$-module. What's the multiplication on $B_1$ and what does it have to do with the multiplication on $B$?2012-06-09
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    @QiaochuYuan sorry, it should be "finitely generated $A$-module", I edited.2012-06-09
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    That's curious. I just took a look and the exercise doesn't seem to specify any compatibility between the $B_0$-module structure and the $A$-module structure. Does the exercise work if you assume that $A$ is a subring of $B_0$? (Well, in any case, Qing Liu is a user here so presumably he'll clear things up soon.)2012-06-09
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    @QiaochuYuan I don't know, I can only handle the case $A=B_0$. I also checked his errata on website, but get no luck.2012-06-09

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I think the author just forgot to add this kind of assumption. Otherwise it is even unclear that $B_d$ has a structure of $A$-module. So you can suppose $B_0$ is an $A$-algebra and $B_1$ is finitely generated over $A$.

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    Thank you! Then, by looking at $D_+(f_i)$, where $\{f_i\}_i$ generate $B_1$ over $A$, we get $Proj(A\oplus B_1\oplus...)=Proj(B)$, and we thus conclude the exercise by embeding in $Proj(A[T_i]_i)$, right?2012-06-10
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    @M.N.Yes, this is correct.2012-06-10