Every projective scheme over $A$ is a closed subscheme of $\mathbb P^n_A$

Question

I asked a similar question here, but was quickly told that my hypotheses were not strong enough, so I am asking a new question with the correct hypothesises.

Let $X$ be a projective scheme over $A$, that is $X=\operatorname{Proj}R$, where $R$ is a graded ring with $R_0=A$. We also take $R$ to be finitely generated as an $A$ algebra, i.e. equivalently the irrelevant ideal $R_+$ is a finitely generated ideal. (As secondary question, is there some condition on $X$ as a scheme, which is maybe stronger, that forces $R_+$ to be finitely generated? Quasi-compactness is not strong enough as this only implies the irrelevant ideal is the radical of a finitely generated ideal... perhaps a Noetherian condition on $X$?)

With the above (hopefully strong enough hypotheses) I want to show that there exists a closed embedding $X\hookrightarrow\mathbb P^n_A$ for some $m$. Since $R$ is a finitely generated $A$ algebra, we have that $R\cong A[x_0,\dots, x_n]/I$ for some ideal $I$. If I can show that $I$ is homogenous then we have that $\operatorname{Proj} R=\operatorname{Proj}A[x_0,\dots, x_n]/I\cong \mathbb V(I)\subset \mathbb P^n_A$, and so we are done.

However, I see no reason why $I$ can't be homogenous, and $R$ is equipped with some grading which does not come from $A$. So to some up, my two questions are: is there a reason $I$ should be homogenous? Is there a condition on $X$ that forces the irrelevant ideal to be finitely generated?

Answer 1

If $r_1,\cdots,r_n$ generate $R_+$ as an ideal of $R$, and $r_i=\sum_{j=1}^{m_i} r_{ij}$ for $\deg r_{ij} = j$ is the decomposition in to homogeneous parts, then $r_{11},\cdots,r_{1m_1},\cdots,r_{nm_n}$ form a homogeneous generating set for $R_+$ and so $R_+$ is finitely generated as a homogeneous ideal.

But the fun's not done: given a homogeneous generating set for $R_+$, it may not be the case that all of these generators are in degree 1. Instead, we may need to pass to a Veronese subring $R^{(d)}=\bigoplus_{n\geq 0} R_{dn}$ for some $d$. This can always be done in such a way that $R^{(d)}$ is finitely generated as a graded ring by its degree-one piece (ref) and we also have $\operatorname{Proj} R\cong \operatorname{Proj} R^{(d)}$ (see for instance here). From here, you can choose a graded surjection $A[x_0,\cdots,x_N]\to R^{(d)}$ with all $x_i$ in degree one, and this will get you the closed immersion you're after.

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There's still more fun to be had in addressing your secondary question. In fact, one can show that Proj of a graded ring is only determined by what happens in large enough degrees (see here). So you can't force that $R_+$ is finitely generated - for instance, if I altered some perfectly nice graded ring $R$ with $R_0=A$ by replacing its degree one piece by $R_1\oplus M$ for some infinitely-generated $A$-module $M$ and declared the multiplication on this new $R$ to be that $a\cdot m = am$ for $a\in A=R_0$ while $r_dm=0$ for any $r_d$ homogeneous of positive degree, I'd still get the same Proj but now I have a non-finitely-generated $R_+$.