# Graded ring constructed from an integral domain

The aim of this construction is to obtain a projective curve $$\operatorname{Proj}(A)$$ from a given affine curve $$\operatorname{Spec}(R)$$. Example: from $$\mathbb{A}^1_k$$ to $$\mathbb{P}^1_k$$

But I want to consider a more general situation as follows.

Let $$R$$ be an integral domain with a degree function $$\deg: R\to \mathbb{N}$$, satisfying the usual degree properties as in polynomial rings. We can construct a graded ring $$A = \bigoplus_{d=0}^\infty R_d$$ by setting $$R_d = \{r\in R: \deg(r)\leq d\}.$$ Now, suppose $$t = \gcd\{\deg(r):r\in R\}$$ and $$R_0$$ is a field. Moreover, there exists a positive integer $$r$$ such that $$\dim_{R_0} (R_{(n+1)t}/R_{nt})\leq r$$ for all $$n$$ with equality for large enough $$n$$.

Now, I want to show that $$R$$ and $$A$$ are finitely generated $$R_0$$-algebra. Put $$X = \operatorname{Proj} A$$. Do we have $$\dim X = \dim A-1$$?

An example would be $$R = k[x]$$ polynomial ring over a field $$k$$. Then, $$R_0 = k$$ and $$R_d = \{\text{polynomials of degree }\leq d\text{ in }x\}\cong \{\text{homogeneous polynomials of degree }d \text{ in }x_0,x_1\}$$. In this case, $$t=1$$ and $$\dim_k (R_{n+1}/R_n) = 1$$ for all $$n$$. Moreover, $$A = \bigoplus_{d=0}^\infty R_d \cong k[x_0,x_1]$$ is finitely generated $$R_0$$-algebra. Then, $$X = \mathbb{P}^1_k$$; $$\dim X = 1 = \dim A-1$$.

• If you want it to be a direct sum you need $R_d$ to be the degree $=d$ elements rather than $\le d$. Feb 25, 2023 at 19:50
• No, degree = d is not what I want. Feb 25, 2023 at 20:15
• We can still form direct sum by using deg $\leq d$, like using two copies of $A$, i.e. $A\oplus A$; kind of like this way. Feb 25, 2023 at 20:17
• Ok. I can't imagine why it helps to consider this thing, but anyway there's an easy proof(for finite generation) along the lines of Sandor Kovacs's answer to mathoverflow.net/questions/79959/… I'll type it up when I get time. Didn't think about dim X = dim A -1 yet. Feb 25, 2023 at 20:22
• I put an example just now. You can look at it. Feb 25, 2023 at 20:26

We may assume $$t = 1$$ (by dividing all degrees by $$t$$). We have $$R = \bigoplus_{d=0}^\infty R_d/R_{d-1}.$$ Let $$k = R_0$$ and $$K = Frac(R)$$. Let $$x$$ be a homogeneous element with $$\deg x = d > 0$$. Let $$\dim (R_i/R_{i-1}) = r$$ for $$i \ge N$$. Then multiplication by $$x$$ induces an injection of $$k$$-vector spaces $$(R_i/R_{i-1}) \xrightarrow{\cdot x} (R_{i+d}/R_{i+d-1})$$ and these are of the same dimension for $$i \ge N$$ which implies that this is an isomorphism for $$i \ge N$$. Thus, if we take some finitely many homogeneous generators over $$k$$ of $$R_N$$ together with $$x$$ then these will generate $$R$$ over $$k$$.