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, $X = \mathbb{P}^1_k$; $\dim X = 1 = \dim A-1$.

  • $\begingroup$ If you want it to be a direct sum you need $R_d$ to be the degree $=d$ elements rather than $\le d$. $\endgroup$ Feb 25, 2023 at 19:50
  • $\begingroup$ No, degree = d is not what I want. $\endgroup$ Feb 25, 2023 at 20:15
  • $\begingroup$ 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. $\endgroup$ Feb 25, 2023 at 20:17
  • $\begingroup$ 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. $\endgroup$ Feb 25, 2023 at 20:22
  • $\begingroup$ I put an example just now. You can look at it. $\endgroup$ Feb 25, 2023 at 20:26

1 Answer 1


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 $k$.

Edit: My initial answer was needlessly complicated, I edited out the useless parts.


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