The Stacks project

10.117 Dimension of graded algebras over a field

Here is a basic result.

Lemma 10.117.1. Let $k$ be a field. Let $S$ be a graded $k$-algebra generated over $k$ by finitely many elements of degree $1$. Assume $S_0 = k$. Let $P(T) \in \mathbf{Q}[T]$ be the polynomial such that $\dim (S_ d) = P(d)$ for all $d \gg 0$. See Proposition 10.58.7. Then

  1. The irrelevant ideal $S_{+}$ is a maximal ideal $\mathfrak m$.

  2. Any minimal prime of $S$ is a homogeneous ideal and is contained in $S_{+} = \mathfrak m$.

  3. We have $\dim (S) = \deg (P) + 1 = \dim _ x\mathop{\mathrm{Spec}}(S)$ (with the convention that $\deg (0) = -1$) where $x$ is the point corresponding to the maximal ideal $S_{+} = \mathfrak m$.

  4. The Hilbert function of the local ring $R = S_{\mathfrak m}$ is equal to the Hilbert function of $S$.

Proof. The first statement is obvious. The second follows from Lemma 10.57.8. By (2) every irreducible component passes through $x$. Thus we have $\dim (S) = \dim _ x\mathop{\mathrm{Spec}}(S) = \dim (S_\mathfrak m)$ by Lemma 10.114.5. Since $\mathfrak m^ d/\mathfrak m^{d + 1} \cong \mathfrak m^ dS_\mathfrak m/\mathfrak m^{d + 1}S_\mathfrak m$ we see that the Hilbert function of the local ring $S_\mathfrak m$ is equal to the Hilbert function of $S$, which is (4). We conclude the last equality of (3) by Proposition 10.60.9. $\square$

Comments (5)

Comment #6300 by PS on

Don't you need some condition for to exist?

Comment #6412 by on

The polynomial exists by the reference given.

Comment #6448 by PS on

The reference has a condition.

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