
## 10.116 Dimension of graded algebras over a field

Here is a basic result.

Lemma 10.116.1. Let $k$ be a field. Let $S$ be a finitely generated graded algebra over $k$. 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 10.57.7. 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 10.56.8. The equality $\dim (S) = \dim _ x\mathop{\mathrm{Spec}}(S)$ follows from the fact that every irreducible component passes through $x$ according to (2). Hence we may compute this dimension as the dimension of the local ring $R = S_{\mathfrak m}$ with $\mathfrak m = S_{+}$ by Lemma 10.113.6. Since $\mathfrak m^ d/\mathfrak m^{d + 1} \cong \mathfrak m^ dR/\mathfrak m^{d + 1}R$ we see that the Hilbert function of the local ring $R$ is equal to the Hilbert function of $S$, which is (4). We conclude the last equality of (3) by Proposition 10.59.8. $\square$

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