The Stacks project

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10.115 Dimension of finite type algebras over fields, reprise

This section is a continuation of Section 10.113. In this section we establish the connection between dimension and transcendence degree over the ground field for finite type domains over a field.

Lemma 10.115.1. Let $k$ be a field. Let $S$ be a finite type $k$ algebra which is an integral domain. Let $K$ be the field of fractions of $S$. Let $r = \text{trdeg}(K/k)$ be the transcendence degree of $K$ over $k$. Then $\dim (S) = r$. Moreover, the local ring of $S$ at every maximal ideal has dimension $r$.

Proof. We may write $S = k[x_1, \ldots , x_ n]/\mathfrak p$. By Lemma 10.113.3 all local rings of $S$ at maximal ideals have the same dimension. Apply Lemma 10.114.4. We get a finite injective ring map

\[ k[y_1, \ldots , y_ d] \to S \]

with $d = \dim (S)$. Clearly, $k(y_1, \ldots , y_ d) \subset K$ is a finite extension and we win. $\square$

Lemma 10.115.2. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $\mathfrak q \subset \mathfrak q' \subset S$ be distinct prime ideals. Then $\text{trdeg}_ k\ \kappa (\mathfrak q') < \text{trdeg}_ k\ \kappa (\mathfrak q)$.

Proof. By Lemma 10.115.1 we have $\dim V(\mathfrak q) = \text{trdeg}_ k\ \kappa (\mathfrak q)$ and similarly for $\mathfrak q'$. Hence the result follows as the strict inclusion $V(\mathfrak q') \subset V(\mathfrak q)$ implies a strict inequality of dimensions. $\square$

The following lemma generalizes Lemma 10.113.6.

Lemma 10.115.3. Let $k$ be a field. Let $S$ be a finite type $k$ algebra. Let $X = \mathop{\mathrm{Spec}}(S)$. Let $\mathfrak p \subset S$ be a prime ideal, and let $x \in X$ be the corresponding point. Then we have

\[ \dim _ x(X) = \dim (S_{\mathfrak p}) + \text{trdeg}_ k\ \kappa (\mathfrak p). \]

Proof. By Lemma 10.115.1 we know that $r = \text{trdeg}_ k\ \kappa (\mathfrak p)$ is equal to the dimension of $V(\mathfrak p)$. Pick any maximal chain of primes $\mathfrak p \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ r$ starting with $\mathfrak p$ in $S$. This has length $r$ by Lemma 10.113.4. Let $\mathfrak q_ j$, $j \in J$ be the minimal primes of $S$ which are contained in $\mathfrak p$. These correspond $1-1$ to minimal primes in $S_{\mathfrak p}$ via the rule $\mathfrak q_ j \mapsto \mathfrak q_ jS_{\mathfrak p}$. By Lemma 10.113.5 we know that $\dim _ x(X)$ is equal to the maximum of the dimensions of the rings $S/\mathfrak q_ j$. For each $j$ pick a maximal chain of primes $\mathfrak q_ j \subset \mathfrak p'_1 \subset \ldots \subset \mathfrak p'_{s(j)} = \mathfrak p$. Then $\dim (S_{\mathfrak p}) = \max _{j \in J} s(j)$. Now, each chain

\[ \mathfrak q_ i \subset \mathfrak p'_1 \subset \ldots \subset \mathfrak p'_{s(j)} = \mathfrak p \subset \mathfrak p_1 \subset \ldots \subset \mathfrak p_ r \]

is a maximal chain in $S/\mathfrak q_ j$, and by what was said before we have $\dim _ x(X) = \max _{j \in J} r + s(j)$. The lemma follows. $\square$

The following lemma says that the codimension of one finite type Spec in another is the difference of heights.

Lemma 10.115.4. Let $k$ be a field. Let $S' \to S$ be a surjection of finite type $k$ algebras. Let $\mathfrak p \subset S$ be a prime ideal, and let $\mathfrak p'$ be the corresponding prime ideal of $S'$. Let $X = \mathop{\mathrm{Spec}}(S)$, resp. $X' = \mathop{\mathrm{Spec}}(S')$, and let $x \in X$, resp. $x'\in X'$ be the point corresponding to $\mathfrak p$, resp. $\mathfrak p'$. Then

\[ \dim _{x'} X' - \dim _ x X = \text{height}(\mathfrak p') - \text{height}(\mathfrak p). \]

Proof. Immediate from Lemma 10.115.3. $\square$

Lemma 10.115.5. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $k \subset K$ be a field extension. Then $\dim (S) = \dim (K \otimes _ k S)$.

Proof. By Lemma 10.114.4 there exists a finite injective map $k[y_1, \ldots , y_ d] \to S$ with $d = \dim (S)$. Since $K$ is flat over $k$ we also get a finite injective map $K[y_1, \ldots , y_ d] \to K \otimes _ k S$. The result follows from Lemma 10.111.4. $\square$

Lemma 10.115.6. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Set $X = \mathop{\mathrm{Spec}}(S)$. Let $k \subset K$ be a field extension. Set $S_ K = K \otimes _ k S$, and $X_ K = \mathop{\mathrm{Spec}}(S_ K)$. Let $\mathfrak q \subset S$ be a prime corresponding to $x \in X$ and let $\mathfrak q_ K \subset S_ K$ be a prime corresponding to $x_ K \in X_ K$ lying over $\mathfrak q$. Then $\dim _ x X = \dim _{x_ K} X_ K$.

Proof. Choose a presentation $S = k[x_1, \ldots , x_ n]/I$. This gives a presentation $K \otimes _ k S = K[x_1, \ldots , x_ n]/(K \otimes _ k I)$. Let $\mathfrak q_ K' \subset K[x_1, \ldots , x_ n]$, resp. $\mathfrak q' \subset k[x_1, \ldots , x_ n]$ be the corresponding primes. Consider the following commutative diagram of Noetherian local rings

\[ \xymatrix{ K[x_1, \ldots , x_ n]_{\mathfrak q_ K'} \ar[r] & (K \otimes _ k S)_{\mathfrak q_ K} \\ k[x_1, \ldots , x_ n]_{\mathfrak q'} \ar[r] \ar[u] & S_{\mathfrak q} \ar[u] } \]

Both vertical arrows are flat because they are localizations of the flat ring maps $S \to S_ K$ and $k[x_1, \ldots , x_ n] \to K[x_1, \ldots , x_ n]$. Moreover, the vertical arrows have the same fibre rings. Hence, we see from Lemma 10.111.7 that $\text{height}(\mathfrak q') - \text{height}(\mathfrak q) = \text{height}(\mathfrak q_ K') - \text{height}(\mathfrak q_ K)$. Denote $x' \in X' = \mathop{\mathrm{Spec}}(k[x_1, \ldots , x_ n])$ and $x'_ K \in X'_ K = \mathop{\mathrm{Spec}}(K[x_1, \ldots , x_ n])$ the points corresponding to $\mathfrak q'$ and $\mathfrak q_ K'$. By Lemma 10.115.4 and what we showed above we have

\begin{eqnarray*} n - \dim _ x X & = & \dim _{x'} X' - \dim _ x X \\ & = & \text{height}(\mathfrak q') - \text{height}(\mathfrak q) \\ & = & \text{height}(\mathfrak q_ K') - \text{height}(\mathfrak q_ K) \\ & = & \dim _{x'_ K} X'_ K - \dim _{x_ K} X_ K \\ & = & n - \dim _{x_ K} X_ K \end{eqnarray*}

and the lemma follows. $\square$

Lemma 10.115.7. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $k \subset K$ be a field extension. Set $S_ K = K \otimes _ k S$. Let $\mathfrak q \subset S$ be a prime and let $\mathfrak q_ K \subset S_ K$ be a prime lying over $\mathfrak q$. Then

\[ \dim (S_ K \otimes _ S \kappa (\mathfrak q))_{\mathfrak q_ K} = \dim (S_ K)_{\mathfrak q_ K} - \dim S_\mathfrak q = \text{trdeg}_ k \kappa (\mathfrak q) - \text{trdeg}_ K \kappa (\mathfrak q_ K) \]

Moreover, given $\mathfrak q$ we can always choose $\mathfrak q_ K$ such that the number above is zero.

Proof. Observe that $S_\mathfrak q \to (S_ K)_{\mathfrak q_ K}$ is a flat local homomorphism of local Noetherian rings with special fibre $(S_ K \otimes _ S \kappa (\mathfrak q))_{\mathfrak q_ K}$. Hence the first equality by Lemma 10.111.7. The second equality follows from the fact that we have $\dim _ x X = \dim _{x_ K} X_ K$ with notation as in Lemma 10.115.6 and we have $\dim _ x X = \dim S_\mathfrak q + \text{trdeg}_ k \kappa (\mathfrak q)$ by Lemma 10.115.3 and similarly for $\dim _{x_ K} X_ K$. If we choose $\mathfrak q_ K$ minimal over $\mathfrak q S_ K$, then the dimension of the fibre ring will be zero. $\square$


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