The Stacks project

10.116 Dimension of finite type algebras over fields, reprise

This section is a continuation of Section 10.114. In this section we establish the connection between dimension and transcendence degree over the ground field for finite type domains over a field.

Lemma 10.116.1. 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 10.114.3 all local rings of $S$ at maximal ideals have the same dimension. Apply Lemma 10.115.4. 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 10.116.2. 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 10.116.1 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 10.114.6.

Lemma 10.116.3. 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 10.116.1 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 10.114.4. 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 10.114.5 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 10.116.4. 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 10.116.3. $\square$

Lemma 10.116.5. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $K/k$ be a field extension. Then $\dim (S) = \dim (K \otimes _ k S)$.

Proof. By Lemma 10.115.4 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 10.112.4. $\square$

Lemma 10.116.6. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Set $X = \mathop{\mathrm{Spec}}(S)$. Let $K/k$ 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 10.112.7 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 10.116.4 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 10.116.7. Let $k$ be a field. Let $S$ be a finite type $k$-algebra. Let $K/k$ 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 10.112.7. The second equality follows from the fact that we have $\dim _ x X = \dim _{x_ K} X_ K$ with notation as in Lemma 10.116.6 and we have $\dim _ x X = \dim S_\mathfrak q + \text{trdeg}_ k \kappa (\mathfrak q)$ by Lemma 10.116.3 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. $\square$


Comments (2)

Comment #5456 by Brian Shih on

Should it be d instead of r in the proposition 00P0?

Comment #5674 by on

Since the proof isn't wrong, I think it can stay the way it is for now.


Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 07NB. Beware of the difference between the letter 'O' and the digit '0'.