Let \(X\) be an integral scheme of finite type over a field \(k\), not necessarily algebraically closed. Prove the following, where for rings \(\krulldim\) always means the Krull dimension.
For any closed point \(P \in X\), \(\krulldim X = \krulldim \OO_P\).
Let \(K(X)\) be the function field of \(X\). Then \[ \krulldim X = \trdeg\qty{K(X)/k}. \]
If \(Y\) is a closed subset of \(X\), then \[ \codim(Y, X) = \inf \ts{\krulldim \OO_{P, X} \st P \in Y}. \]
If \(Y\) is a closed subset of \(X\), then \[ \krulldim Y + \codim(Y, X) = \krulldim X. \]
If \(U\) is a nonempty open subset of \(X\), then \(\krulldim U = \krulldim X\).
If \(k \subseteq k'\) is a field extension, then every irreducible component of \(X' = \fiberprod{X}{k}{k'}\) has dimension equal to \(\krulldim X\).
Solution
We use the following standard affine dimension theorem.
If \(A\) is a finitely generated integral \(k\)-algebra with fraction field \(K\), then \[ \dim A=\operatorname{trdeg}_kK, \] every maximal localization \(A_{\mathfrak m}\) has this same dimension, and for every prime \(\mathfrak p\subseteq A\), \[ \boxed{ \dim A_{\mathfrak p} +\operatorname{trdeg}_k\kappa(\mathfrak p) =\operatorname{trdeg}_kK.} \] The first two assertions follow from Noether normalization and going up; the last is the dimension formula for finite-type algebras over a field. These are Stacks Project, Tags 00P0 and 00P1.
1. Every nonempty affine open\[ U=\Spec A\subseteq X \]has\[ \dim U = \operatorname{trdeg}_kK(X). \]
Since \(X\) is integral, \(A\) is a domain. Since \(X\) is of finite type over \(k\), Hartshorne II.3.3(c) shows that \(A\) is a finitely generated \(k\)-algebra.
By Hartshorne II.3.6, \[ \operatorname{Frac}(A)\cong K(X). \] The affine dimension theorem therefore gives \[ \dim U=\dim A=\operatorname{trdeg}_kK(X). \]
2. Put\[ r=\operatorname{trdeg}_kK(X). \]Then\[ \dim X\le r. \]
Take any strict chain of irreducible closed subsets \[ Z_0\subsetneq Z_1\subsetneq\cdots\subsetneq Z_n \] in \(X\).
The closed subscheme \(Z_0\) with its reduced induced structure is of finite type over \(k\). By Hartshorne II.3.14, it contains a closed point \(P\). Since \(Z_0\) is closed in \(X\), this point is closed in \(X\) as well.
Choose an affine open neighborhood \[ U\subseteq X \] of \(P\). Each intersection \[ Z_i\cap U \] is nonempty and irreducible. Moreover the inclusions remain strict. Indeed, if \[ Z_i\cap U=Z_{i+1}\cap U, \] then the nonempty open subset \(Z_{i+1}\cap U\) of the irreducible space \(Z_{i+1}\) would lie in the proper closed subset \(Z_i\), impossible.
Thus \[ Z_0\cap U\subsetneq\cdots\subsetneq Z_n\cap U \] is a chain of length \(n\) in \(U\). By 1, \[ n\le\dim U=r. \] Taking the supremum over all chains gives \(\dim X\le r\).
3. One has\[ \dim X\ge r. \]
Any nonempty affine open \(U\subseteq X\) is a subspace of \(X\), so every chain of irreducible closed subsets in \(U\) gives a chain of the same length in \(X\) after taking closures in \(X\). Hence \[ \dim U\le\dim X. \] By 1, \(\dim U=r\).
4. Therefore\[ \boxed{ \dim X=\operatorname{trdeg}_kK(X).} \]
3. This proves part (b).
5. Let \(P\in X\) be a closed point. Then\[ \boxed{\dim\mathcal O_{X,P}=\dim X.} \]
Choose an affine neighborhood \[ U=\Spec A \] of \(P\), and let \(\mathfrak m\subseteq A\) be the corresponding maximal ideal. Then \[ \mathcal O_{X,P}=A_{\mathfrak m}. \] By 1, the fraction field of \(A\) is \(K(X)\) and \[ \operatorname{trdeg}_kK(X)=\dim X. \] The maximal-localization part of the affine dimension theorem gives \[ \dim A_{\mathfrak m} = \operatorname{trdeg}_kK(X) = \dim X. \] This proves part (a).
6. If \(P\in X\) and\[ Z=\overline{\{P\}}, \]then\[ \boxed{ \codim(Z,X)=\dim\mathcal O_{X,P}.} \]
The point \(P\) is the generic point of the irreducible closed subset \(Z\). By the definition of codimension in a scheme, chains of irreducible closed subsets from \(Z\) up to an irreducible component of \(X\) correspond in an affine neighborhood of \(P\) to chains of prime ideals contained in the prime representing \(P\). Their supremal length is the height of that prime, namely \[ \dim\mathcal O_{X,P}. \] Since \(X\) is integral it has one irreducible component, so this is exactly \(\codim(Z,X)\).
7. For every closed subset \(Y\subseteq X\),\[ \boxed{ \codim(Y,X) = \inf_{P\in Y}\dim\mathcal O_{X,P}.} \]
By definition, \[ \codim(Y,X) \] is the infimum of \(\codim(Z,X)\) over irreducible closed subsets \(Z\subseteq Y\).
Every point \(P\in Y\) gives such an irreducible closed subset \[ \overline{\{P\}}\subseteq Y, \] and 6 identifies its codimension with \(\dim\mathcal O_{X,P}\).
Conversely every irreducible closed \(Z\subseteq Y\) has a generic point \(P\in Y\), and \[ Z=\overline{\{P\}}. \] Thus the two sets of numbers over which the infima are taken are identical. This proves part (c).
8. Let \(Z\subseteq X\) be irreducible closed with generic point \(\zeta\). Then\[ \boxed{ \dim Z+\codim(Z,X)=\dim X.} \]
Give \(Z\) its reduced induced structure. It is an integral scheme of finite type over \(k\), with function field \[ K(Z)=\kappa(\zeta). \] By 4 applied to \(Z\), \[ \dim Z=\operatorname{trdeg}_k\kappa(\zeta). \] By 6, \[ \codim(Z,X)=\dim\mathcal O_{X,\zeta}. \]
Choose an affine neighborhood \(\Spec A\) of \(\zeta\), corresponding to a prime \(\mathfrak p\subseteq A\). The affine dimension formula gives \[ \dim A_{\mathfrak p} +\operatorname{trdeg}_k\kappa(\mathfrak p) =\operatorname{trdeg}_kK(X). \] Using 4 for \(X\) converts this exactly into the displayed identity.
9. For every closed subset \(Y\subseteq X\),\[ \boxed{ \dim Y+\codim(Y,X)=\dim X.} \]
Let \[ Y_1,\ldots,Y_m \] be the irreducible components of \(Y\). Then \[ \dim Y=\max_i\dim Y_i. \]
By 8, \[ \codim(Y_i,X)=\dim X-\dim Y_i. \] Any irreducible closed subset of \(Y\) is contained in some \(Y_i\), and its dimension is at most \(\dim Y_i\), so 8 shows that its codimension is at least \(\codim(Y_i,X)\). Hence \[ \codim(Y,X) = \min_i\codim(Y_i,X) = \dim X-\max_i\dim Y_i. \] This proves part (d).
10. If \(U\subseteq X\) is nonempty open, then\[ \boxed{\dim U=\dim X.} \]
Choose a nonempty affine open \[ V\subseteq U. \] Since \(V\) is also a nonempty affine open of \(X\), 1 and 4 give \[ \dim V=\dim X. \] As subspaces, \[ V\subseteq U\subseteq X, \] so \[ \dim V\le\dim U\le\dim X. \] Therefore all three dimensions are equal. This proves part (e).
11. Let \(k\subseteq k'\) be a field extension and put\[ X'=X\times_k k'. \]Every irreducible component of \(X'\) maps dominantly to \(X\).
The projection \[ \pi:X'\to X \] is flat, because it is obtained by base change from the flat field extension \(k\to k'\).
Let \(C\) be an irreducible component of \(X'\) with generic point \(\xi_C\). Work in an affine neighborhood \[ U=\Spec A\subseteq X \] of the image of \(\xi_C\). On the corresponding affine base change \[ U'=\Spec(A\otimes_k k'), \] the generic point of \(C\) is represented by a minimal prime \(\mathfrak q\). Flatness gives going down, so the contraction of the minimal prime \(\mathfrak q\) is a minimal prime of the domain \(A\), hence is \((0)\).
Thus \(\xi_C\) maps to the generic point of \(X\), and \(C\to X\) is dominant.
12. Choose a nonempty affine open\[ U=\Spec A\subseteq X \]and put\[ d=\dim X. \]There is a finite injective Noether-normalization map\[ k[t_1,\ldots,t_d]\hookrightarrow A. \]After base change,\[ k'[t_1,\ldots,t_d] \hookrightarrow A'=A\otimes_k k' \]is again finite and injective.
By 4, \[ d=\operatorname{trdeg}_kK(X)=\operatorname{trdeg}_k\operatorname{Frac}(A). \] Noether normalization gives algebraically independent elements \(t_1,\ldots,t_d\in A\) such that \(A\) is finite over the polynomial subring \(k[t_1,\ldots,t_d]\).
Tensoring with the flat \(k\)-module \(k'\) preserves injectivity, and tensoring a finite module with \(k'\) remains finite. This gives the asserted map.
13. Every irreducible component of\[ U'=U\times_k k'=\Spec A' \]has dimension \(d\).
12. The contraction\[ \mathfrak q\cap R \]is a minimal prime of \(R\) under an integral extension. Since \(R\) is a domain,\[ \mathfrak q\cap R=(0). \]
Therefore \[ R\hookrightarrow A'/\mathfrak q \] is an integral finite extension. Integral extensions preserve Krull dimension, so \[ \dim(A'/\mathfrak q) = \dim R =d. \] This is the dimension of the irreducible component \(V(\mathfrak q)\).
14. Every irreducible component \(C\) of \(X'\) has dimension \(d=\dim X\).
By 11, \(C\) contains a point mapping to the generic point of \(X\), so it meets the inverse image \(U'\) of every nonempty open \(U\subseteq X\). Hence \[ C\cap U' \] is a nonempty open subset of \(C\). It is an irreducible component of \(U'\): if it were properly contained in a larger irreducible closed subset of \(U'\), its closure in \(X'\) would be an irreducible closed subset properly containing the component \(C\).
By 13, \[ \dim(C\cap U')=d. \] Give \(C\) its reduced induced structure. It is an integral scheme of finite type over \(k'\). Part (e), already proved in 10, gives \[ \dim C=\dim(C\cap U')=d. \] This proves part (f).
15. Q.E.D.
Step 5 proves part (a), 4 proves part (b), 7 proves part (c), 9 proves part (d), 10 proves part (e), and 14 proves part (f).