스킴
Dimension
The definition of scheme dimension and its relation to Krull dimension of local rings
This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.
Dimension of Schemes
We now define the dimension of a scheme.
Definition 1 The dimension of a scheme \(X\) is defined as the Krull dimension of the topological space \(X\). ([Topology] §Dimension, ⁋Definition 10)
Then, by the Galois correspondence of §Spectra, ⁋Proposition 16, we know that the dimension of \(\Spec A\) as a scheme equals the dimension of \(A\) as a ring. ([Commutative Algebra] §Dimension, ⁋Definition 1) Moreover, by definition one can show that \(\Spec A\) and \(\Spec A/\mathfrak{N}(A)\) are homeomorphic, so \(\dim A=\dim A/\mathfrak{N}(A)\). That is, reducedness does not affect dimension.
On the other hand, by the same reasoning as in [Topology] §Dimension, ⁋Proposition 13, the following holds.
Proposition 2 For any scheme \(X\), the condition \(\dim X=n\) is equivalent to the existence of an affine open covering \((U_i)\) of \(X\) such that \(\dim U_i\leq n\) for all \(i\), with equality holding for at least one \(i\).
Meanwhile, we saw in §Properties of Scheme Morphisms, ⁋Proposition 14 that a finite morphism is an integral morphism of finite type, and in §Fiber Product of Schemes, ⁋Proposition 14 that any finite morphism is quasi-finite. In general, there exist integral morphisms that are not of finite type, so we cannot yet say anything about the fibers of integral morphisms.
Example 3 For example, consider the algebraic closure \(\overline{\mathbb{Q}}\) of \(\mathbb{Q}\). Then \(\mathbb{Q} \rightarrow \overline{\mathbb{Q}}\) is integral, so the scheme morphism \(\Spec \overline{\mathbb{Q}} \rightarrow \Spec \mathbb{Q}\) is integral.
On the other hand, by §Fiber Product of Schemes, ⁋Proposition 15, integral morphisms are preserved under base change, so the base change of \(\Spec \overline{\mathbb{Q}} \rightarrow \Spec \mathbb{Q}\) along itself, namely \(\Spec \overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}} \rightarrow \Spec \overline{\mathbb{Q}}\), is also integral. However, the prime ideals of \(\overline{\mathbb{Q}}\otimes_\mathbb{Q} \overline{\mathbb{Q}}\) are in bijection with \(\Gal(\overline{\mathbb{Q}}/\mathbb{Q})\), so \(\Spec\overline{\mathbb{Q}}\otimes_\mathbb{Q} \overline{\mathbb{Q}}\) is an infinite set, and thus \(\Spec \overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}} \rightarrow \Spec \overline{\mathbb{Q}}\) is not a quasi-finite morphism, hence not a finite morphism.
Nevertheless, the following holds.
Proposition 4 Every fiber of an integral morphism \(\varphi: X \rightarrow Y\) has dimension \(0\).
Proof
By definition, the fiber over a point \(y\) of \(Y\) is given by the base change of \(\varphi\) along the inclusion map \(\Spec \kappa(y) \rightarrow Y\):
\[\varphi^{-1}(y)=X\times_Y\Spec \kappa(y)\]Since integral morphisms are preserved under base change,
\[\varphi^{-1}(y)=X\times_Y\Spec \kappa(y) \rightarrow \Spec \kappa(y)\]is an integral morphism, and by definition an integral morphism is affine. Thus it suffices to show that for any integral morphism \(\Spec B \rightarrow \Spec \mathbb{K}\), we have \(\dim \Spec B=\dim B=0\). That is, we must show that for any integral extension \(\mathbb{K} \rightarrow B\), there cannot exist a chain of prime ideals of \(B\)
\[\mathfrak{q}_1\subsetneq \mathfrak{q}_2\]This is a consequence of [Commutative Algebra] §Integral Extensions and Ideals, ⁋Corollary 4.
Since [Commutative Algebra] §Integral Extensions and Ideals, ⁋Corollary 4 used in the proof above also holds for any integral extension \(A\hookrightarrow B\), more generally the following holds.
Proposition 5 For any integral extension \(\phi:A \rightarrow B\),
\[\dim\Spec A=\dim\Spec B\]always holds.
In particular, for any integral domain \(A\) and its normalization \(\tilde{A}\), the dimensions of \(\Spec \tilde{A}\) and \(\Spec A\) are always equal. This holds more generally for the normalization of a scheme, but the normalization of schemes will be treated in a separate post.
Definition 6 For an irreducible subset \(Y\) of a topological space \(X\), the codimension of \(Y\) in \(X\), denoted \(\codim_XY\), is defined as the supremum of the lengths of strictly descending chains of irreducible closed subsets of \(X\)
\[A_n\supsetneq A_{n-1}\supsetneq\cdots\supsetneq A_0=\cl_X(Y)\]Then one can verify that the codimension of a prime ideal \(\mathfrak{p}\) of a ring \(A\) ([Commutative Algebra] §Dimension, ⁋Definition 2) equals the codimension of the point \(\mathfrak{p}\) in \(\Spec A\).
Proposition 7 For an irreducible closed subset \(Y\) of \(X\) and its generic point \(y\), we have \(\codim Y=\dim \mathcal{O}_{X, y}\).
Proof
Since \(Y\) has generic point \(y\), by definition \(\codim_XY\) equals \(\codim_X\{y\}\). Now choose any affine open subset \(U\cong\Spec A\) containing \(y\), and let \(y\in U\) correspond to \(\mathfrak{p}_y\in \Spec A\) under this isomorphism. Then by [Topology] §Dimension, ⁋Proposition 14, we know that there is a bijection between the irreducible closed subsets of \(X\) meeting \(U\) and the irreducible closed subsets of \(U\). That is, \(\codim_X\{y\}=\codim_U \mathfrak{p}_y\). Now we obtain the desired result from §Spectra, ⁋Proposition 16.
More generally, after defining codimension in [Commutative Algebra] §Dimension, ⁋Definition 2, we proved the inequality
\[\dim \mathfrak{a}+\codim \mathfrak{a}\leq \dim A\]Using [Topology] §Dimension, ⁋Proposition 14 in place of [Commutative Algebra] §Localization, ⁋Proposition 8 used there, one can verify that for a scheme \(X\) and an irreducible closed subset \(Y\) of \(X\), the following inequality holds:
\[\dim Y+\codim_XY\leq \dim X\]However, as before, equality does not hold in general.
Noether Normalization
We now prove the following important result.
Theorem 8 (Noether normalization lemma) Let \(\mathbb{K}\) be an arbitrary field and \(A\) a finitely generated \(\mathbb{K}\)-algebra. If \(A\) is an integral domain and
\[\trdeg_\mathbb{K}\Frac(A)=n\]then there exist elements \(x_1,\ldots, x_n\) of \(A\) that are algebraically independent and such that \(A\) is a finite \(\mathbb{K}[x_1,\ldots, x_n]\)-module.
Proof
From the assumption that \(A\) is a finitely generated \(\mathbb{K}\)-algebra, we can write
\[A=\mathbb{K}[y_1,\ldots, y_m]/\mathfrak{p}\]Then the images of \(y_1,\ldots, y_m\) in \(\Frac(A)\) generate \(\Frac(A)\) as a field extension of \(\mathbb{K}\), so we must have \(m\geq n\).
Now if \(m=n\), then the \(y_i\) are exactly the desired elements, and there is nothing more to prove. To prove the given statement, assume \(m>n\) and suppose the theorem holds for any \(k\) with \(n\leq k< m\). Then by the assumption \(m>n\), the elements \(y_1,\ldots, y_m\) are algebraically dependent. That is, there exists an \(m\)-variable polynomial over \(\mathbb{K}\)
\[f(\x_1,\ldots, \x_m)=\sum \alpha_{d_1d_2\cdots d_m}\x_1^{d_1}\cdots\x_m^{d_m}\in \mathbb{K}[\x_1,\ldots, \x_m]\tag{$\ast$}\]satisfying
\[f(y_1,\ldots, y_m)=0\]Now for integers \(r_1,\ldots, r_{m-1}\), define elements \(z_1,\ldots, z_{m-1}\) by
\[z_1=y_1-y_m^{r_1},\quad z_2=y_2-y_m^{r_2},\quad\ldots\quad,\quad z_{m-1}=y_{m-1}-y_m^{r_{m-1}}\]Then by definition
\[f(z_1+y_m^{r_1},\ldots, z_{m-1}+y_m^{r_{m-1}}, y_m)=0\tag{$\ast\ast$}\]holds. Now substitute
\[\x_1=z_1+y_m^{r_1},\quad \ldots\quad,\quad \x_{m-1}=z_{m-1}+y_m^{r_{m-1}},\quad \x_m=y_m\]into each monomial \(\alpha_{d_1d_2\cdots d_m}\x_1^{d_1}\cdots\x_m^{d_m}\) constituting \(f\) in (\(\ast\)) and expand. The result will be a power of \(y_m\) with constant coefficient
\[\alpha_{d_1d_2\cdots d_m}y_m^{r_1d_1+\cdots+r_{m-1}d_{m-1}+d_m}\]and other terms involving the \(z_k\). Now if we choose \(r_1,\ldots, r_{m-1}\) sufficiently large, we can make such a term the leading term, and thus the above equation (\(\ast\ast\)) shows that \(y_m\) is integrally dependent on \(z_1,\ldots, z_{m-1}\). On the other hand, for the \(\mathbb{K}\)-subalgebra \(A'\) of \(A\) generated by \(z_1,\ldots, z_{m-1}\), that is, the \(\mathbb{K}\)-subalgebra \(A'\) of \(A\) containing the coefficients when (\(\ast\ast\)) is viewed as a polynomial in \(y_m\), the inductive hypothesis gives elements \(x_1,\ldots, x_n\in A\) satisfying the desired condition. Now \(A\) is a finite \(A'\)-module by the above argument, and \(A'\) is a finite \(\mathbb{K}[x_1,\ldots, x_n]\)-module by the inductive hypothesis, so we obtain the desired result.
Geometrically, setting \(A=\mathbb{K}[y_1,\ldots, y_m]/\mathfrak{p}\) means that \(\Spec A\) is an integral closed subscheme of the affine space \(\mathbb{A}^m_\mathbb{K}\), so the finite ring homomorphism \(\mathbb{K}[x_1,\ldots, x_n] \rightarrow \mathbb{K}[y_1,\ldots, y_m]/\mathfrak{p}\) obtained from the above theorem corresponds geometrically to finding a finite scheme morphism \(\Spec A \rightarrow \Spec \mathbb{K}[x_1,\ldots, x_n]\). Now since the finite extension \(\mathbb{K}[x_1,\ldots, x_n] \rightarrow A\) is an integral extension, by Proposition 5 we have \(\dim A=\dim \mathbb{K}[x_1,\ldots, x_n]\), and thus by [Commutative Algebra] §System of Parameters, ⁋Corollary 11 we obtain the following result.
Proposition 9 Let \(\mathbb{K}\) be an arbitrary field and \(A\) a finitely generated \(\mathbb{K}\)-algebra. If \(A\) is an integral domain, then \(\dim\Spec A=\trdeg_\mathbb{K} \Frac(A)\).
The most important results used in the above statements are of course those from [Commutative Algebra] §Integral Extensions and Ideals. As a classical result paired with this, there is the going-down theorem, which holds for integral extensions over normal domains ([AM]). Using this, we obtain the following.
Proposition 10 Let \(\mathbb{K}\) be an arbitrary field and \(A\) a finitely generated \(\mathbb{K}\)-algebra. If \(A\) is an integral domain and \(f\in A\) is a non-unit, then \(\dim A/(f)=\dim A-1\).
Principal Ideal Theorem
Earlier we saw that for any affine integral \(\mathbb{K}\)-scheme \(X=\Spec A\), the closed subscheme \(Z(f)\) defined by a non-unit \(f\) of \(A\) has dimension one less than that of \(A\). This is clearly a useful result, but we can examine the result in the following more general case as well.
Proposition 11 For a locally noetherian scheme \(X\) and a function \(f\) on \(X\), each irreducible component of \(Z(f)\) has codimension \(0\) or \(1\).
References
[AM] M. F. Atiyah and I. G. Macdonald, Introduction to commutative algebra, Addison-Wesley, 1969.
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