This post was translated from Korean by LLM (Kimi). The translation may contain errors or awkward sentences. The Korean original is the source of truth.
Dimension of Schemes
In this post we 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, from 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 \(\dim A=\dim A/\mathfrak{N}(A)\) holds. In other words, reducedness does not affect the dimension.
On the other hand, for the same reason as [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 \(U_i\), 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 Products, ⁋Proposition 14 we saw that any finite morphism is quasi-finite. In general, there exist morphisms that are integral but not of finite type, and therefore we cannot yet speak about the fibers of an integral morphism.
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 Products, ⁋Proposition 15, integral morphisms are preserved under base change, so the base change of this along \(\Spec \overline{\mathbb{Q}} \rightarrow \Spec \mathbb{Q}\), namely \(\Spec \overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}} \rightarrow \Spec \overline{\mathbb{Q}}\), is also integral. However, the prime ideals of \(\overline{\mathbb{Q}}\otimes_\mathbb{Q} \overline{\mathbb{Q}}\) correspond bijectively to \(\Gal(\overline{\mathbb{Q}}/\mathbb{Q})\), so \(\Spec\overline{\mathbb{Q}}\otimes_\mathbb{Q} \overline{\mathbb{Q}}\) is an infinite set, and therefore \(\Spec \overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}} \rightarrow \Spec \overline{\mathbb{Q}}\) is not a quasi-finite morphism, hence not a finite morphism.
However, the following holds.
Proposition 4 Any fiber of an integral morphism \(\varphi: X \rightarrow Y\) is always \(0\)-dimensional.
Proof
By definition, the fiber over a point \(y\) of \(Y\) is the base change of \(\varphi\) along the inclusion map \(\Spec \kappa(y) \rightarrow Y\)
\[\varphi^{-1}(y)=X\times_Y\Spec \kappa(y)\]and 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 since an integral morphism is an affine morphism by definition, 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, for any integral extension \(\mathbb{K} \rightarrow B\), we must show that there cannot exist a chain of prime ideals of \(B\)
\[\mathfrak{q}_1\subsetneq \mathfrak{q}_2\]of the form \(\mathfrak{q}_1\subsetneq \mathfrak{q}_2\). 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 \(A\hookrightarrow B\), the following more general statement 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 the same. This continues to hold even after we define the normalization of a scheme \(X\). (##ref##)
Definition 6 For an irreducible subset \(Y\) of a topological space \(X\), the codimension \(\codim_XY\) of \(Y\) in \(X\) is defined as the supremum of the lengths of strictly descending chains of irreducible closed subsets of \(X\)
\[A_n\supsetneq A_{n-1}\supsetneq\cdots\supsetneq A_0=\cl_X(Y)\]ending at \(A_0=\cl_X(Y)\).
Then the codimension of a prime ideal \(\mathfrak{p}\) of a ring \(A\) ([Commutative Algebra] §Dimension, ⁋Definition 2) coincides with 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 \(\codim Y=\dim \mathscr{O}_{X, y}\).
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 from [Topology] §Dimension, ⁋Proposition 14, we know that there is a one-to-one correspondence between irreducible closed subsets of \(X\) meeting \(U\) and irreducible closed subsets of \(U\). Hence \(\codim_X\{y\}=\codim_U \mathfrak{p}_y\). Now the desired result follows from §Spectra, ⁋Proposition 16.
More generally, after defining codimension in [Commutative Algebra] §Dimension, ⁋Definition 2, we proved the inequality
\[\dim \mathfrak{a}+\codim \mathfrak{a}\leq \dim A\]and using [Topology] §Dimension, ⁋Proposition 14 in place of [Commutative Algebra] §Localization, ⁋Proposition 8 used there, we can verify that for a scheme \(X\) and an irreducible closed subset \(Y\) of \(X\), the inequality
\[\dim Y+\codim_XY\leq \dim X\]holds. However, just as before, equality does not hold in general.
Noether Normalization
Now we prove the following important result.
Theorem 8 (Noether normalization lemma) Let \(\mathbb{K}\) be an arbitrary field and let \(A\) be a finitely generated \(\mathbb{K}\)-algebra. If \(A\) is an integral domain and
\[\trdeg_\mathbb{K}\Frac(A)=n\]then there exist suitable 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, so there is nothing more to prove. To prove the claim, suppose \(m>n\) and assume the theorem holds for any \(k\) satisfying \(n\leq k< m\). Then the assumption \(m>n\) implies that \(y_1,\ldots, y_m\) are algebraically dependent. That is, there exists an \(m\)-variable polynomial with coefficients in \(\mathbb{K}\)
\[f(y_1,\ldots, y_m)=0\]satisfying
\[f(\x_1,\ldots, \x_m)=\sum \alpha_{d_1d_2\cdots d_m}\x_1^{d_1}\cdots\x_m^{d_m}\in \mathbb{K}[\x_1,\ldots, \x_m]\tag{$\ast$}\]such that \(f(y_1,\ldots, y_m)=0\). 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 \(\alpha_{d_1d_2\cdots d_m}\x_1^{d_1}\cdots\x_m^{d_m}\) appearing in \(f\) in (\(\ast\)) and expand. The result consists of a power of \(y_m\) whose coefficient is a constant term
\[\alpha_{d_1d_2\cdots d_m}y_m^{r_1d_1+\cdots+r_{m-1}d_{m-1}+d_m}\]plus the remaining terms involving \(z_k\). If we choose \(r_1,\ldots, r_{m-1}\) sufficiently large, such a term becomes 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 of \(A\) in which the coefficients exist when viewing (\(\ast\ast\)) as a polynomial in \(y_m\), the inductive hypothesis implies the existence of \(x_1,\ldots, x_n\in A\) satisfying the desired conditions. Now \(A\) is a finite \(A'\)-module by the argument above, 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 as a result of the above theorem corresponds geometrically to finding a finite scheme morphism \(\Spec A \rightarrow \Spec \mathbb{K}[x_1,\ldots, x_n]\). Now the finite extension \(\mathbb{K}[x_1,\ldots, x_n] \rightarrow A\) is an integral extension, so by Proposition 5 we have \(\dim A=\dim \mathbb{K}[x_1,\ldots, x_n]\), and thus by [Commutative Algebra] §System of Parameters, ⁋Corollary 10 we obtain the following result.
Proposition 9 Let \(\mathbb{K}\) be an arbitrary field and let \(A\) be a finitely generated \(\mathbb{K}\)-algebra. If \(A\) is an integral domain, then \(\dim\Spec A=\trdeg_\mathbb{K} \Frac(A)\) holds.
The most important results used in the above statements are of course those from [Commutative Algebra] §Integral Extensions and Ideals. Using a similar type of result, ##ref##, we obtain the following.
Proposition 10 Let \(\mathbb{K}\) be an arbitrary field and let \(A\) be a finitely generated \(\mathbb{K}\)-algebra. If \(A\) is an integral domain and \(f\in A\) is a non-unit, then \(\dim A/(f)=\dim A-1\) holds.
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 \(f\in A\) has dimension one less than \(A\). This is clearly a useful result, but we can also examine the result in the following more general case.
Proposition 11 For a locally Noetherian scheme \(X\) and a function \(f\) on \(X\), each irreducible component of \(Z(f)\) has codimension \(0\) or codimension \(1\).
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