Dimension

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)

From the Galois correspondence of [Spectrums] §Spectrum, ⁋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. That is, reducedness does not affect dimension.

On the other hand, by the same reasoning as in [Topology] §Dimension, ⁋Proposition 15, the following holds.

Proposition 2 For any scheme \(X\) and integer \(n\geq 0\), 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 \(i\).

Proof

For any chain of irreducible closed subsets of \(X\)

\[Y_0\subsetneq Y_1\subsetneq\cdots\subsetneq Y_r\]

the generic point \(\eta_0\) of the smallest term \(Y_0\) is a point of \(X\), so it belongs to some \(U_i\) by the covering \((U_i)\). Since every term of the chain meets \(U_i\), considering the inclusion-preserving bijection of [Topology] §Dimension, ⁋Proposition 15, it corresponds to a chain of the same length inside \(U_i\). Conversely, any chain in \(U_i\) lifts to \(X\) by taking closures, so \(\dim X\geq\dim U_i\), and therefore \(\dim X=\sup_i\dim U_i\), which is equivalent to the condition of the proposition.

What the proof actually yields is \(\dim X=\sup_i\dim U_i\), and without the assumption that \(\dim X\) is finite, there may not exist an \(i\) for which equality holds. For example, \(X=\coprod_{d\geq 0}\mathbb{A}^d_\mathbb{K}\) is infinite-dimensional, but each affine open subset meets only finitely many components, so all are finite-dimensional.

On the other hand, we saw in [Properties of Scheme Morphisms] §Properties of Scheme Morphisms, ⁋Proposition 15 that a finite morphism is an integral morphism of finite type, and in [Fiber Products] §Fiber Products, ⁋Proposition 15 that any finite morphism is quasi-finite. In general, there exist integral morphisms that are not of finite type, so until now we have not been able to say anything about the fibers of integral morphisms.

Example 3 For example, consider the algebraic closure \(\overline{\mathbb{Q}}\) of \(\mathbb{Q}\). Every element of \(\overline{\mathbb{Q}}\) is algebraic over \(\mathbb{Q}\), hence integral, and therefore \(\mathbb{Q} \rightarrow \overline{\mathbb{Q}}\) is an integral extension, from which the scheme morphism \(\varphi:\Spec \overline{\mathbb{Q}} \rightarrow \Spec \mathbb{Q}\) is also an integral morphism.

Now base change \(\varphi\) along \(\Spec\overline{\mathbb{Q}}\rightarrow\Spec\mathbb{Q}\) to obtain the following pullback diagram

and at this time the left vertical map

\[\Spec(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}})\rightarrow \Spec \overline{\mathbb{Q}}\]

is also integral by [Fiber Products] §Fiber Products, ⁋Proposition 16.

To examine this map, let us look concretely at the ring homomorphism \(\overline{\mathbb{Q}}\rightarrow \overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}}\). Viewing a section of the above map of schemes is the same as viewing a retraction of this map, which comes from the following surjective ring homomorphism

\[\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}}\rightarrow\overline{\mathbb{Q}},\qquad a\otimes b\mapsto a\sigma(b)\]

Specifically, the kernel \(\mathfrak{p}_\sigma\) of this ring homomorphism is a maximal ideal, so it defines a point of \(\Spec(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}})\), and if \(\sigma\neq\tau\), choosing \(b\in\overline{\mathbb{Q}}\) with \(\sigma(b)\neq\tau(b)\), we have \(1\otimes b-\sigma(b)\otimes 1\in\mathfrak{p}_\sigma\) but it does not belong to \(\mathfrak{p}_\tau\), so \(\mathfrak{p}_\sigma\neq\mathfrak{p}_\tau\). Therefore \(\Spec(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}})\) has at least as many points as \(\Gal(\overline{\mathbb{Q}}/\mathbb{Q})\), that is, infinitely many points, and \(\Spec(\overline{\mathbb{Q}}\otimes_\mathbb{Q}\overline{\mathbb{Q}})\rightarrow\Spec\overline{\mathbb{Q}}\) is not a quasi-finite morphism, hence not a finite morphism.

Or consider the simpler example \(\Spec \mathbb{C}\rightarrow \Spec \mathbb{R}\). Since \(\mathbb{R}\) and \(\mathbb{C}\) are both fields, \(\Spec\mathbb{C}\) and \(\Spec\mathbb{R}\) each consist of a single point, so this map itself is a trivial map from one point to one point. However, pulling this back along \(\Spec \mathbb{C}\rightarrow \Spec \mathbb{R}\) to construct a similar function as in the example above

\[\Spec(\mathbb{C}\otimes_\mathbb{R} \mathbb{C}) \rightarrow \Spec \mathbb{C}\]

we find that \(\mathbb{C}\otimes_\mathbb{R}\mathbb{C}\) is no longer a field. Indeed, since \(\Spec\mathbb{C}=\Spec\mathbb{R}[\x]/(\x^2+1)\),

\[\mathbb{C}\otimes_\mathbb{R} \mathbb{C}\cong \mathbb{C}\otimes_\mathbb{R} \frac{\mathbb{R}[\x]}{(\x^2+1)}\cong \frac{\mathbb{C}[\x]}{(\x^2+1)}\]

and \(\x^2+1\) factors in \(\mathbb{C}\) as a product of two linear terms \(\x^2+1=(\x-i)(\x+i)\), and since \((\x-i)\) and \((\x+i)\) are comaximal, by [Ring Theory] §Chinese Remainder Theorem, ⁋Proposition 6

\[\frac{\mathbb{C}[\x]}{((\x-i)(\x+i))}\cong\frac{\mathbb{C}[\x]}{(\x-i)}\times\frac{\mathbb{C}[\x]}{(\x+i)}\cong\mathbb{C}\times\mathbb{C}\]

Thinking in the language of the Galois group examined in the example above, this decomposition arises because the two factors \(\mathbb{C}[\x]/(\x-i)\) and \(\mathbb{C}[\x]/(\x+i)\) correspond to the two automorphisms of \(\mathbb{C}\rightarrow \mathbb{C}\) fixing \(\mathbb{R}\), that is, the two elements of \(\Gal(\mathbb{C}/\mathbb{R})\), and the same phenomenon occurs in Example 3 for \(\mathbb{Q}\rightarrow \overline{\mathbb{Q}}\). The only difference is that since \(\Gal(\overline{\mathbb{Q}}/\mathbb{Q})\) is infinite, the fiber consists of infinitely many points rather than two.

Nevertheless, this example suggests some kind of finiteness for the fibers of an integral morphism: for example, since \(\Gal(\overline{\mathbb{Q}}/\mathbb{Q})\) is a profinite group ([Field Theory] §Properties of Galois Groups, ⁋Proposition 5), it is \(0\)-dimensional. This is a fact that holds for any integral morphism.

Proposition 4 Any nonempty fiber of an integral morphism \(\varphi: X \rightarrow Y\) is always \(0\)-dimensional.

Proof

By definition, the fiber at a point \(y\) of \(Y\) is given by the base change of \(\varphi\) along the inclusion map \(\Spec \kappa(y) \rightarrow Y\) for the residue field \(\kappa(y)\) from [Schemes] §Schemes, ⁋Definition 5

\[\varphi^{-1}(y)=X\times_Y\Spec \kappa(y)\]

and since integral morphisms are preserved under base change ([Fiber Products] §Fiber Products, ⁋Proposition 16)

\[\varphi^{-1}(y)=X\times_Y\Spec \kappa(y) \rightarrow \Spec \kappa(y)\]

is an integral morphism, and since an integral morphism is by definition an affine morphism, it suffices to show that for an integral morphism \(\Spec B \rightarrow \Spec \kappa(y)\), we have \(\dim \Spec B=\dim B=0\). That is, we must show that for any integral extension \(\kappa(y) \rightarrow B\), there cannot exist a chain of prime ideals of \(B\)

\[\mathfrak{q}_1\subsetneq \mathfrak{q}_2\]

This is the result of [Commutative Algebra] §Integral Extensions and Ideals, ⁋Corollary 4.

Geometrically, this proposition shows that each fiber of an integral morphism has no positive dimension.

The [Commutative Algebra] §Integral Extensions and Ideals, ⁋Corollary 4 used in the proof of the above proposition also holds for any integral extension \(A\hookrightarrow B\). By this, contracting a prime ideal chain of \(B\) to \(A\) remains strict, so \(\dim B\leq\dim A\), and conversely by lying over and going up from [Commutative Algebra] §Integral Extensions and Ideals, ⁋Proposition 1, a prime ideal chain of \(A\) lifts to \(B\), so \(\dim A\leq\dim B\). Therefore 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 \(\tilde{A}\), since the extension \(A\hookrightarrow\tilde{A}\) is integral, by Proposition 5 we have \(\dim\Spec\tilde{A}=\dim\Spec A\). Here the normalization \(\tilde{A}\) is the extension obtained by enlarging \(A\) to become integrally closed inside its field of fractions \(\Frac(A)\), that is, by adjoining to \(A\) all elements of \(\Frac(A)\) that are integral over \(A\). ([Commutative Algebra] §Integral Extension, ⁋Definition 3) By definition \(A\subseteq\tilde{A}\subseteq\Frac(A)\), so \(\Frac(\tilde{A})=\Frac(A)\), that is, normalization preserves the function field of \(A\).

Example 6 In the above discussion we saw that normalization preserves the function field. Geometrically, when \(A\) is the coordinate ring of an affine variety over \(\mathbb{K}\), this means that the two spaces obtained by normalization are birational. ([Algebraic Varieties] §Rational Maps, ⁋Proposition 10) That is, normalization coincides with the original space outside certain loci that are negligible.

Where normalization actually differs is the locus where \(A\) is not integrally closed, that is, the non-normal locus. This is contained in the singular locus but is generally not equal to it. For example, the quadric cone \(\mathbb{K}[\x,\y,\z]/(\x\y-\z^2)\) is a \(2\)-dimensional domain singular at the origin but normal, so normalization becomes the identity map and the singular point remains. However, in the case of curves, that is, in dimension \(1\), a normal local ring is exactly a regular local ring, so the non-normal locus coincides with the singular locus. As a representative example, consider the cusp from [Algebraic Varieties] §Tangent Spaces and Smoothness, ⁋Example 7

\[A=\mathbb{K}[\x,\y]/(\y^2-\x^3)\cong\mathbb{K}[t^2,t^3]\]

To see the field of fractions of \(A\), using \(t=\y/\x\) we can verify that \(\Frac(A)=\mathbb{K}(t)\), and at this time the element \(t\in\Frac(A)\) satisfies \(t^2=\x\in A\), so it is integral over \(A\). Therefore the extension obtained by adjoining \(t\)

\[A[t]=\mathbb{K}[t^2,t^3,t]=\mathbb{K}[t]\]

is an integral extension of \(A\), and since \(A[t]\) is a UFD, by [Commutative Algebra] §Integral Extension, ⁋Proposition 9 it is integrally closed, so this is the normalization \(\tilde{A}\).

Now let us see geometrically what this means. We first need to examine the map between spaces given by the above integral extension \(A\rightarrow A[t]\)

\[\Spec A[t]\rightarrow \Spec A\]

First, looking at the origin \(\mathfrak{m}=(t^2,t^3)\in\Spec A\), which is the singular point of the curve \(\Spec A\), the fiber of the above map at this point is given by the following pullback diagram

that is, by the following map

\[\Spec(A[t]\otimes_A A/\mathfrak{m})=\Spec(A[t]/(t^2,t^3))=\Spec(A[t]/(t^2))\]

That is, the fiber itself becomes a single point, but the scheme structure given on it is non-reduced, whereas the origin \(\Spec A/\mathfrak{m}\) of \(\Spec A\) is a reduced single point as the spectrum of a field, so the above fiber cannot be equal to this point. On the other hand, on the open set \(D(\x)\) excluding the origin, \(\x=t^2\) becomes invertible, so \(t=t^3\cdot(t^2)^{-1}\) comes in and

\[A[\x^{-1}]=\tilde{A}[\x^{-1}]\]

so the two schemes are completely the same away from the origin.

To examine what happens at the origin a bit more algebraically, let us look at the local ring. First, the preimage of the origin \(\mathfrak{m}\) of \(\Spec A\) is by definition the prime ideals of \(A[t]\) containing \(\mathfrak{m}\), and the prime ideals containing \(\mathfrak{m}A[t]=(t^2)\) are the radical \((t)\) of this ideal. Then the local ring of \(A[t]\) at the origin \((t)\) is

\[A[t]_{(t)}=\mathbb{K}[t]_{(t)}\]

whereas the local ring at the origin of the original curve \(\Spec A\) is

\[A_{\mathfrak{m}}=\mathbb{K}[t^2,t^3]_{(t^2, t^3)}\]

Comparing these reveals algebraically what normalization does at the origin. \(A_{\mathfrak{m}}\) is a \(1\)-dimensional local ring that cannot have its maximal ideal generated by a single element and requires two elements \(t^2\) and \(t^3\), so it is not a regular local ring. ([Commutative Algebra] §Dimension, ⁋Definition 12) Indeed \(\mathfrak{m}/\mathfrak{m}^2\) is a \(2\)-dimensional vector space generated by the images of \(t^2\) and \(t^3\), which is the same phenomenon as the tangent space at the cusp’s origin being computed as \(2\)-dimensional, larger than the dimension of the curve, in [Algebraic Varieties] §Tangent Spaces and Smoothness, ⁋Example 7. On the other hand, the local ring \(A[t]_{(t)}=\mathbb{K}[t]_{(t)}\) of the normalization is a regular local ring whose maximal ideal is generated by the single element \(t\). That is, normalization replaces the singular local ring \(A_{\mathfrak{m}}\) with the regular local ring \(A[t]_{(t)}\), smoothing out the cusp.

For any integral scheme \(X\), normalization can be defined in the same way. Cover \(X\) by affine opens \(\Spec A_i\). Since \(X\) is integral, it has a unique generic point \(\xi\), which corresponds in each \(\Spec A_i\) to the minimal prime \((0)\) of the domain \(A_i\), so that its stalk becomes \(\Frac(A_i)\). Since the stalk \(\mathcal{O}_{X,\xi}\) is the same no matter which affine open we compute it in, all \(\Frac(A_i)\) coincide as a common function field \(K(X)\) (§Properties of Scheme Morphisms, §§Rational Maps), and we can take the normalization \(\tilde{A}_i\) of each \(A_i\) inside \(K(X)\). At this time, since normalization commutes with localization ([Commutative Algebra] §Integral Extension, ⁋Proposition 12), the restrictions of each \(\Spec\tilde{A}_i\) to the overlaps \(\Spec A_i\cap\Spec A_j\) agree with each other, and therefore they glue together into a single scheme \(\tilde{X}\) defining the normalization morphism \(\tilde{X}\rightarrow X\). This morphism is an integral morphism since affine-locally \(A_i\hookrightarrow\tilde{A}_i\) is an integral extension, and by Proposition 5 we have \(\dim\Spec\tilde{A}_i=\dim\Spec A_i\) in each piece, so from Proposition 2 we obtain \(\dim\tilde{X}=\dim X\).

We now define codimension.

Definition 7 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 \(X\)

\[Z_n\supsetneq Z_{n-1}\supsetneq\cdots\supsetneq Z_0=\cl_X(Y)\]

Then one can verify that the codimension of a prime ideal \(\mathfrak{p}\) of a ring \(A\) equals the codimension of the point \(\mathfrak{p}\) in \(\Spec A\). ([Commutative Algebra] §Dimension, ⁋Definition 2)

Proposition 8 For an irreducible closed subset \(Y\) of \(X\) and its generic point \(\eta\), \(\codim_X Y=\dim \mathcal{O}_{X,\eta}\) holds.

Proof

Since \(Y\) has generic point \(\eta\), by definition \(\codim_XY\) and \(\codim_X\{\eta\}\) are equal. Now choose any affine open subset \(U\cong\Spec A\) containing \(\eta\), and let \(\eta\in U\) correspond to \(\mathfrak{p}_\eta\in \Spec A\) under this isomorphism. Then from [Topology] §Dimension, ⁋Proposition 15 we know that there is a one-to-one correspondence between irreducible closed subsets of \(X\) meeting \(U\) and irreducible closed subsets of \(U\). That is, \(\codim_X\{\eta\}=\codim_U \mathfrak{p}_\eta\). Now we obtain the desired result from [Spectrums] §Spectrum, ⁋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 15 in place of [Commutative Algebra] §Localization, ⁋Proposition 8, one can verify that for a scheme \(X\) and an irreducible closed subset \(Y\) of \(X\), the following inequality

\[\dim Y+\codim_XY\leq \dim X\]

holds. However, similarly, equality does not hold in general.

Noether Normalization

We now prove the following important result.

Theorem 9 (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 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 since the images of these \(y_1,\ldots, y_m\) in \(\Frac(A)\) generate \(\Frac(A)\) as a field extension of \(\mathbb{K}\), we must have \(m\geq n\).

Now if \(m=n\), the images of the \(y_i\) form a transcendence basis of \(\Frac(A)\), so in particular they are algebraically independent, and therefore \(\mathfrak{p}=0\), that is, \(A=\mathbb{K}[y_1,\ldots, y_n]\). Indeed, a nonzero element of \(\mathfrak{p}\) would give a nontrivial algebraic relation among the \(y_i\), causing the transcendence degree to become less than \(n\). That is, in this case the \(y_i\) are exactly the desired elements, so there is nothing more to prove. Now to prove the given claim, assume \(m>n\), and suppose the theorem holds for any \(k\) with \(n\leq k< m\). Then from the assumption \(m>n\), the elements \(y_1,\ldots, y_m\) are algebraically dependent. That is, there exists a polynomial \(f\) in \(m\) variables with coefficients in \(\mathbb{K}\)

\[f(\x_1,\ldots, \x_m)=\sum \alpha_{d_1d_2\cdots d_m}\x_1^{d_1}\cdots\x_m^{d_m}\in \mathbb{K}[\x_1,\ldots, \x_m]\tag{$\ast$}\]

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 substituting

\[\x_1=z_1+y_m^{r_1},\quad \ldots\quad,\quad \x_{m-1}=z_{m-1}+y_m^{r_{m-1}},\quad \x_m=y_m\]

into each monomial \(\alpha_{d_1d_2\cdots d_m}\x_1^{d_1}\cdots\x_m^{d_m}\) constituting \(f\) in (\(\ast\)) and expanding, 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 \(z_k\). Now choosing an integer \(r\) larger than the maximum of the exponents \(d_j\) actually appearing in \(f\) and setting \(r_i=r^i\), by the uniqueness of base-\(r\) expansion the exponents

\[r_1d_1+\cdots+r_{m-1}d_{m-1}+d_m=d_m+d_1r+\cdots+d_{m-1}r^{m-1}\]

take distinct values for different monomials of \(f\), so exactly one such term remains as the leading term. Its coefficient is a nonzero element of \(\mathbb{K}\), so we can divide both sides by it, and therefore the above equality (\(\ast\ast\)) shows that \(y_m\) is integrally dependent on \(z_1,\ldots, z_{m-1}\).

On the other hand, let \(A'\) be the \(\mathbb{K}\)-subalgebra of \(A\) generated by \(z_1,\ldots, z_{m-1}\), that is, the \(\mathbb{K}\)-subalgebra \(A'\) of \(A\) where the coefficients exist when viewing (\(\ast\ast\)) as a polynomial in \(y_m\) with one variable. By the above argument \(A\) is a finite \(A'\)-module, and therefore \(\Frac(A)\) is an algebraic extension of \(\Frac(A')\), so \(\trdeg_\mathbb{K}\Frac(A')=n\). Then since \(A'\) is an integral domain generated by \(m-1\) elements, by the inductive hypothesis there exist \(x_1,\ldots, x_n\in A'\) satisfying the desired condition, and since \(A'\) is a finite \(\mathbb{K}[x_1,\ldots, x_n]\)-module, \(A\) is also a finite \(\mathbb{K}[x_1,\ldots, x_n]\)-module.

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 the result of the above theorem is the same as finding a finite scheme morphism \(\Spec A \rightarrow \Spec \mathbb{K}[x_1,\ldots, x_n]\) geometrically. Now since the finite extension \(\mathbb{K}[x_1,\ldots, x_n] \rightarrow A\) is an integral extension, by Proposition 5 we have \(\dim A=\dim \mathbb{K}[x_1,\ldots, x_n]\), so by [Commutative Algebra] §System of Parameters, ⁋Corollary 11 we obtain the following result.

Proposition 10 Let \(\mathbb{K}\) be an arbitrary field and \(A\) a finitely generated \(\mathbb{K}\)-algebra. If \(A\) is an integral domain, then \(\dim\Spec A=\trdeg_\mathbb{K} \Frac(A)\) holds.

The results from [Commutative Algebra] §Integral Extensions and Ideals are of course the most importantly used results in the above claims. On the other hand, using the dimension formula [Commutative Algebra] §Noether Normalization, ⁋Theorem 4, we obtain the following.

Proposition 12 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 nonzero non-unit, then \(\dim A/(f)=\dim A-1\) holds.

Proof

Choose a minimal prime \(\mathfrak{p}\) of \(A\) containing \((f)\). By [Commutative Algebra] §Dimension, ⁋Theorem 6 we have \(\operatorname{ht}\mathfrak{p}\leq 1\), and since \(A\) is a domain and \(f\neq 0\), from \((0)\subsetneq\mathfrak{p}\) we have \(\operatorname{ht}\mathfrak{p}\geq 1\). Therefore \(\operatorname{ht}\mathfrak{p}=1\), and by the dimension formula [Commutative Algebra] §Noether Normalization, ⁋Theorem 4 we have \(\dim A/\mathfrak{p}=\dim A-1\). Now \(\dim A/(f)\) is given as the maximum of \(\dim A/\mathfrak{p}\) over the minimal primes \(\mathfrak{p}\) of \((f)\), and since all minimal primes have height \(1\) as above, all these values are \(\dim A-1\), and therefore \(\dim A/(f)=\dim A-1\).

Principal Ideal Theorem

Earlier we saw that for a finite type affine integral \(\mathbb{K}\)-scheme \(X=\Spec A\), the closed subscheme \(Z(f)\) defined by a nonzero non-unit \(f\) of \(A\) has dimension one less than \(X\). This is clearly a useful result, but we can examine its consequences in more general cases as follows.

Proposition 13 For a locally Noetherian scheme \(X\) and a function \(f\) on \(X\), the irreducible components of \(Z(f)\) have codimension \(0\) or \(1\).

Proof

Let \(W\) be an irreducible component of \(Z(f)\) and \(w\) the generic point of \(W\). Now choose an affine open subset \(U\cong\Spec A\) containing \(w\); since \(X\) is locally Noetherian we can take \(A\) to be a Noetherian ring, and let \(w\) correspond to \(\mathfrak{p}\in\Spec A\) under this isomorphism. By the correspondence of [Topology] §Dimension, ⁋Proposition 15, \(W\cap U\) is an irreducible component of \(Z(f\vert_U)\), so \(\mathfrak{p}\) is a minimal prime ideal containing the principal ideal generated by \(f\vert_U\in A\). Therefore by [Commutative Algebra] §Dimension, ⁋Theorem 6 we have \(\codim\mathfrak{p}\leq 1\).

On the other hand, since the stalk depends only on an open neighborhood of \(w\), we have \(\mathcal{O}_{U,w}=\mathcal{O}_{X,w}\), and since \(W\) and \(W\cap U\) are irreducible closed subsets of \(X\) and \(U\) respectively both having \(w\) as generic point, applying Proposition 8 twice yields

\[\codim_XW=\dim\mathcal{O}_{X,w}=\dim\mathcal{O}_{U,w}=\codim_U(W\cap U)\]

Now since the codimension of the point \(\mathfrak{p}\) in \(\Spec A\) equals \(\codim\mathfrak{p}\) in the ring \(A\) ([Commutative Algebra] §Dimension, ⁋Definition 2), we eventually have \(\codim_XW=\codim\mathfrak{p}\leq 1\).

The condition \(\codim_XW=0\) means that \(W\) is an irreducible component of \(X\) itself, that is, \(f\) vanishes identically on that component. Therefore if \(f\) does not vanish identically on any irreducible component of \(X\), then all components of \(Z(f)\) have codimension exactly \(1\), and this is the role played by the assumptions in Proposition 12 that \(A\) is an integral domain and \(f\) is nonzero.

---

References

[AM] M. F. Atiyah and I. G. Macdonald, Introduction to commutative algebra, Addison-Wesley, 1969.
[Vak] R. Vakil, The rising sea: Foundation of algebraic geometry. Available online.