Proof. This is proven in [Reference Eisenbud28, Proposition 7.12] for any ring R which is complete with respect to some filtration $\mathfrak {m}_i$.

Question 2.8. Does the converse of Proposition 2.7 hold? That is, is the initial ideal $\operatorname {\mathrm {in}}(\mathfrak {a})$ finitely generated for any finitely generated ideal $\mathfrak {a}$ of $\widehat {P}$? Alternatively, is any finitely generated $\mathfrak {a}$ already closed inside $\widehat {P}$?

Proof. First, note that for every $n \ge 1$ the composition $K \to \widehat {A} \to \widehat {A}/\widehat {\mathfrak {m}^n}$ is injective (since K maps isomorphically to the residue field $k = \widehat {A}/\widehat {\mathfrak {m}}$) and hence gives an embedding $K \subset A/\mathfrak {m}^n$ via the natural isomorphism $\widehat {A}/\widehat {\mathfrak {m}^n} \simeq A/\mathfrak {m}^n$. Letting $P = k[x_i \mid i \in I]$ and $\mathfrak {n} = (x_i \mid i \in I) \subset P$, we have compatible homomorphisms $\tau _n \colon P/\mathfrak {n}^n \to A/\mathfrak {m}^n$ such that $\tau _n(k) = K$ and $\tau _n(x_i) = a_i+\mathfrak {m}^n$. Taking limits, these maps define $\tau $ and determine it uniquely. By construction, $\operatorname {\mathrm {gr}}(\tau )$ is surjective, and hence $\tau $ is surjective by [Reference Bourbaki7, Chapter III, Section 2.8, Corollary 2]. Since $\tau $ induces an isomorphism at the level of continuous Zariski cotangent spaces, we see that it is an efficient formal embedding. For the last statement, notice that if $\tau $ is an efficient formal embedding, then clearly $K = \tau (k)$ is a formal coefficient field and $\tau (x_i)$, $i\in I$, are formal coordinates.

Proof. See [Reference Grothendieck32, Chapter ${0}_{\mathsf{IV}}$, Proposition 19.3.10].

Proof. Note that a basis for the topology on $S \otimes _k k[[t_i \mid i \in I]]$ is given by the filtration

$$ \begin{align*} \mathfrak{m}_n:=\sum_{d+e=n} \mathfrak{n}^d + \big((t_i \mid i\in I)^e\big)\widehat{\phantom{t}}. \end{align*} $$

Thus it follows that for the associated graded rings we have

$$ \begin{align*} \operatorname{\mathrm{gr}}\left(S \hat{\otimes}_k k[[t_i \mid i\in I]]\right)\simeq\operatorname{\mathrm{gr}}(S \otimes_k k[t_i \mid i\in I])\simeq\operatorname{\mathrm{gr}}(S)\otimes_k k[t_i \mid i\in I]. \end{align*} $$

The map $\varphi $ induces a $\operatorname {\mathrm {gr}}(S)$-algebra map

$$ \begin{align*} \operatorname{\mathrm{gr}}(S)[t_i \mid i\in I] \to \operatorname{\mathrm{gr}}(S)[z_i \mid i\in I], \end{align*} $$

which by assumption is an isomorphism. Thus we can use [Reference Artin3, Lemma 10.23] to see that $\varphi $ is bijective. It is easy to check that $\varphi ^{-1}$ is continuous, and we are done.

Proof of Proposition 3.9

Let $k_0$ be the prime field contained in $\widehat {A}$. Notice that k is formally smooth over $k_0$. We apply Proposition 3.10 in the situation in which $C = \widehat {P}$, $\mathcal {I} = \ker (\tau )$ and $u \colon k \to \widehat {A} = C/\mathcal {I}$ is the map such that $u(k) = K'$. Notice that $\mathcal {I} = \tau ^{-1}(0)$ is closed because $\widehat {A}$ is separated, and that $\{\mathcal {I}^n\}$ tends to zero because $\mathcal {I}^n \subset \widehat {\mathfrak {n}^n}$. We get a map $v \colon k \to \widehat {P}$ verifying $\tau (v(k)) = K'$.

Since $\tau $ is surjective, there exist power series $f_i \in \widehat {P}$ such that $\tau (f_i) = a_i'$. The map $\varphi $ is given by $\varphi (x_i' ) = f_i$ and $\varphi \rvert _k = v$. Lemma 3.12 shows that $\varphi $ is an isomorphism.

Proof. Let $\tau \colon \widehat {P} \to \widehat {A}$ be a formal embedding, and write $P = k[x_i \mid i \in I]$ and $\mathfrak {n} = (x_i \mid i\in I) \subset P$. Since $\tau $ is continuous, we have that $\tau (\widehat {\mathfrak {n}}^c ) \subset \widehat {\mathfrak {m}}$ for some c. As $\widehat {\mathfrak {m}}$ is maximal, this forces $\tau (\widehat {\mathfrak {n}} ) \subset \widehat {\mathfrak {m}}$, and continuity gives $\tau (\widehat {\mathfrak {n}^n} ) \subset \widehat {\mathfrak {m}^n}$ for all n. Hence we get an induced map at the level of graded rings $\operatorname {\mathrm {gr}}(\tau ) \colon P \to \operatorname {\mathrm {gr}}(A)$. Since $\tau $ is surjective, $\operatorname {\mathrm {gr}}(\tau )$ is also surjective and $\tau (\widehat {\mathfrak {n}^n} ) = \widehat {\mathfrak {m}^n}$ for every n. In particular, $\tau $ induces a surjection at the level of Zariski cotangent spaces $\mathfrak {n}/\mathfrak {n}^2 \to \mathfrak {m}/\mathfrak {m}^2$, and we see that $\mathrm {edim}(A) \leq \dim \widehat {P}$. If $\tau $ is an efficient formal embedding, then the map $\mathfrak {n}/\mathfrak {n}^2 \to \mathfrak {m}/\mathfrak {m}^2$ is an isomorphism and we have $\mathrm {edim}(A) = \dim \widehat {P}$.

Proof. We may assume that X is projective and embedded in ${\mathbb P}^n$ for some $n> d$. Denote by $\bar {k}$ the algebraic closure of k and write $\bar {X}:=X\times _k \operatorname {\mathrm {Spec}} (\bar {k} )$ and $\bar {x}$ for the $\bar {k}$-point on $\bar {X}$ corresponding to x. As $\mathcal {O}_{\bar {X},\bar {x}}$ is not a regular ring, we have

$$ \begin{align*} \dim_{\bar{x}}\left(\bar{X}\right)<\mathrm{edim}\left(\bar{X},\bar{x}\right)=\mathrm{edim}(X,x)=d. \end{align*} $$

Suppose we can find a linear projection $\pi \colon {\mathbb P}^n \to {\mathbb P}^d$ defined over k such that if $\bar {Y}$ denotes the scheme-theoretic image of $\bar {X}$ under $\pi $ and $\bar {y}=\pi (\bar {x})$, then the induced map $\mathcal {O}_{\bar {Y},\bar {y}}\to \mathcal {O}_{\bar {X},\bar {x}}$ is an isomorphism. Since $\bar {Y}=Y\times _k \operatorname {\mathrm {Spec}} (\bar {k} )$, where Y is the scheme-theoretic image of X under the linear projection centred at x, we obtain a map $\mathcal {O}_{Y,y}\to \mathcal {O}_{X,x}$ whose base change to $\bar {k}$ gives the foregoing map. Thus, by faithfully flat descent, we get that $\mathcal {O}_{Y,y}\simeq \mathcal {O}_{X,x}$.

Now, in order to prove the claim, let $\bar {T}\subset {\mathbb P}^n_{\bar {k}}$ be the unique linear space passing through $\bar {x}$ whose tangent space at $\bar {x}$ agrees with that of $\bar {X}$. Furthermore, let $\bar {S}$ be the closure of the set of all lines connecting $\bar {z}$ with $\bar {x}$, where $\bar {z}\in \bar {X}$, $\bar {z}\neq \bar {x}$. Note that $\dim (\bar {S} )=\dim (\bar {X} )+1\leq d$. Consider now the closure $\bar {Z}$ of the set $\bar {X}\cup \bar {T} \cup \bar {S}$, equipped with its reduced scheme structure. Since $\dim (\bar {S} )\leq d$ the set of all linear spaces $\bar {L}$ with $\bar {L}\cap \bar {Z}=\emptyset $ is open inside $\operatorname {\mathrm {Gr}}(n-d-1,n)\times _k \operatorname {\mathrm {Spec}} (\bar {k} )$. The preimage of this set in $\operatorname {\mathrm {Gr}}(n-d-1,n)$ is a nonempty open set, and since k is infinite, it has a k-rational point, which we denote by L. Hence we have that the corresponding projection $\pi _L\colon {\mathbb P}^n\to {\mathbb P}^d$, defined over k, satisfies $\pi _{\bar {L}}^{-1} (\bar {y} ) \cap \bar {X} = \{\bar {x} \}$ set-theoretically, where $\bar {y}$ corresponds to the k-point $y:=\pi _L(x)$. Writing $\bar {Y}:=\pi _{\bar {L}} (\bar {X} )$, we get that the map of local rings $\mathcal {O}_{\bar {Y},\bar {y}}\to \mathcal {O}_{\bar {X},\bar {x}}$ is injective and finite. Since $\bar {L}\cap \bar {T}=\emptyset $, the tangent spaces of $\bar {x}$ and $\bar {y}$ are isomorphic and thus $\mathfrak {m}_{\bar {y}}\mathcal {O}_{\bar {X},\bar {x}}=\mathfrak {m}_{\bar {x}}$. The claim now follows from the Nakayama lemma.

Proof. This is immediate from the definitions. Compare with [Reference Bourbaki7, Chapter III, Section 3.1, Theorem 1] or [Reference Matsumura36, Theorem 8.5] or [Reference Artin3, Lemma 10.8], but notice that no finite generation hypotheses are needed for the statement of the lemma.

Proof. Let $\mathfrak {a}$ be a finitely generated ideal of A. Since A is coherent, there exists an exact sequence

Moreover, since the Artin–Rees property holds for $\ker \varphi \subset A^q$, the $\mathfrak {m}$-adic topology on $\ker \varphi $ agrees with the one induced by the inclusion $\ker \varphi \subset A^q$. From [Reference Bourbaki7, Chapter III, Section 2.12, Lemma 2] or [Reference Artin3, Lemma 10.3], the sequence remains exact after taking $\mathfrak {m}$-adic completions, and we have a commutative diagram

with exact rows. Since taking completion commutes with finite direct sums, the map $\mathfrak {a} \otimes _A \widehat {A} \to \widehat {\mathfrak {a}}$ is an isomorphism. As the natural map $\widehat {\mathfrak {a}} \to \widehat {A}$ is an injection, the flatness of $A \to \widehat {A}$ follows from [Reference Matsumura36, Theorem 7.7]. The fact that $\mathfrak {a} \otimes _A \widehat {A} \to \widehat {\mathfrak {a}}$ is an isomorphism also shows that $\mathfrak {a}\widehat {A} = \widehat {\mathfrak {a}}$.

Proof. Let E be a finitely generated free P-module and $F \subseteq E$ a finitely generated submodule. Assume that E is freely generated by $e_1, \dotsc , e_s$.

Given any subset $J \subseteq I$, we write $P_J := S[x_i \mid i \in J]$, and for any ideal $\mathfrak {a} \subseteq P$ we denote $\mathfrak {a}_J := \mathfrak {a} \cap P_J$. We define $E_J := P_J \cdot e_1 \oplus \dotsb \oplus P_J \cdot e_s$, and for any P-submodule $G \subseteq E$ we write $G_J := E_J \cap G$. Note that $P, \mathfrak {m}, \mathfrak {a}, E, G$ are the colimits of $P_J, \mathfrak {m}_J, \mathfrak {a}_J, E_J, G_J$ for $J \subseteq I$ finite. We have

$$ \begin{align*} G_J \cap G^{\prime}_J = (G \cap G')_J, \qquad \mathfrak{a}_J G_J \subseteq (\mathfrak{a} G)_J, \qquad \mathfrak{a}_J E_J = (\mathfrak{a} E)_J, \qquad (\mathfrak{m}_J)^n = (\mathfrak{m}^n)_J. \end{align*} $$

In particular, for all $n, d \in {\mathbb N}$ with $n> d$, we have

$$ \begin{align*} \mathfrak{m}_J^n E_J \cap F_J = (\mathfrak{m}^n E \cap F)_J \qquad\text{and}\qquad \mathfrak{m}_J^{n-d}\left(\mathfrak{m}_J^d E_J \cap F_J\right) \subseteq \left(\mathfrak{m}^{n-d}\left(\mathfrak{m}^d E \cap F\right)\right)_J. \end{align*} $$

Assume that F is generated by $f_1, \dotsc , f_r$. Then there exists a finite set $L \subset I$ such that $f_1,\dotsc ,f_r \in F_L$, and for any J with $L \subseteq J \subseteq I$ we have $F_J = P_J \cdot f_1 + \dotsb + P_J \cdot f_r = P_J \cdot F_L$.

Since $P_L$ is Noetherian, it has the Artin–Rees property with respect to $\mathfrak {m}_L$, and hence there exists a $c \in {\mathbb N}$ such that

$$ \begin{align*} \mathfrak{m}_L^n E_L \cap F_L = \mathfrak{m}_L^{n-c}\left(\mathfrak{m}_L^c E_L \cap F_L\right) \end{align*} $$

for all $n> c$. The smallest such c is the Artin–Rees index of $F_L \subseteq E_L$. Since for any finite set J with $L \subseteq J \subset I$ we have $F_J = P_J \cdot F_L$, we can apply Lemma 4.6, and we see that the Artin–Rees index of $F_J \subseteq E_J$ is again c. This implies that

$$ \begin{align*} (\mathfrak{m}^n E \cap F)_J \subseteq (\mathfrak{m}^{n-c}(\mathfrak{m}^c E \cap F))_J. \end{align*} $$

Taking the colimit for all finite $J \subset I$, we get

$$ \begin{align*} \mathfrak{m}^n E \cap F \subseteq \mathfrak{m}^{n-c}(\mathfrak{m}^c E \cap F). \end{align*} $$

The reversed inclusion is immediate, and the theorem follows.

Proof. Let $c_0$ and c be the Artin–Rees indexes of $F_0 \subseteq E_0$ and $F \subseteq E$. As in Lemma 4.2, consider the Rees algebras

$$ \begin{align*} A_0^* = \bigoplus_{n \geq 0} \mathfrak{m}_0^n \qquad\text{and}\qquad A^* = \bigoplus_{n \geq 0} \mathfrak{m}^n \end{align*} $$

and the graded modules

$$ \begin{align*} F^*_0 = \bigoplus_{n \geq 0} \mathfrak{m}_0^n E_0 \cap F_0 \qquad\text{and}\qquad F^* = \bigoplus_{n \geq 0} \mathfrak{m}^n E \cap F. \end{align*} $$

Then $F_0$ is generated in degree $\leq c_0$ as a graded $A_0$-algebra (and not in any lower degree), and similarly for F.

Any element $f \in \mathfrak {m}^n E \cap F$ can be written as $f = \sum _{i=0}^n f_i z^{n-i}$, where $f_i \in \mathfrak {m}_0^i E_0 \cap F_0$. In particular, $F^*$ is generated by $F^*_0$ as an $A^*$-algebra, and therefore $c \leq c_0$. Conversely, if $F^*$ is generated by homogeneous elements $f^{(1)}, \dotsc , f^{(r)}$ with $f^{(j)} = \sum _i f_i^{(j)} z^{n_j -i}$, then $F^*_0$ is generated by $f^{(1)}_{n_1}, \ldots , f^{(r)}_{n_r}$. We see that $c_0 \leq c$, and the result follows.

Proof. Observe that a basis for the topology on $T[x_i \mid i\in I]$ is given by

$$ \begin{align*} \mathfrak{n}^m[x_i \mid i\in I]+(x_i \mid i\in I)^n,\quad m,n\in{\mathbb N}, \end{align*} $$

which is easily seen to be equivalent to the $\mathfrak {m}$-adic one, where $\mathfrak {m}:=(x_i \mid i\in I)+\mathfrak {n}$. As $T[x_i \mid i \in I]$ is coherent (see, for example, [Reference Glaz30, Theorem 6.2.2]), the first assertion follows from Theorem 4.5 and Proposition 4.4. Regarding the last two assertions, (1) follows by observing that $S \hat {\otimes }_S S[x_i \mid i\in I] = S[[x_i \mid i\in I]]$, and (2) by taking $T = S [ [x_j \mid j\in J ] ]$ with the $(x_j \mid j\in J)$-adic topology and observing that the given inclusion factors as

$$ \begin{align*} S\left[\left[x_j \mid j\in J\right]\right] \to S\left[\left[x_j \mid j\in J\right]\right][x_i \mid i\in I \setminus J] \to S[[x_i \mid i\in I]], \end{align*} $$

and so is flat.

Proof. Let $f_1,\dotsc ,f_r\in \mathfrak {a}$ be such that $\operatorname {\mathrm {in}}(f_1),\dotsc ,\operatorname {\mathrm {in}}(f_r)$ generate $\operatorname {\mathrm {in}}(\mathfrak {a})$. Since $\operatorname {\mathrm {in}}(\mathfrak {a})=\operatorname {\mathrm {in}} (\widehat {\mathfrak {a}} )$, we can apply Proposition 2.7 to see that $\widehat {\mathfrak {a}}=(f_1,\dotsc ,f_r)\widehat {P}$. By Corollary 4.8, the map $P_{\mathfrak {m}}\to \widehat {P}$ is faithfully flat and thus $\mathfrak {a}\subset \widehat {\mathfrak {a}}\cap P_{\mathfrak {m}}=(f_1,\dotsc ,f_r)P_{\mathfrak {m}}$. The other inclusion is trivial, so $\mathfrak {a}=(f_1,\dotsc ,f_r)$.

Proof. Choose a local monomial order $\prec $ compatible with the standard filtration on $\widehat {P}$; for example, take the order defined by $x^a \prec x^b$ if $x^b <_{\operatorname {\mathrm {grlex}}} x^a$, where $<_{\operatorname {\mathrm {grlex}}}$ denotes the usual graded lexicographic order. Then $\prec $ restricts to a local monomial order $\prec _J$ on $\widehat {P}_J$. Choose a standard basis $S = \{f_1,\dotsc ,f_r\}$ of $\mathfrak {a}_J$ with respect to $\prec _J$. By [Reference Becker4, Theorem 4.1], this is equivalent to S being closed under s-series. Note that [Reference Becker4, Theorem 4.1] extends directly to the case of infinitely many indeterminates, and thus it follows immediately that S is a standard basis of $\mathfrak {a}$ with respect to $\prec $. Clearly we have $\operatorname {\mathrm {in}}(\mathfrak {a}) = (\operatorname {\mathrm {in}}(f_1),\dotsc ,\operatorname {\mathrm {in}}(f_r))$, which proves the claim.

Proof. Notice that if $\mathfrak {p} \subset \widehat {P}$ is a prime ideal, then $\mathfrak {p}_J \subset \widehat {P}_J$ remains prime. Moreover, by Corollary 4.8 we have that $\widehat {P}_J\to \widehat {P}$ is faithfully flat and thus $\mathfrak {q} = (\mathfrak {q}\widehat {P} ) \cap \widehat {P}_J$ for any ideal $\mathfrak {q} \subset \widehat {P}_J$. It is therefore sufficient to show that for every prime ideal $\mathfrak {q} \subset \widehat {P}_J$, the extension $\mathfrak {q} \widehat {P}$ is prime. By Remark 2.4 we have an injection $\widehat {P} \to (\widehat {P}_J )^{{\mathbb N}^{ (I \setminus J )}}$. Since J is finite, $\widehat {P}_J$ is Noetherian and $\mathfrak {q}$ is finitely generated. This implies that $\mathfrak {q} (\widehat {P}_J )^{{\mathbb N}^{ (I \setminus J )}} = \mathfrak {q}^{{\mathbb N}^{ (I \setminus J )}}$ – that is, the elements of the extension $\mathfrak {q} (\widehat {P}_J )^{{\mathbb N}^{ (I \setminus J )}}$ are precisely the Bourbaki power series that, when expanded in the variables indexed by $I \setminus J$, have coefficients in $\mathfrak {q}$. Therefore we have an injection

$$ \begin{align*} \widehat{P}/\mathfrak{q} \widehat{P} \hookrightarrow \left(\widehat{P}_J/\mathfrak{q}\right)^{{\mathbb N}^{\left(I \setminus J\right)}}, \end{align*} $$

and the ring in the right-hand side is clearly a domain. Thus $\mathfrak {q} \widehat {P}$ is prime.

Proof. From Proposition 5.7 we can assume that $\mathfrak {a} = \mathfrak {p} = \mathfrak {p}_J \widehat {P}$ is a prime ideal. Notice that $\mathfrak {p} \subset \mathfrak {b} := (x_j \mid j\in J )$. From Proposition 5.13, the localisation $\widehat {P}_{\mathfrak {b}}$ is Noetherian, and therefore $\widehat {P}_{\mathfrak {p}}$, which is a further localisation of $\widehat {P}_{\mathfrak {b}}$, is also Noetherian. By Corollary 4.8, the extension $\widehat {P}_J \subset \widehat {P}$ is flat, and therefore the extension $ (\widehat {P}_J )_{\mathfrak {p}_J} \subset \widehat {P}_{\mathfrak {p}}$ is also flat. Since $\widehat {P}_{\mathfrak {p}}$ is Noetherian, it follows from [Reference Matsumura36, Theorem 15.1] that $\operatorname {\mathrm {ht}}(\mathfrak {p}) = \operatorname {\mathrm {ht}}(\mathfrak {p}_J)$.

Proof. As discussed in Remark 2.5, since J is finite we have an isomorphism

$$ \begin{align*} \widehat{P} \simeq k[[x_i \mid i\in I\setminus J]]\left[\left[x_j \mid j\in J\right]\right]. \end{align*} $$

The proposition now follows from the next lemma:

Proof. If f is already $y_{n+1}$-regular, then we are done. If not, then pick any monomial of the form $x_{i_1}^{d_1}\dotsm x_{i_r}^{d_r} y_1^{e_1}\dotsm y_{n+1}^{e_{n+1}}$ appearing in the expansion of f. Then decompose f as

$$ \begin{align*} f=f'+f'', \quad f'\in k[[x_{i_1},\dotsc,x_{i_r},y_1,\dotsc,y_{n+1}]], \end{align*} $$

such that $f'$ cannot be decomposed further. By [Reference Asgharzadeh and Tousi1, Lemma 6.11] there exist new coordinates $x^{\prime }_{i_j} = x_{i_j}+y_{n+1}^{a_j}$, $y^{\prime }_l = y_l+y_{n+1}^{b_l}$ and $y^{\prime }_{n+1} := y_{n+1}$ such that $f' (x^{\prime }_{i_j},y^{\prime }_l )$ is $y^{\prime }_{n+1}$-regular. We may extend this change of coordinates trivially to a continuous automorphism $\varphi \colon \widehat {P}_{n+1} \to \widehat {P}_{n+1}$ by setting $x^{\prime }_i=x_i$ for all indices i that are different from $i_j$ for all j. Then clearly $\varphi (f)$ is $y_{n+1}$-regular and $\varphi $ fixes $\mathfrak {b}_{n+1}=(y_1,\dotsc ,y_{n+1})$.

Proof of Lemma 5.14

We prove the lemma by induction on n. Let $Q_n:= (\widehat {P}_n )_{\mathfrak {b}_n}$. Clearly $Q_0\simeq \operatorname {\mathrm {Quot}} (\widehat {P} )$, so let us assume that $Q_n$ is Noetherian. We have injections $Q_n \to Q_{n+1}$. Let $\mathfrak {a}$ be an ideal of $Q_{n+1}$ and let $f\in \mathfrak {a}$, $f\neq 0$. After multiplication by a unit, we may assume $f\in \widehat {P}_{n+1}$; by Lemma 5.16 we may also assume f is y-regular. Consider the ideal $\mathfrak {a}':=\mathfrak {a}\cap Q_n[y_{n+1}]$. Since $Q_n$ is Noetherian, so is $Q_n[y_{n+1}]$, and thus there exist $f_1,\dotsc ,f_r\in Q_n[y_{n+1}]$ that generate $\mathfrak {a}'$. We claim that $\mathfrak {a}=(f,f_1,\dotsc ,f_r)Q_{n+1}$.

Let $g\in \mathfrak {a}$. By Theorem 5.15, there exist a unit $u\in Q_{n+1}$, $q\in \widehat {P}_{n+1}$ and $r\in \widehat {P}_n[y_{n+1}]$ such that $u g=q f+r$. Since $r\in \mathfrak {a}'$, we can find $v_1,\dotsc ,v_r\in Q_n$ such that $r=\sum _{j=1}^r v_j f_j$. Hence we have $g=u^{-1}q f+\sum _{j=1}^r u^{-1}v_j f_j$, which proves our claim.

Proof. This follows from the fact that for every polynomial ring $P = k[x_i \mid i \in I]$ and every ideal $\mathfrak {a} \subset P$, we have $\dim (P) = \dim (P/\mathfrak {a}) + \operatorname {\mathrm {ht}}(\mathfrak {a})$ (compare Remark 5.12). For the last assertion we use the fact that $\dim (\operatorname {\mathrm {gr}}(A)) = \dim (A)$ if A is Noetherian.

Proof of Proposition 6.6

We have the commutative diagram

The existence of $\sigma $ follows from the fact that $\operatorname {\mathrm {Im}}(\pi ) \otimes _{k_0}k = \operatorname {\mathrm {Sym}}_k(\operatorname {\mathrm {Im}}(\varphi ^*))$. The map $\iota $ is a linear extension of polynomial rings, and hence is faithfully flat. Since $\iota ^{-1}(\ker (\gamma )) = \ker (\sigma )$, we see that

$$ \begin{align*} \operatorname{\mathrm{ht}}(\ker(\gamma)) \geq \operatorname{\mathrm{ht}}(\ker(\sigma)) \end{align*} $$

by the going-down theorem. On the other hand,

$$ \begin{align*} \operatorname{\mathrm{ht}}(\ker(\sigma)) = \operatorname{\mathrm{rank}}(\varphi^*) - \dim(\operatorname{\mathrm{Im}}(\sigma)). \end{align*} $$

Since the inclusion $\operatorname {\mathrm {Im}}(\sigma ) \subset \operatorname {\mathrm {Im}}(\operatorname {\mathrm {gr}}(\varphi )) \otimes _{k_0}k$ is an inclusion of Noetherian local rings with the same residue field, and $\operatorname {\mathrm {Im}}(\operatorname {\mathrm {gr}}(\varphi ))$ is a quotient of $\operatorname {\mathrm {gr}}(B)$, we have

$$ \begin{align*} \dim(\operatorname{\mathrm{Im}}(\sigma)) \le \dim(\operatorname{\mathrm{gr}}(B)). \end{align*} $$

Combining these formulas, we get

$$ \begin{align*} \operatorname{\mathrm{ht}}(\ker(\gamma)) \geq \operatorname{\mathrm{rank}}(\varphi^*) - \dim(\operatorname{\mathrm{gr}}(B)). \end{align*} $$

To conclude, notice that $\dim (\operatorname {\mathrm {gr}}(B)) = \dim (B)$ if B is Noetherian.

Proof. First, observe that $A'$ is a local $k'$-algebra with maximal ideal $\mathfrak {m}' = \mathfrak {m} \otimes _k k'$. We have $\mathrm {ecodim}(A) = \operatorname {\mathrm {ht}}(\ker (\gamma ))$, where

$$ \begin{align*} \gamma \colon \operatorname{\mathrm{Sym}}_k\left(\mathfrak{m}/\mathfrak{m}^2\right) \to \operatorname{\mathrm{gr}}(A) \end{align*} $$

is defined, as at the beginning. Since for every n we have $(\mathfrak {m}')^n/(\mathfrak {m}')^{n+1} = \mathfrak {m}^n/\mathfrak {n}^{n+1} \otimes _k k'$, we see that $\gamma $ induces, by base change, the analogous map

$$ \begin{align*} \gamma' \colon \operatorname{\mathrm{Sym}}_k\left(\mathfrak{m}'/(\mathfrak{m}')^2\right) \to \operatorname{\mathrm{gr}}(A'). \end{align*} $$

The next lemma gives $\operatorname {\mathrm {ht}}(\ker (\gamma ')) = \operatorname {\mathrm {ht}}(\ker (\gamma ))$, and the assertion follows.

Proof. For short, let $\mathfrak {a}' = \mathfrak {a} P'$. If I is finite, then the lemma follows from dimension theory. In general, suppose by contradiction that $\operatorname {\mathrm {ht}}(\mathfrak {a}) \ne \operatorname {\mathrm {ht}}(\mathfrak {a}')$. Then we can find a finite subset $J \subset I$ such that $\operatorname {\mathrm {ht}}(\mathfrak {a}_J) \ne \operatorname {\mathrm {ht}} (\mathfrak {a}^{\prime }_J )$ (compare Remark 5.12). Since $\mathfrak {a}^{\prime }_J = \mathfrak {a}_J P^{\prime }_J$, this contradicts the finite-dimensional case.

Proof. Given two formal embeddings $\tau \colon \widehat {P} \to \widehat {A}$ and $\tau ' \colon \widehat {P}' \to \widehat {A}$ with $\tau $ efficient, by Remark 3.13 there is a surjection $\varphi \colon \widehat {P}' \to \widehat {P}$ such that $\tau ' = \tau \circ \varphi $, and hence $\operatorname {\mathrm {ht}}(\ker (\tau ')) \ge \operatorname {\mathrm {ht}}(\ker (\tau ))$.

Proof. Consider an efficient formal embedding $\tau \colon \widehat {P} = k[[x_i \mid i\in I]] \to \widehat {A}$. Note that $\dim (\widehat {P} ) = \mathrm {edim}(A)$ by Proposition 3.14. The first formula follows from the simple fact that $\dim (\widehat {P} ) \ge \operatorname {\mathrm {ht}}(\ker (\tau )) + \dim (\widehat {P}/\ker (\tau ) )$. If A has finite embedding dimension, then the set I is finite, and equality holds in the formula because a power series ring in finitely many variables is catenary of dimension equal to the number of variables. The second formula follows from the first and the fact that $\dim (A) = \dim (\widehat {A} )$ if A is Noetherian.

Proof. By Propositions 7.4 and 6.2, it suffices to show that $\dim (\operatorname {\mathrm {gr}}(A)) = \dim (\widehat {A} )$. By Proposition 3.7, the completion $\widehat {A}$ is the quotient of a power series ring in finitely many variables, and therefore is Noetherian and carries the $\widehat {\mathfrak {m}}$–adic topology. The result now follows from [Reference Matsumura36, Theorem 15.7] and the identification $\operatorname {\mathrm {gr}} (\widehat {A} ) \simeq \operatorname {\mathrm {gr}}(A)$.

Proof. Since B is Noetherian, we have $\mathrm {edim}(B) < \infty $. By Corollary 7.5, we have that $\mathrm {ecodim}(B) = \mathrm {fcodim}(B)$. Now we make use of the fact that $A \simeq k[[x_i \mid i \in I]]/\mathfrak {a}$ with $\mathfrak {a}$ of finite definition. Note that there exists $J \subset I$ finite such that $\mathfrak {a}$ is the extension of $\mathfrak {a}_J := \mathfrak {a} \cap k [ [x_j \mid j \in J ] ]$ and $\widehat {B} \simeq k [ [x_j \mid j \in J ] ]/\mathfrak {a}_J$. We may assume that the surjection $\tau _{B} \colon k [ [x_j \mid j \in J ] ] \to \widehat {B}$ is an efficient formal embedding; then so is $\tau _{A} \colon k[[x_i \mid i \in I]] \to \widehat {A}$. By Remark 5.3 and Theorem 5.10, it follows that $\mathrm {fcodim}(B) = \mathrm {fcodim}(A)$. It remains to show that $\mathrm {ecodim}(B) = \mathrm {ecodim}(A)$.

To that end, note that $\operatorname {\mathrm {gr}}(\tau _A)$ factors through the natural surjection $\operatorname {\mathrm {Sym}}_k (\mathfrak {m}/\mathfrak {m}^2 ) \to \operatorname {\mathrm {gr}}(A)$, and similarly for $\operatorname {\mathrm {gr}}(\tau _B)$. We have the commutative diagram

and thus the claim follows by Lemma 5.6.

Proof. Fix an efficient formal embedding $\tau \colon \widehat {P} \to \widehat {A}$, and let $\operatorname {\mathrm {gr}}(\tau ) \colon P \to \operatorname {\mathrm {gr}}(A)$ be the induced map on associated graded rings (as in Remark 2.6, we identify $\operatorname {\mathrm {gr}} (\widehat {P} ) = \operatorname {\mathrm {gr}}(P) = P$). As explained in Remark 3.8, $P \simeq \operatorname {\mathrm {Sym}}_k (\mathfrak {m}/\mathfrak {m}^2 )$ and $\operatorname {\mathrm {gr}}(\tau )$ gets identified with the canonical surjection $\gamma $. In particular, it is enough to show that

$$ \begin{align*} \operatorname{\mathrm{ht}}(\ker(\tau)) \geq \operatorname{\mathrm{ht}}(\ker(\operatorname{\mathrm{gr}}(\tau))). \end{align*} $$

Write $\mathfrak {a} = \ker (\tau )$. By [Reference Bourbaki7, Chapter III, Section 2.4, Proposition 2], we have $\ker (\operatorname {\mathrm {gr}}(\tau )) = \operatorname {\mathrm {in}}(\mathfrak {a})$. To conclude, it is therefore enough to prove that

$$ \begin{align*} \operatorname{\mathrm{ht}}(\mathfrak{a}) \ge \operatorname{\mathrm{ht}}(\operatorname{\mathrm{in}}(\mathfrak{a})). \end{align*} $$

This follows from the next proposition.

Proof. If $\operatorname {\mathrm {ht}}(\operatorname {\mathrm {in}}(\mathfrak {a})) < \infty $, then $\operatorname {\mathrm {in}}(\mathfrak {a})$ is finitely generated and there exists a finite subset $J \subset I$ such that $\operatorname {\mathrm {in}}(\mathfrak {a})_J$ has the same height as $\operatorname {\mathrm {in}}(\mathfrak {a})$ (see Remark 5.12). Otherwise, if $\operatorname {\mathrm {ht}}(\operatorname {\mathrm {in}}(\mathfrak {a})) = \infty $, then we can pick a finite $J \subset I$ such that the height of $\operatorname {\mathrm {in}}(\mathfrak {a})_J$ is arbitrary large. Define $c := \operatorname {\mathrm {ht}}(\operatorname {\mathrm {in}}(\mathfrak {a})_J)$.

By [Reference Bruns and Herzog17, Proposition 1.5.11], we can fix homogeneous elements $g_1,\dotsc ,g_c \in \operatorname {\mathrm {in}}(\mathfrak {a})_J$ forming a regular sequence in $P_J$. By the definition of initial ideal, there are elements $f_1,\dotsc ,f_c \in \mathfrak {a}$ such that $\operatorname {\mathrm {in}}(f_i) = g_i$ for all i. Let $R = R (\widehat {P} ) := \bigoplus _{n \in {\mathbb Z}} \widehat {\mathfrak {m}^n}u^{-n}$ be the extended Rees algebra of $\widehat {P}$, where we set $\widehat {\mathfrak {m}^n} = \widehat {P}$ whenever $n < 0$. For every i, let $\widetilde {f}_i := u^{-\operatorname {\mathrm {ord}} (f_i )}f_i \in R$. Note that $\widetilde {f}_i\rvert _{u=0} = \operatorname {\mathrm {in}}(f_i) = g_i$ via the identification $R/uR \simeq P$.

We claim that for every $1 \le r \le c$, the elements $\widetilde {f}_1,\dotsc ,\widetilde {f}_r$ form a regular sequence in R and $R/ (\widetilde {f}_1,\dotsc ,\widetilde {f}_r )$ is flat over $k[u]$. We argue by induction on r, the assertion being clear if $r=0$. Setting for short $B := R/ (\widetilde {f}_1,\dotsc ,\widetilde {f}_{r-1} )$, we know by induction that B is flat over $k[u]$. Assume that there exists $h\in \widehat {P}$ with $h=\sum _{i=1}^{r-1} a_if_i$ and $\operatorname {\mathrm {in}}(h)$ not divisible by $g_1,\dotsc ,g_{r-1}$. Writing

$$ \begin{align*} \sum_{i=1}^{r-1} u^{\operatorname{\mathrm{ord}}\left(f_i\right)}a_i \widetilde{f}_i=u^{\operatorname{\mathrm{ord}}(h)}\widetilde{h}, \end{align*} $$

Lemma 7.10 yields that B has torsion over $k[u]$, which gives a contradiction. Thus $\operatorname {\mathrm {in}}(f_1,\dotsc ,f_{r-1})=(g_1,\dotsc ,g_{r-1})$ and B is isomorphic to the algebra $\bigoplus _{n \in {\mathbb Z}} \mathfrak {b}_n u^{-n}$, where $\mathfrak {b}_n := (\widehat {\mathfrak {m}^n} + (f_1,\dotsc ,f_{r-1}) )/(f_1,\dotsc ,f_{r-1})$ for $n \ge 0$ and $\mathfrak {b}_n = B$ for $n < 0$. It follows by Proposition 2.7 that $\bigcap _{n \ge 1} \mathfrak {b}_n = \{0\}$. Then Lemma 7.11 implies that B is $(u)$-adically separated, and Lemma 7.12 (with $t = u$) implies that the class b of $\widetilde {f}_r$ in B is a regular element and $B/bB$ is flat over $k[u]$.

The natural isomorphism $R/(u-1)R \simeq \widehat {P}$ sends $\widetilde {f}_i$ to $f_i$, and hence we see by Lemma 7.12 (with $t = u-1$) that $f_1,\dotsc ,f_c$ form a regular sequence in $\widehat {P}$. This implies that $\operatorname {\mathrm {depth}} (\mathfrak {a},\widehat {P} ) \ge c$. We conclude using the fact that $\operatorname {\mathrm {ht}}(\mathfrak {a}) \ge \operatorname {\mathrm {depth}} (\mathfrak {a},\widehat {P} )$ (see, for example, [Reference Atiyah and Macdonald2, Proposition 2.3 and Lemma 3.2]).

Proof. Given $\widetilde {h}=\sum b_i \widetilde {f}_i$, we may assume that $b_i$ is homogeneous in R – that is, of the form $b_i=u^{-\operatorname {\mathrm {ord}}(h)+\operatorname {\mathrm {ord}} (f_i )}a_i$, with $a_i\in \widehat {P}$. But then $\operatorname {\mathrm {ord}}(a_i)\geq \operatorname {\mathrm {ord}}(h)-\operatorname {\mathrm {ord}}(f_i)$ and the claim follows.

Proof. Let $a \in R(A)$ be any element. Write $a = \sum _{i=p}^q a_i u^{-i}$ for some $a_i \in \widehat {P}$ and $p,q \in {\mathbb Z}$. By the definition of Rees algebra, we have $a_i \in \mathfrak {a}_i$ for all i. The condition that $a \in u^nR(A)$ is equivalent to having $a_i \in \mathfrak {a}_{n+i}$ for all i. If $a \in \bigcap _{n \ge 1} u^nR(A)$, then we have $a_i \in \bigcap _{n \ge 1}\mathfrak {a}_n = \{0\}$ for all i, and hence $a = 0$.

Proof. The proof is an adaptation of the proof of [Reference Matsumura36, Theorem 22.5]. The implication (1) $\Rightarrow $ (2) follows by the snake lemma applied to the commutative diagram

after observing that the map $B \to B$ given by multiplication by b is injective, since b is regular, and so is the map $B/bB \to B/bB$ given by multiplication by t, since $B/bB$ is flat over $k[t]$.

In order to prove the implication (2) $\Rightarrow $ (1) when B is $(t)$-adically separated, suppose $x \in B$ is an element such that $bx = 0$. Then $\bar {b}\bar {x} = 0$ in $B/tB$, and hence $\bar {x} = 0$. This means that $x \in tB$. Suppose $x \in t^nB$ for some positive integer n, and write $x = t^ny$ in B. Then $t^n(by) = bx = 0$ and hence $by = 0$, since B is flat over $k[t]$. This implies that $y \in tB$, and hence $x \in t^{n+1}B$. Therefore $x \in \bigcap _{n \ge 1} t^nB$, and since B is $(t)$-adically separated, this means that $x = 0$. This proves that b is a regular element. To conclude that $B/bB$ is flat over $k[t]$, we just compute $\operatorname {\mathrm {Tor}}^{k[t]}_1(k,B/bB) = 0$ from the exact sequence $0 \to B \to B \to B/bB \to 0$ and apply [Reference Matsumura36, Theorem 22.3].

Question 7.13. We do not know of any example where the inequality in Theorem 7.8 is strict. The question whether $\mathrm {ecodim}(A)=\mathrm {fcodim}(A)$ holds for all equicharacteristic local rings $(A,\mathfrak {m},k)$ is, to our knowledge, still open.

Proof. By assumption, we have that $\operatorname {\mathrm {ord}}_\alpha (\operatorname {\mathrm {Fitt}}^d (\Omega _{X/k} ) )<\infty $, and by taking a general linear projection we can ensure that $\operatorname {\mathrm {ord}}_\alpha (\operatorname {\mathrm {Fitt}}^d (\Omega _{X/k} ) )=\operatorname {\mathrm {ord}}_\alpha (\operatorname {\mathrm {Fitt}}^0 (\Omega _{X/Y} ) )$.

Since k is perfect, we have a commutative diagram with exact rows

The main step is to understand the map $\varphi $. As in the proof of [Reference de Fernex and Docampo21, Theorem 9.2], denote for short $B_L := L[[t]]$ and $P_L := L(\:\!\!(t)\:\!\!)/tL[[t]]$.

Note that, by [Reference de Fernex and Docampo21, Theorem 5.3], there are natural isomorphisms

$$ \begin{align*} \Omega_{X_{\infty}/k} \otimes_{\mathcal{O}_{X_{\infty}}} L \simeq \Omega_{X/k} \otimes_{\mathcal{O}_X} P_L \end{align*} $$

and

$$ \begin{align*} \Omega_{Y_{\infty}/k} \otimes_{\mathcal{O}_{Y_{\infty}}} L \simeq \Omega_{Y/k} \otimes_{\mathcal{O}_Y} P_L. \end{align*} $$

We will use these isomorphisms to study $\varphi $.

By pulling back the terms of the exact sequence

$$ \begin{align*} \Omega_{Y/k} \otimes_{O_Y} \mathcal{O}_X \to \Omega_{X/k} \to \Omega_{X/Y} \to 0 \end{align*} $$

along $\alpha $, we obtain the exact sequence

$$ \begin{align*} \Omega_{Y/k} \otimes_{\mathcal{O}_Y} B_L \to \Omega_{X/k} \otimes_{\mathcal{O}_X} B_L \to \Omega_{X/Y} \otimes_{\mathcal{O}_X} B_L \to 0. \end{align*} $$

Since Y is smooth, we see that the term $F_Y := \Omega _{Y/k} \otimes _{\mathcal {O}_Y} B_L$ is a free $B_L$-module. Write $\Omega _{X/k} \otimes _{\mathcal {O}_X} B_L = F_X \oplus T_X$, where $F_X$ is free and $T_X$ is torsion. Since $\operatorname {\mathrm {ord}}_\alpha (\operatorname {\mathrm {Fitt}}^0 (\Omega _{X/Y} ) ) < \infty $, the term $T_{X/Y} := \Omega _{X/Y} \otimes _{\mathcal {O}_X} B_L$ is a torsion $B_L$-module, and we get an exact sequence

$$ \begin{align*} 0 \to F_Y \to F_X \oplus T_X \to T_{X/Y} \to 0. \end{align*} $$

Since $P_L$ is a divisible $B_L$-module, tensoring with $P_L$ kills torsion, and hence this sequence gives the exact sequence

$$ \begin{align*} 0 \to \operatorname{\mathrm{Tor}}_1^{B_L}(T_X,P_L) \to \operatorname{\mathrm{Tor}}_1^{B_L}\left(T_{X/Y},P_L\right) \to F_Y \otimes_{B_L}P_L \xrightarrow{\,\varphi'\,} F_X \otimes_{B_L}P_L \to 0. \end{align*} $$

Note that $\varphi ' = \varphi $ under the aforementioned isomorphisms. We have $\operatorname {\mathrm {Tor}}_1^{B_L}(T_X,P_L) \simeq T_X$, and this has dimension $\operatorname {\mathrm {ord}}_\alpha (\operatorname {\mathrm {Fitt}}^d (\Omega _{X/k} ) )$ over L. Similarly, $\operatorname {\mathrm {Tor}}_1^{B_L} (T_{X/Y},P_L ) \simeq T_{X/Y}$ has dimension $\operatorname {\mathrm {ord}}_\alpha (\operatorname {\mathrm {Fitt}}^0 (\Omega _{X/Y} ) )$ over L. Since these two dimensions are equal, the map $\varphi '$ in the sequence is an isomorphism. We conclude that $\varphi $ is an isomorphism.

The surjectivity of $\varphi $ implies that $\delta $ is surjective, and the injectivity of $\delta $ follows from our assumption that either (1) or (2) holds. We conclude that $(T_\alpha f_{\infty })^*$ is an isomorphism.

Proof. This follows from the diagram

and the fact that the top horizontal and left vertical arrows are injections.

Proof. We can assume without loss of generality that X is affine. Given a map

$$ \begin{align*} f \colon X \to Y := {\mathbb A}^d, \end{align*} $$

we let $\beta := f_{\infty }(\alpha ) \in Y_{\infty }$. For every n, we denote by $\alpha _n \in X_n$ and $\beta _n \in Y_n$ the images of $\alpha $ and $\beta $ at the respective n-jet schemes. It is convenient, within this proof, to change notation from before to let $A_{\infty } := \mathcal {O}_{X_{\infty },\alpha }$ and $B_{\infty } := \mathcal {O}_{Y_{\infty },\beta }$, and denote by $\mathfrak {m}_{\infty } \subset A_{\infty }$ and $\mathfrak {n}_{\infty } \subset B_{\infty }$ the respective maximal ideals and by $L_{\infty } := A_{\infty }/\mathfrak {m}_{\infty }$ and $L_{\infty }' := B_{\infty }/\mathfrak {n}_{\infty }$ the residue fields. Similarly, for every $n \in {\mathbb N}$, we let $A_n := \mathcal {O}_{X_n,\alpha _n}$ and $B_n := \mathcal {O}_{Y_n,\beta _n}$, and denote by $\mathfrak {m}_n \subset A_n$ and $\mathfrak {n}_n \subset B_n$ the respective maximal ideals and by $L_n := A_n/\mathfrak {m}_n$ and $L_n' := B_n/\mathfrak {n}_n$ the residue fields.

Note that we have direct systems $\{A_n \to A_{n+1} \mid n \in {\mathbb N}\}$ and $\{B_n \subset B_{n+1} \mid n \in {\mathbb N} \}$, and $A_{\infty } = \operatorname {\mathrm {\varinjlim }}_n A_n$ and $B_{\infty } = \operatorname {\mathrm {\varinjlim }}_n B_n$. Moreover, we have commutative diagrams

where $\mathfrak {n}_n = \varphi _n^{-1}(\mathfrak {m}_n)$, $\mathfrak {n}_n = \mathfrak {n}_{\infty } \cap B_n$ and $\mathfrak {m}_n = \mathfrak {m}_{\infty } \cap A_n$.

For every $n \in {\mathbb N} \cup \{\infty \}$, let

$$ \begin{align*} d\varphi_n \colon \mathfrak{n}_n/\mathfrak{n}_n^2 \otimes_{L^{\prime}_n}L_n \to \mathfrak{m}_n/\mathfrak{m}_n^2 \end{align*} $$

be the induced $L_n$-linear map.

We pick f as in Theorem 8.1. For every $n \in {\mathbb N} \cup \{\infty \}$, there is an associated map of graded rings $\operatorname {\mathrm {gr}}(\varphi _n) \colon \operatorname {\mathrm {gr}}(B_n) \to \operatorname {\mathrm {gr}}(A_n)$. We denote by

$$ \begin{align*} \psi_n \colon \operatorname{\mathrm{gr}}(B_n) \otimes_{L^{\prime}_n}L_{\infty} \to \operatorname{\mathrm{gr}}(A_n) \otimes_{L_n}L_{\infty} \end{align*} $$

the map induced by $\operatorname {\mathrm {gr}}(\varphi _n)$ by the indicated base changes.

Note that $\mathfrak {m}_{\infty } = \operatorname {\mathrm {\varinjlim }}_n \mathfrak {m}_n$ and hence $\mathfrak {m}_{\infty }^r = \operatorname {\mathrm {\varinjlim }}_n \mathfrak {m}_n^r$ for all r. Indeed, if $a \in \mathfrak {m}_{\infty }^r$ for some $r \ge 2$, then we can write $a = a_1\dotsm a_r$ with $a_i \in \mathfrak {m}_{\infty }$; then we can pick n such that $a_i \in \mathfrak {m}_n$ for all i, and hence $a \in \mathfrak {m}_n^r$. It follows that

$$ \begin{align*} \operatorname{\mathrm{gr}}(A_{\infty}) = \operatorname{\mathrm{\varinjlim}}_n \operatorname{\mathrm{gr}}(A_n) \otimes_{L_n}L_{\infty}, \end{align*} $$

and similarly we have

$$ \begin{align*} \operatorname{\mathrm{gr}}(B_{\infty}) = \operatorname{\mathrm{\varinjlim}}_n \operatorname{\mathrm{gr}}(B_n) \otimes_{L^{\prime}_n}L^{\prime}_{\infty}. \end{align*} $$

Since $\mathfrak {n}_n^r = \varphi _n^{-1} (\mathfrak {m}_n^r )$ for all r, for every n we have a commutative diagram

For short, let $R_n := \operatorname {\mathrm {gr}}(A_n) \otimes _{L_n}L_{\infty }$, $S_n := \operatorname {\mathrm {gr}}(B_n) \otimes _{L^{\prime }_n}L_{\infty }$ and $K_n := \ker (\psi _n)$.

Proof. First, note that $K_{\infty } = \operatorname {\mathrm {\varinjlim }}_n K_n$. Indeed, the inclusion $K_{\infty } \supset \operatorname {\mathrm {\varinjlim }}_n K_n$ is clear, and conversely, if $b \in K_{\infty }$ and we fix $n \in {\mathbb N}$ such that $b \in S_n$, then $\psi _m(b)$ is in the kernel of $R_m \to R_{\infty }$ for all $m \ge n$ and hence, since each $\psi _m(b)$ maps to $\psi _{m+1}(b)$ via $R_m \to R_{m+1}$, it follows that $\psi _m(b)$ is zero for $m \gg n$, which means that $b \in K_m$ for $m \gg n$.

We are now ready to prove that

$$ \begin{align*} \operatorname{\mathrm{ht}}(K_{\infty}) = \limsup_n\operatorname{\mathrm{ht}}(K_n). \end{align*} $$

Note that the maps $S_n \to S_{\infty }$ are extensions of polynomial rings over the same field $L_{\infty }$. Thus they are faithfully flat, and hence for every prime $\mathfrak {p}_n \subset S_n$ its extension $\mathfrak {p}_n S_{\infty }$ is prime.

Consider first the case where $\operatorname {\mathrm {ht}}(K_{\infty }) < \infty $ and let $\mathfrak {p} \subset S_{\infty }$ be a minimal prime over $K_{\infty }$ with $\operatorname {\mathrm {ht}}(\mathfrak {p})=\operatorname {\mathrm {ht}}(K_{\infty })$. By Remark 5.12 we have that $\mathfrak {p}$ is finitely generated by elements $f_1,\dotsc ,f_r\in S_{\infty }$. For each $n>0$, let $\mathfrak {p}_n$ be any minimal prime over $K_n$ contained in $\mathfrak {p}\cap S_n$. Then $\mathfrak {p}':=\operatorname {\mathrm {\varinjlim }}_n \mathfrak {p}_n$ is a prime of $S_{\infty }$ with $K_{\infty } \subset \mathfrak {p}' \subset \mathfrak {p}$, so $\mathfrak {p}'=\mathfrak {p}$. Let $n_1>0$ be such that $f_1,\dotsc ,f_r\in \mathfrak {p}_{n_1}$; then $\mathfrak {p}_n=\mathfrak {p}\cap S_n$ for $n\geq n_1$. Given any chain of primes

$$ \begin{align*} (0) = \mathfrak{q}_0 \subsetneq \mathfrak{q}_1 \subsetneq \dotsb \subsetneq \mathfrak{q}_t = \mathfrak{p} \subset S_{\infty}, \end{align*} $$

pick $s_i \in \mathfrak {q}_i \setminus \mathfrak {q}_{i-1}$ and fix $n_2$ such that $s_1,\dotsc ,s_t \in S_{n_2}$. Then for every $n \ge \max \{n_1,n_2\}$ we get a chain of primes

$$ \begin{align*} (0) = \mathfrak{q}_0 \cap S_n \subsetneq \mathfrak{q}_1 \cap S_n \subsetneq \dotsb \subsetneq \mathfrak{q}_t \cap S_n = \mathfrak{p}_n. \end{align*} $$

Thus $\operatorname {\mathrm {ht}}(K_{\infty }) \le \limsup _n \operatorname {\mathrm {ht}}(K_n)$. The other inequality follows by the going-down theorem applied to $S_n \to S_{\infty }$.

If $\operatorname {\mathrm {ht}}(K_{\infty }) = \infty $, then, since $\operatorname {\mathrm {ht}}(K_{\infty }) \ge \operatorname {\mathrm {ht}}(K_nS_{\infty })$, a similar argument shows that the sequence $\{\operatorname {\mathrm {ht}}(K_n)\}_n$ is unbounded.

We can now finish the proof of the theorem. Since $B_n$ is formally smooth, for every $n \in {\mathbb N} \cup \{\infty \}$ the natural map

$$ \begin{align*} \operatorname{\mathrm{Sym}}_{L_n'}\left(\mathfrak{n}_n/\mathfrak{n}_n^2\right) \to \operatorname{\mathrm{gr}}(B_n) \end{align*} $$

is an isomorphism (see Remark 7.3 for the case $n=\infty $). Furthermore, the diagram

is commutative.

By Theorem 8.1, the map $\sigma _{\infty }$ is an isomorphism, and hence

$$ \begin{align*} \operatorname{\mathrm{ht}}(K_{\infty}) = \operatorname{\mathrm{ht}}(\ker(\gamma_{\infty})) = \mathrm{ecodim}(A_{\infty}). \end{align*} $$

Similarly, by Corollary 8.2, $\sigma _n$ is an injective $L_{\infty }$-linear map of polynomial rings, and we have

$$ \begin{align*} \operatorname{\mathrm{ht}}(K_n) \le \operatorname{\mathrm{ht}}(\ker(\gamma_n)) = \mathrm{ecodim}(A_n). \end{align*} $$

Then we conclude by Lemma 8.4.