IND-ÉTALE VS FORMALLY ÉTALE

(1) Let $d$ be the Krull dimension of $A$. Noether normalization (see [Sta, 00OY]) guarantees a module finite inclusion $k[x_{1},\ldots,x_{d}]\hookrightarrow A$. So the $k$-transcendence cardinality of $A$ is at least $d$. We show that the $k$-transcendence cardinality of $A$ is at most $d$. For that, we use the next lemma to reduce the problem to the case where $A$ is a domain.

Now given a $k$-algebra inclusion $\phi:k[x_{1},\ldots,x_{n}]\hookrightarrow A$, take $S=\phi(k[x_{1},\ldots,x_{n}]\setminus 0)$. Using Lemma 2.5, choose a minimal prime $\mathfrak{p}$ of $A$ such that $S\cap\mathfrak{p}=\emptyset$. This means the composition of $\phi$ with the quotient $A\rightarrow\frac{A}{\mathfrak{p}}$ is also injective. The last injection gives an injection of the fraction fields $k(x_{1},\ldots,x_{n})\hookrightarrow\text{Frac}(\frac{A}{\mathfrak{p}})$. So $n$ is at most the $k$-transcendence degree of $\text{Frac}(\frac{A}{\mathfrak{p}})$. Since the $k$-transcendence degree of $\text{Frac}(\frac{A}{\mathfrak{p}})$ is the Krull dimension of $\frac{A}{\mathfrak{p}}$ and the Krull dimension of $\frac{A}{\mathfrak{p}}$ is at most $d$ ([Ser00, Prop. 14]), $n\leq d$.

(2) The Krull dimension of any finite type $k$-subalgebra $B$ of $A$ is at most the $k$-transcendence degree of $B$. Since the $k$-transcendence degree of $B$ is at most $n$, we are done.

We now prove that the Krull dimension of $A$ is at most $n$. Let $\mathfrak{p_{0}}\subseteq\mathfrak{p_{1}}\subseteq\ldots\subseteq\mathfrak{p_{m}}$ be chain of prime ideals of $A$- where each containment is strict. For each $j\geq 1$, choose $x_{j}\in\mathfrak{p_{j}}\setminus\mathfrak{p_{j-1}}.$ Let $B$ be the $k$-subalgebra of $A$ generated by $x_{1},\ldots,x_{m}$. So we have a chain of prime ideals in $B$ with strict containments,

So $m$ is at most the Krull dimension of $B$. Since the Krull dimension of $B$ is at most $n$ by part $1$, the Krull dimension of $A$ is at most $n$.

(3) We show that if the $k$-transcendence cardinality of $A$ is $n\in\mathbb{N}$, then the $k$-transcendence cardinality of $B$ is also $n$. Given any $k$-algebra inclusion $\phi:k[x_{1},\ldots,x_{m}]\hookrightarrow B$, we can choose finite type $k$-subalgebras $B^{\prime}\subseteq B$ and $A^{\prime}\subseteq A$, such that $\text{Im}(\phi)\subseteq B^{\prime}$, $A^{\prime}\subseteq B^{\prime}$ and $A^{\prime}\hookrightarrow B^{\prime}$ is module finite. The choice can be made as follows: for each $j$, $1\leq j\leq m$, there is a nonzero monic polynomial $F_{j}\in A[t]$ such that $F_{j}(\phi(x_{j}))=0$. Take $A^{\prime}$ to be the $k$-subalgebra of $A$ generated by the coefficients of $F_{j}$’s where $j$ varies. Take $B^{\prime}$ to be $A^{\prime}$-subalgebra of $B$ generated by all the $\phi(x_{j})$’s.

Now, by (1), the $k$-transcendence degree of $B^{\prime}$ is the Krull dimension of $B^{\prime}$. Since $A^{\prime}\subseteq B^{\prime}$ is module finite, the Krull dimension of $A^{\prime}$ and $B^{\prime}$ are the same. By (2), the Krull dimension of $A^{\prime}$ is at most $n$. So the $k$-transcendence degree of $B^{\prime}$ is at most $n$. Thus $m\leq n$, proving that the $k$-transcendence cardinality of $B$ is also at most $n$. Again since $A\subseteq B$, the $k$-transcendence cardinality of $B$ is at least $n$.

If the $k$-transcendence cardinality of $B$ is finite, the $k$- transcendence cardinality of $A$ is also finite as $A\subseteq B$. Moreover the $k$-transcendence cardinalities of $A$ and $B$ coincide by the argument above.

(4) Since there is a $k$-algebra surjection $A\rightarrow\phi(A)$, the $k$-transcendence cardinality of $\phi(A)$ is at most $n$. Since $\phi(A)\subseteq C$ is module finite, by (3), the $k$-transcendence cardinality of $C$ is at most that of $\phi(A)$ and the later is at most $n$.

(5) Given a set $S$ of cardinality $J$ and a $k$-algebra injection $\phi:k[\{t_{j}\,|\,j\in S\}]\hookrightarrow A$, the intersection of the image of $\phi$ and the nilradical of $A$ is zero. Thus composing $\phi$ with the surjection $A\rightarrow A_{\text{red}}$ also gives an injection. Thus the transcendence cardinality of $A_{\text{red}}$ is at least that of $A$. Given a set $S$ and a $k$-algebra injection $\psi:k[\{x_{j}\,|\,j\in S\}]\rightarrow A_{\text{red}}$, lift $\psi$ to a $k$-algebra map $k[\{x_{j}\,|\,j\in S\}]\rightarrow A$; the lift is necessarily injective. So the transcendence cardinality of $A_{\text{red}}$ is at most that of $A$.

(6) Suppose that the transcendence cardinality of $A$ is $n\in\mathbb{N}$. It is enough to show that the transcendence cardinality of $\text{Frac}(A)$ is at most $n$. By contradiction, assume that $\text{Frac}(A)$ contains elements $\frac{a_{1}}{b_{1}},\frac{a_{2}}{b_{2}},\ldots,\frac{a_{n+1}}{b_{n+1}}$, which are algebraically independent over $k$, where all $a_{i},b_{i}$’s are in $A$. Then the subalgebra $k[a_{1},\ldots,a_{n+1},b_{1},\ldots,b_{n+1}]\subseteq A$ has transcendence degree at least ${n+1}$ as its fraction field contains $k[\frac{a_{1}}{b_{1}},\frac{a_{2}}{b_{2}},\ldots,\frac{a_{n+1}}{b_{n+1}}]$. Since the transcendence degree of $A$ is $n$, we get a contradiction.

(7) This assertion follows from Proposition 2.6 proven below.

∎

We first prove Proposition 2.6 assuming that $A$ is a field and then deduce the general case from the field case in a few steps.

Assume that $A$ is a field. Pick a subset $\{x_{i}\}_{i\in I}$ of $A$ such that $\{\iota(t_{1}),\ldots,\iota(t_{s})\}\cup\{x_{i}\}_{i\in I}$ is a $k$-transcendence basis of $A$; for example, $\{x_{i}\}_{i\in I}$ can be chosen to be a $k(\iota(t_{1}),\ldots,\iota(t_{s}))$-transcendence basis of $A$; see [Sta, Tag 030F]. Set $L$ to be smallest subfield of $A$ containing $k$ and $\{\iota(t_{1}),\ldots,\iota(t_{s})\}\cup\{x_{i}\}_{i\in I}$. For any finite field extension $L^{\prime}\supseteq L$ where $L^{\prime}\subseteq A$, since $L\subseteq L^{\prime}$ is separable, we have an isomorphism,

see [Liu02, Chapter 6, Lemma 1.13]. Varying $L^{\prime}$ over finite extensions of $L$ such that $L^{\prime}\subseteq A$, we get a direct system of isomorphisms from Equation 1; taking the direct limit of this direct system of isomorphisms we get an isomorphism

To get Equation 2, we have used that formation of modules of Kähler differentials commute with taking direct limit (see [Sta, Tag 00RM]) and $A$ is the direct limit of the fields $L^{\prime}$. Since $\Omega_{L/k}$ is isomorphic to the free $L$-module with basis $\{d\iota(t_{1}),\ldots,d\iota(t_{s})\}\cup\{dx_{i}\}_{i\in I}$, Equation 2 implies that $\Omega_{A/k}$ is a free $A$-module with basis $\{d\iota(t_{1}),\ldots,d\iota(t_{s})\}\cup\{dx_{i}\}_{i\in I}$. Hence $d\iota(t_{1})\wedge d\iota(t_{2})\wedge\ldots\wedge d\iota(t_{s})$ is a nonzero element of $\wedge^{s}_{A}\Omega_{A/k}$.

Given a $k$-algebra $A$ as in Proposition 2.6, which is not necessarily a field, set $A^{\prime}$ to be $A$ modulo the nilradical of $A$. Then the composition $k[t_{1},\ldots,t_{s}]\xhookrightarrow{\iota}A\rightarrow A^{\prime}$ is also injective; denote the composition by $\phi$. Set $S=\phi(k[t_{1},\ldots,t_{s}]\setminus{0})$. We note that $S$ is a multiplicative set. Using Lemma 2.5 we can choose a minimal prime $\mathfrak{p}$ of $A^{\prime}$ such that $S\subseteq A^{\prime}-\mathfrak{p}$. So the image of any nonzero element of $k[t_{1},\ldots,t_{s}]$ under the composition $k[t_{1},\ldots,t_{s}]\xrightarrow{\phi}A^{\prime}\rightarrow A^{\prime}_{\mathfrak{p}}$ is a unit; hence the composition is also injective. Denote the last composition by $\psi$. We have a commutative diagram,

where the unlabelled downward arrow is induced by the canonical map $A\rightarrow A^{\prime}_{\mathfrak{p}}$. We want to show that $\wedge^{s}d\iota(dt_{1}\wedge\ldots\wedge dt_{s})$ is nonzero. To that end, first note that $A^{\prime}_{\mathfrak{p}}$ is a field: since $\mathfrak{p}$ is minimal, the only prime ideal of $A^{\prime}_{\mathfrak{p}}$ namely $\mathfrak{p}A^{\prime}_{\mathfrak{p}}$ coincides with the nilradical of $A^{\prime}_{\mathfrak{p}}$, which is zero as $A^{\prime}$ and hence $A^{\prime}_{\mathfrak{p}}$ is reduced. Now by the field case of Proposition 2.6, $\wedge^{s}d\psi(dt_{1}\wedge\ldots\wedge dt_{s})$ is nonzero. The commutativity of diagram 3 implies that $\wedge^{s}d\psi(dt_{1}\wedge\ldots\wedge dt_{s})$ is the image of $\wedge^{s}d\iota(dt_{1}\wedge\ldots\wedge dt_{s})$. So $\wedge^{s}d\iota(dt_{1}\wedge\ldots\wedge dt_{s})$ must be a nonzero element of $\wedge^{s}_{A}\Omega_{A/k}$. ∎

Contrary to the assertion of Corollary 2.9, assume that for $a\in A$ the $k$-algebra map from the polynomial ring $k[t]$ to $A$ sending $t$ to $a$ is injective. Now Proposition 2.6 implies that $da\in\Omega_{A/k}$ is nonzero, contradicting our hypothesis $\Omega_{A/k}=0$. ∎

We shall show that any finitely generated $k$-subalgebra of $A$ is étale over $k$; this will prove Theorem 2.10 since $A$ is the directed union of all finitely generated $k$-subalgebras.
Fix a finitely generated $k$-subalgebra $B$ of $A$. The ring $B$ is integral over $k$ as $A$ is integral over $k$ by Corollary 2.9. Therefore $B$ has Krull dimension zero. Hence every minimal prime of $B$ is maximal. Since $B$ is noetherian, $B$ has only finitely many minimal primes and hence $B$ has only finitely many maximal ideals – say $\mathfrak{m}_{1},\ldots,\mathfrak{m}_{r}$. By the Chinese remainder theorem, we have

Since $A$ is reduced, so is $B$. Hence $\cap_{i=1}^{r}\mathfrak{m}_{i}=0$. Thus from Equation 4, we get that $B\cong\frac{B}{\mathfrak{m}_{1}}\times\ldots\frac{B}{\mathfrak{m}_{r}}$. Since $k$ has characteristic zero and $B$ is finite type over $k$, for each $i$, $1\leq i\leq r$, $B/\mathfrak{m}_{i}$ is a finite, separable field extension of $k$, so $\Omega_{\frac{B}{\mathfrak{m}_{i}}/k}=0$; see [Sta, Tag090W]). Finally we conclude $\Omega_{B/k}=0$, since as abelian groups

Thus $B$ is étale over $k,$ as desired. ∎

See [Mor19, Proposition 2.28] and [BMS19, Section 2.2]. ∎

We note that $\mathrm{HH}_{1}(A/k)=\mathrm{Tor}_{1}^{A\otimes_{k}A}(A,A)\simeq\Omega_{A/k}$. Therefore, our hypothesis implies that $\Omega_{A/k}=0.$ Since $A$ is reduced, it follows from Theorem 2.10 that $A$ is in fact ind-étale and therefore $\mathbb{L}_{A/k}$ is exact. So $\mathbb{L}_{A/k}$ is isomorphic to $0$ when viewed as an object of $D(A)$. Let $\text{Fil}^{n}_{\text{HKR}}(\mathrm{HH}(A/k))$ denote the HKR filtration on $\mathrm{HH}(A/k).$ Since the $i$-th graded piece for the HKR filtration is zero for $i\geq 1$ by Proposition 3.4,, we see that

for $n\geq 1$. Thus, we have an exact triangle

Using the fact that the HKR filtration is complete, we argue that $\mathrm{HH}(A/k)\simeq A[0]$; see for e.g., [BMS19, Definition 5.1] for the definition of a filtered object in the derived category being complete. Indeed, from the completeness of the HKR filtration and Equation 5, it follows that

However, by the exact triangle above, that implies that $\mathrm{HH}(A/k)\simeq A[0].$ This finishes the proof. ∎

By Proposition 2.6, the $k$-transcendence cardinality of $A$ is $\leq d.$ It then follows from (2), Proposition 2.3 that any finite type $k$-subalgebra of $A$ must have Krull dimension $\leq d.$ Therefore, by using [Cor+08, Theorem 6.2] and taking filtered colimits over all finite type $k$-subalgebras of $A$, we obtain the desired conclusion. ∎

One observes that $A$ being $K_{1}$-regular implies that $A$ is reduced. In order to see this, we note some well-known general constructions. For any commutative ring $R,$ taking determinant induces a natural group homomorphism $\text{det}:\mathrm{GL}(R)\to R^{\times},$ which factors to give a map

Here $R^{\times}$ is the abelian group of units of $R$. Note that there is also a natural map $R^{\times}=\mathrm{GL}_{1}(R)\to K_{1}(A)$ which admits a section provided by $\text{det}:K_{1}(R)\to R^{\times}.$ Coming back to our situation, the $K_{1}$-regularity of $A$ in particular implies that the map $K_{1}(A)\to K_{1}(A[t])$ induced by the natural inclusion $A\hookrightarrow A[t]$ is an isomorphism. We have the following commutative diagram where the vertical arrows are given by the ‘det’ maps and the horizontal maps are induced by the inclusion $A\hookrightarrow A[t].$

Since the vertical maps and the bottom horizontal maps are surjective, the inclusion $A^{\times}\hookrightarrow(A[t])^{\times}$ is also surjective. If there were a nonzero nilpotent element $a\in A,$ the element $1+at\in(A[t])^{\times}$ would not come from $A^{\times}.$ Thus we conclude that $A$ is reduced. Theorem 2.10 now implies that $A$ is ind-étale. For the last part of the proposition, we again note that the $K$-groups commute with taking direct limits and étale algebras are $K_{n}$-regular for all $n,$ which yields the claim. ∎