Relative Cartier divisors and Laurent polynomial extensions

Abstract

If $i:A\subset B$ is a commutative ring extension, we show that the group ${\mathcal I}(A,B)$ of invertible A-submodules of B is contracted in the sense of Bass, with $L{\mathcal I}(A,B)=H^0_{\mathrm {et}}(A,i_*{\mathbb Z}/{\mathbb Z})$. This gives a canonical decomposition for ${\mathcal I}(A[t,\frac{1}{t}],B[t,\frac{1}{t}])$.

1 Introduction

Let $A\subset B$ be a ring extension. The group ${\mathcal I}(A, B)$ of invertible A-submodules of B is related to the Picard groups and the units groups of A and B by the exact sequence

$$\begin{aligned} 1\rightarrow U(A) \rightarrow U(B) \rightarrow {\mathcal I}(A,B) \rightarrow {\text {Pic}}\, A \rightarrow {\text {Pic}}\, B. \end{aligned}$$

(See [6, §2].) Replacing $A\subset B$ with $A[t]\subset B[t]$ and $A[t,1/t]\subset B[t,1/t]$ yields similar exact sequences. Following Bass [1], each functor F on rings defines functors NF and LF so that $F(A[t])=F(A)\oplus NF(A)$ and for certain functors like $F=U$ and ${\text {Pic}}$, called contracted functors, we even have a natural decomposition

$$\begin{aligned} F(A[t,1/t])\cong F(A)\oplus NF(A)\oplus NF(A)\oplus LF(A). \end{aligned}$$

The decompositions of U(A[t, 1 / t]) and ${\text {Pic}}A[t,1/t]$ are given in [1, XII.7.8] and [11]. We can define $N{\mathcal I}(A,B)$ and $L{\mathcal I}(A,B)$ in the same way. Here is our main result.

Theorem 1.1

Given a commutative ring extension $f:A\subset B$, ${\mathcal I}$ is a contracted functor with $L{\mathcal I}(A,B)=H^0_{\mathrm {et}}({\text {Spec}}A,f_*{\mathbb Z}/{\mathbb Z})$. In particular, there is a natural decomposition

$$\begin{aligned} {\mathcal I}(A[t,1/t],B[t,1/t]) \cong {\mathcal I}(A,B)\oplus N{\mathcal I}(A,B)\oplus N{\mathcal I}(A,B) \oplus L{\mathcal I}(A,B), \end{aligned}$$

In addition, $L{\mathcal I}(A,B)=L{\mathcal I}(A[t],B[t])=L{\mathcal I}(A[t,1/t],B[t,1/t]).$

This is proven in Theorem 5.1 and Proposition 3.4 below. Here ${\mathbb Z}$ is regarded as the constant étale sheaf on both ${\text {Spec}}A$ and ${\text {Spec}}B$, and $f_*{\mathbb Z}$ is the direct image sheaf on ${\text {Spec}}A$. The group $L{\mathcal I}(A,B)$ also equals the Nisnevich cohomology group $H^0_{\mathrm {nis}}({\text {Spec}}A,f_*{\mathbb Z}/{\mathbb Z})$, but differs from the Zariski cohomology group $H^0_{\mathrm {zar}}({\text {Spec}}A,f_*{\mathbb Z}/{\mathbb Z})$; see Example 5.5.1.

For convenience, let us write A[T] for $A[t_1,1/t_1,\dots ,t_n,1/t_n]$. As pointed out by Bass [1], we can iterate the operations N and L to get decompositions of ${\mathcal I}(A[T],B[T])$ using components $N^iL^j{\mathcal I}(A,B)$ for $1\le i,j\le n$. Since our Main Theorem says that $NL{\mathcal I}=L^2{\mathcal I}=0$, most of these terms are unnecessary.

Corollary 1.2

For every ring extension $A\subset B$, ${\mathcal I}(A[T],B[T])$ is the direct sum of ${\mathcal I}(A,B)$, n terms of the form $L{\mathcal I}(A,B)$ and $2^i\left( {\begin{array}{c}n\\ i\end{array}}\right)$ terms of the form $N^i{\mathcal I}(A,B)$, $1\!\le \! i\!\le \! n$.

Since we know from [8] that $N{\mathcal I}(A,B)=0$ is equivalent to A being seminormal in B (Definition 6.5) we can further conclude:

Corollary 1.3

For $A\subset B$, the following are equivalent:

• (1) ${\mathcal I}(A,B)={\mathcal I}(A[t,1/t],B[t,1/t])$;

• (2) $H_{\mathrm {et}}^0({\text {Spec}}(A),f_*{\mathbb Z}/{\mathbb Z})=0$ and A is seminormal in B.

It is immediate from our Main Theorem that $L{\mathcal I}(A,B)$ is a torsionfree group (we give a simple proof in Corollary 3.6); it is free abelian of finite rank when A is pseudo-geometric and finite dimensional (Proposition 3.7).

A secondary goal of this paper is to give simple techniques for determining $L{\mathcal I}(A,B)$. For example, we may assume A and B are reduced as $L{\mathcal I}(A,B)\cong L{\mathcal I}(A_{\mathrm {red}},B_{\mathrm {red}})$ (Theorem 4.1). The following special case of Proposition 6.2 gives an elementary criterion for the vanishing of $L{\mathcal I}(A,B)$. We say that an extension B / A is connected if for every prime ideal $\wp$ of A, the ring $B_{\wp }/\wp B_\wp$ is connected.

Proposition 1.4

If B / A is finite and B is connected over A then $L{\mathcal I}(A,B)=0$.

We also show that $L{\mathcal I}(A,B)=0$ can only happen if the extension $A\subset B$ is anodal in the sense of Asanuma (Theorem 6.8). The converse is true for integral, birational extensions of 1-dimensional domains, but Example 6.9 (taken from [11, 3.5]) shows that being integral, birational and anodal is not sufficient in higher dimension.

This paper is laid out as follows. In Sect. 2, we define contracted functors on extensions and recall some basic theory. In Sect. 3, we define ${\mathcal I}(A,B)$ and prove (in Proposition 3.4) that $NL{\mathcal I}=LL{\mathcal I}=0$. Section 4 gives some basic properties of ${\mathcal I}(A,B)$. The rest of Theorem 1.1 is proven in Sect. 5, and Sect. 6 describes some conditions under which $L{\mathcal I}(A,B)$ vanishes.

2 Contracted functors

All of the rings we consider are commutative with 1, and all ring homomorphisms are unitary. The category of ring extensions has objects $f:A\hookrightarrow B$; a morphism from f to $f':A'\hookrightarrow B'$ is a morphism $B\rightarrow B'$ sending A to $A'$.

In [1, XII], Bass defined the notion of a contracted functor from rings to abelian groups. This has a natural translation into the setting of ring extensions, which we now lay out. Given an indeterminate t, we write f[t] for the polynomial ring extension $A[t]\hookrightarrow B[t]$ and write f[t, 1 / t] for the Laurent polynomial extension $A[t,1/t]\hookrightarrow B[t,1/t]$.

Definition 2.1

Let F be a functor from ring extensions to abelian groups. We write LF(A, B) or LF(f) for the cokernel of the map $F(f[t])\times F(f[1/t]){\;\buildrel \pm \over \longrightarrow }\; F(f[t,1/t])$ which is the difference of the maps induced by applying F to the morphisms $f\rightarrow f[t]$ and $f\rightarrow f[1/t]$. We write Seq(F, f) for the following sequence (where $\Delta$ is the diagonal map):

$$\begin{aligned} 1 \rightarrow F(f) {\;\buildrel \Delta \over \longrightarrow }\; F(f[t])\times F(f[1/t]){\;\buildrel \pm \over \longrightarrow }\; F(f[t,1/t]) \rightarrow LF(f)\rightarrow 1. \end{aligned}$$

We say F is acyclic if Seq(F, f) is exact for every ring extension f. We say that F is contracted if Seq(F, f) is naturally split exact, i.e., if there is a map $h(f): LF(f)\rightarrow F(f[t,1/t])$.

Following Bass [1, XII], we write NF(f) or $N_tF(f)$ for the kernel of the map $F(f[t])\rightarrow F(f)$ induced by $t\mapsto 1$. This map is split by the map $F(f\rightarrow f[t])$ induced by $B\subset B[t]$, and we have a natural decomposition $F(f[t])=F(f)\oplus NF(f)$. Thus Seq(F, f) is quasi-isomorphic to the sequence

$$\begin{aligned} 1 \rightarrow F(f)\oplus N_tF(f)\oplus N_{1/t}F(f) \rightarrow F(f[t,1/t])\rightarrow LF(f)\rightarrow 1. \end{aligned}$$

If F is a functor from rings to abelian groups, we can define functors $s^*F(f)=F(A)$ and $t^*F(f)=F(B)$ by composing with the source and target functors from ring extensions to rings sending $f:A\hookrightarrow B$ to $s(f)=A$ and $t(f)=B$. If F is contracted in Bass’ sense then $s^*F$ and $t^*F$ are contracted in the sense of Definition 2.1.

It should be clear to the reader that the notion of contracted functor also makes sense for functors from many categories (such as commutative rings, schemes, ring extensions, ...) to any abelian category (such as abelian groups, sheaves, modules). When these choices are irrelevant, we will not specify them and merely refer to “contracted functors.”

Lemma 2.2

Let $\eta :F\rightarrow G$ be a morphism of contracted functors. Then $\ker (\eta )$ and $\mathop {\mathrm {coker}\,}(\eta )$ are contracted functors, with $L\ker (\eta )=\ker (L\eta )$ and $L\mathop {\mathrm {coker}\,}(\eta )=\mathop {\mathrm {coker}\,}(L\eta )$.

If $1\rightarrow F\rightarrow G\rightarrow H\rightarrow 1$ is a short exact sequence of functors and F, H are acyclic then G is acyclic and there is a short exact sequence

$$\begin{aligned} 1 \rightarrow LF \rightarrow LG\rightarrow LH \rightarrow 1. \end{aligned}$$

Proof

The first assertion is proven exactly as the corresponding assertion for contracted functors on rings in [1, XII.7.2] or [12, III.4.2]. The second assertion is proven exactly as Carter proved the corresponding assertion in [2, 1.2]. $\square$

Corollary 2.3

If $F_1\rightarrow F_2 \rightarrow G\rightarrow H_1\rightarrow H_2$ is an exact sequence of functors and the $F_i$, $H_i$ are contracted then G is acyclic and there is an exact sequence for every f:

$$\begin{aligned} LF_1(f) \rightarrow LF_2(f) \rightarrow LG(f) \rightarrow LH_1(f)\rightarrow LH_2(f). \end{aligned}$$

Of course, Corollary 2.3 may be iterated to get exact sequences for LLG(f), etc., because the $LF_i$ and $LH_i$ are contracted functors.

Example 2.4

Recall from [1, XII.7.8] [12, III.4.1.3] that the units U form a contracted functor on rings with $LU(A)=H^0({\text {Spec}}A,{\mathbb Z})$ and $LLU=NLU=0$. Similarly, $F(A)=U(A_{\mathrm {red}})$ is a contracted functor and its contraction $LF(A)=LU(A_{\mathrm {red}})$ is isomorphic to LU(A). Define ${U_{\mathrm {nil}}}(A)$ to be the kernel of $U(A)\rightarrow U(A_{\mathrm {red}})$; it is the multiplicative group $(1+\text {nil}(A))^\times$. Since $1\rightarrow {U_{\mathrm {nil}}}(A)\rightarrow U(A)\rightarrow U(A_{\mathrm {red}})\rightarrow 1$ is an exact sequence, Lemma 2.2 implies that ${U_{\mathrm {nil}}}$ is a contracted functor with $L{U_{\mathrm {nil}}}(A)=0$.

Given a commutative ring extension $f:A\hookrightarrow B$, define ${U_{\mathrm {nil}}}(f)$ to be the cokernel of ${U_{\mathrm {nil}}}(A)\rightarrow {U_{\mathrm {nil}}}(B)$. From the exact sequence $1\rightarrow {U_{\mathrm {nil}}}(A)\rightarrow {U_{\mathrm {nil}}}(B)\rightarrow {U_{\mathrm {nil}}}(f)\rightarrow 1$ and Lemma 2.2, we see that ${U_{\mathrm {nil}}}(f)$ is a contracted functor with $L{U_{\mathrm {nil}}}(f)=0$.

Remark 2.4.1

Since $N{U_{\mathrm {nil}}}(f)=(1+t\,{\mathrm {nil}}(B)[t])^\times /(1+t\,{\mathrm {nil}}(A)[t])^\times$, it follows that $N{U_{\mathrm {nil}}}(f)=0$ if and only if ${\mathrm {nil}}(A)={\mathrm {nil}}(B)$. This is trivial if B is reduced.

3 Relative Cartier divisors

Relative Cartier divisors are functors on ring extensions. Recall that the A-submodules of B form a monoid under multiplication, with identity A. An A-submodule $L_1$ of B is said to be invertible if $L_1L_2=A$ for some $L_2$. In particular, $L_1$ is isomorphic to an invertible ideal of A. An invertible A-submodule is also said to be a relative Cartier divisor.

Definition 3.1

Given a ring extension $f:A\hookrightarrow B$, let ${\mathcal I}(f)$ denote the multiplicative group of all A-submodules of B which are invertible. We shall sometimes write ${\mathcal I}(A,B)$ for ${\mathcal I}(f)$. It is easily seen that ${\mathcal I}$ is a functor from the category of ring extensions to abelian groups.

The study of ${\mathcal I}(A,B)$ was initiated by Roberts and Singh in [6].

If ${\mathcal O}^\times _A$ is the Zariski sheaf of units on ${\text {Spec}}(A)$ and $f_*{\mathcal O}^\times _B$ is the direct image sheaf on ${\text {Spec}}(A)$ associated to the units on ${\text {Spec}}(B)$, it is easy to see that

$$\begin{aligned} {\mathcal I}(A,B) \cong H_{\mathrm {zar}}^0({\text {Spec}}A,f_*{\mathcal O}^\times _B/{\mathcal O}^\times _A). \end{aligned}$$

(3.2)

In effect, an invertible A-submodule L can be described by giving an open cover $\{U_i\}$, $U_i={\text {Spec}}(A[1/s_i])$ of ${\text {Spec}}(A)$ and elements $f_i$ of $B[1/s_i]^\times$ (defined modulo $A[1/s_i]^\times$) such that each $f_i/f_j$ is in $A[1/s_is_j]^\times$.

For example, if A is a domain and K is the field of fractions, then ${\mathcal I}(A,K)$ is the group of Cartier divisors and the interpretation of ${\mathcal I}(A,K)$ as $H^0({\text {Spec}}A,f_*{\mathcal O}^\times _K/{\mathcal O}^\times _A)$ is standard. For this reason, we shall call ${\mathcal I}(f)$ the group of relative Cartier divisors.

Since ${\text {Pic}}(A)$ (the Picard group of A) is $H^1({\text {Spec}}A,{\mathcal O}^\times _A)$, and $H^1({\text {Spec}}A,f_*{\mathcal O}^\times _B)$ is a subgroup of ${\text {Pic}}(B)$, the (Zariski or étale) cohomology sequence associated to the exact sequence of sheaves on ${\text {Spec}}A, 1 \rightarrow {\mathcal O}^\times _A \rightarrow f_*{\mathcal O}^\times _B \rightarrow f_*{\mathcal O}^\times _B/{\mathcal O}^\times _A\rightarrow 1$, is the exact sequence mentioned in the Introduction:

$$\begin{aligned} 1\rightarrow U(A) \rightarrow U(B) \rightarrow {\mathcal I}(A,B) \rightarrow {\text {Pic}}\,\,\, A \rightarrow {\text {Pic}}\,\,\, B. \end{aligned}$$

(3.3)

It is clear that this sequence is natural in f. (A more elementary proof of exactness is given in [6, Theorem 2.4]).

Proposition 3.4

The functor ${\mathcal I}$ is acyclic, $NL{\mathcal I}=LL{\mathcal I}=0$ and there is an exact sequence

$$\begin{aligned} 1 \rightarrow LU(A) \rightarrow LU(B) \rightarrow L{\mathcal I}(f) \rightarrow L\!{\text {Pic}}(A) \rightarrow L\!{\text {Pic}}(B). \end{aligned}$$

Proof

The units U and Picard group ${\text {Pic}}$ are contracted functors on rings (see [11, 5.2]). Applying Corollary 2.3 to (3.3), we see that ${\mathcal I}$ is acyclic, and that there are exact sequences

$$\begin{aligned} 1 \rightarrow LU(A) \rightarrow LU(B) \rightarrow&L{\mathcal I}(f) \rightarrow L\!{\text {Pic}}(A) \rightarrow L\!{\text {Pic}}(B), \\ 1 \rightarrow LLU(A) \rightarrow LLU(B) \rightarrow&LL{\mathcal I}(f) \rightarrow LL\!{\text {Pic}}(A) \rightarrow LL\!{\text {Pic}}(B). \end{aligned}$$

Now $NLU=LLU=0$ by Example 2.4, and $LL\!{\text {Pic}}=NL\!{\text {Pic}}=0$ by [11, 7.7]. This yields $LL{\mathcal I}(f)=0$, $LU(A)=LU(A[t])$ and $L\!{\text {Pic}}(A[t])=L\!{\text {Pic}}(A)$. It is immediate that $NL{\mathcal I}(f)=0$. $\square$

Since $L {\text {Pic}}$ vanishes on normal domains [11, 1.5.2], we see that (i) if A is a normal domain then $L{\mathcal I}(f)=0$ if and only if B is connected, and (ii) If B is a normal domain then $L{\mathcal I}(f)=0$ if and only if $L\!{\text {Pic}}(A)=0$. More generally, we have:

Corollary 3.5

If A is connected, and $f:A\hookrightarrow B$ is an extension, then $L{\mathcal I}(f)=0$ if and only if (i) B is connected and (ii) $L\!{\text {Pic}}(A)\rightarrow L\!{\text {Pic}}(B)$ is an injection.

Corollary 3.6

The group $L{\mathcal I}(A, B)$ is always a torsion-free abelian group.

Proof

By [11, 2.3.1], $L\!{\text {Pic}}(A)$ is a torsion-free abelian group. In addition, the image of LU(B) in $L{\mathcal I}(A, B)$ is free abelian by [11, Proposition 1.3]. The fact that $L{\mathcal I}(A, B)$ is torsionfree now follows from Proposition 3.4. $\square$

Recall from [11] that a noetherian ring A is called pseudo-geometric if every reduced finite A-algebra B has finite normalization. For example, any finitely generated algebra over a field or over ${\mathbb Z}$ is pseudo-geometric.

Proposition 3.7

If A is pseudo-geometric and $\dim A\!<\!\infty$, then $L{\mathcal I}(A,B)$ is a free abelian group.

Proof

When A is pseudo-geometric with $\dim A<\infty$, $L\!{\text {Pic}}A$ is a free abelian group by Proposition 2.3 of [11]. So the image of $L{\mathcal I}(f)$ in $L\!{\text {Pic}}(A)$ is a free abelian group. Again, Proposition 3.4 implies that $L{\mathcal I}(A, B)$ is a free abelian group. $\square$

Remark 3.7.1

If A is a 1-dimensional domain, then $L{\mathcal I}(A, B)$ is a free abelian group. This follows from the sequence of Proposition 3.4, and the facts that (i) $L\!{\text {Pic}}(A)$ is a free abelian group [11, 3.4.1], (ii) subgroups of free abelian groups are free and (iii) the image of LU(B) in $L{\mathcal I}(A,B)$ is free abelian [11, Prop. 1.3].

Question 3.8

Is $L{\mathcal I}(A, B)$ always a free abelian group?

For the rest of this paper, it is convenient to adopt scheme-theoretic language. Recall that a morphism of schemes $f:X\rightarrow S$ is affine if the inverse image $f^{-1}U$ of any affine open subset U of S is an affine open subset of X. We will say that an affine morphism is faithful if ${\mathcal O}_S\rightarrow f_*{\mathcal O}_X$ is an injection; if the inverse image of ${\text {Spec}}(A)$ is $f^{-1}U={\text {Spec}}(B)$, this implies that $A\rightarrow B$ is an injection.

Notation 3.9

The category of ring extensions embeds contravariantly into the category of faithful affine morphisms of schemes, $f:X\rightarrow S$; morphisms $f\rightarrow f'$ in this category are compatible pairs of maps $X\rightarrow X'$ and $S\rightarrow S'$. If f is a faithful affine morphism, ${\mathcal I}(f)$ will denote the multiplicative group of all ${\mathcal O}_S$-submodules of $f_*{\mathcal O}_X$ which are invertible.

It is clear that the formal yoga of Sects. 2 and 3 extend to the category of faithful affine morphisms $X\rightarrow S$. Given a faithful affine map $f:X\rightarrow S$, (3.2) easily generalizes to ${\mathcal I}(f)\cong H_{\mathrm {zar}}^0(S,f_*{\mathcal O}^\times _X/{\mathcal O}^\times _S)$. Proposition 3.4 implies that ${\mathcal I}$ is an acyclic functor with $NL{\mathcal I}=LL{\mathcal I}=0$, Corollary 3.6 states that $L{\mathcal I}(f)$ is torsionfree. Remark 2.4.1 is replaced by: $N{U_{\mathrm {nil}}}(f)=0$ if and only if $H^0(S,{\mathrm {nil}}\,{\mathcal O}_S)=H^0(X,{\mathrm {nil}}\,{\mathcal O}_X)$.

4 Basic properties

In this short section, we give a few results that allow us to relate $L{\mathcal I}(f)$ to $L{\mathcal I}$ of other ring extensions. Given a map $f:A\hookrightarrow B$, we write $f_{\mathrm {red}}$ for the evident map $A_{\mathrm {red}}\hookrightarrow B_{\mathrm {red}}$.

Theorem 4.1

The natural map $L{\mathcal I}(A,B) {\;\buildrel \cong \over \longrightarrow }\; L{\mathcal I}(A_{\mathrm {red}},B_{\mathrm {red}})$ is an isomorphism. In addition, there is a natural short exact sequence of functors on ring extensions

$$\begin{aligned} 1\rightarrow {U_{\mathrm {nil}}}(f)\rightarrow {\mathcal I}(f)\rightarrow {\mathcal I}(f_{\mathrm {red}})\rightarrow 0. \end{aligned}$$

Proof

Consider the commutative diagram

The groups ${U_{\mathrm {nil}}}(A)$ and ${U_{\mathrm {nil}}}(f)$ were defined in Example 2.4, where we observed that the two left columns and the top row are short exact sequences; the bottom two rows are the exact sequences (3.3). Since ${U_{\mathrm {nil}}}(A)$ is the intersection of ${U_{\mathrm {nil}}}(B)$ and U(A) in U(B), a diagram chase shows that the third column is exact.

Since $L{U_{\mathrm {nil}}}(f)=0$ by Example 2.4, the isomorphism $L{\mathcal I}(f)\cong L{\mathcal I}(f_{\mathrm {red}})$ follows from the second part of Lemma 2.2, applied to the third column. $\square$

Remark 4.1.1

The first part of Theorem 4.1 extends to $X\rightarrow S$ by our Main Theorem 5.1 below; see 5.3. The second part of Theorem 4.1 can fail for $X\rightarrow S$ as $U(S)\rightarrow U(S_{\mathrm {red}})$ need not be onto.

Corollary 4.2

$N{\mathcal I}(A,B){\;\buildrel \cong \over \longrightarrow }\; N{\mathcal I}(A_{\mathrm {red}},B_{\mathrm {red}})$ if and only if ${\mathrm {nil}}(A)={\mathrm {nil}}(B)$.

Moreover, if ${\mathrm {nil}}(A)\ne {\mathrm {nil}}(B)$ then $N{\mathcal I}(f)\ne 0$.

Proof

Replacing f by f[t] in Theorem 4.1, we have the exact sequence

$$\begin{aligned} 1\rightarrow N{U_{\mathrm {nil}}}(f)\rightarrow N{\mathcal I}(f)\rightarrow N{\mathcal I}(f_{\mathrm {red}})\rightarrow 0. \end{aligned}$$

By Remark 2.4.1, the first term vanishes if and only if ${\mathrm {nil}}(A)={\mathrm {nil}}(B)$. $\square$

Lemma 4.3

([9, §3]) Suppose that $f:A\hookrightarrow B$ and $g:B\hookrightarrow C$ are extensions. Then there is a short exact sequence

$$\begin{aligned} 1\rightarrow {\mathcal I}(A,B) \rightarrow {\mathcal I}(A,C) \rightarrow {\mathcal I}(B,C). \end{aligned}$$

Proof

We have an exact sequence of sheaves on ${\text {Spec}}(A)$:

$$\begin{aligned} 1\rightarrow f_*{\mathcal O}_B^\times /{\mathcal O}_A^\times \rightarrow (fg)_*{\mathcal O}_C^\times /{\mathcal O}_A^\times \rightarrow f_*(g_*{\mathcal O}_C^\times /{\mathcal O}_B^\times ) \end{aligned}$$

Now apply the left exact global sections functor and use (3.2). $\square$

Lemma 4.4

Let ${\mathfrak a}$ be an ideal of B contained in A. Then $L{\mathcal I}(A, B)\cong L{\mathcal I}(A/{\mathfrak a}, B/{\mathfrak a})$.

Proof

Write $\bar{f}$ for $A/{\mathfrak a}\hookrightarrow B/{\mathfrak a}$. By Proposition 2.6 of [6], ${\mathcal I}(A,B)\cong {\mathcal I}(\bar{f})$. Since ${\mathfrak a}[t]$ is an ideal of B[t] contained in A[t], and ${\mathfrak a}[t,1/t]$ is an ideal of B[t, 1 / t] in A[t, 1 / t], the same is true for ${\mathcal I}(A[t], B[t])$ and ${\mathcal I}(A[t,1/t], B[t,1/t])$. The result follows from a comparison of $Seq({\mathcal I},f)$ and $Seq({\mathcal I},\bar{f})$. $\square$

Here is another elementary result about ${\mathcal I}$, which allows us to assume for example that A is noetherian and B is of finite type over A.

Lemma 4.5

${\mathcal I}$ commutes with filtered colimits. That is, if $A\subset B$ is the filtered union of extensions $A_\lambda \subset B_\lambda$ then ${\mathcal I}(A,B)=\varinjlim {\mathcal I}(A_\lambda ,B_\lambda )$ and $L{\mathcal I}(A,B)=\varinjlim L{\mathcal I}(A_\lambda ,B_\lambda )$.

Proof

Since $U(B)=\cup U(B_\lambda )$ and ${\text {Pic}}(B)=\varinjlim {\text {Pic}}(B_\lambda )$, this lemma follows from (3.3), $Seq({\mathcal I},f)$ and the fact that filtered direct limits are exact. $\square$

Proposition 4.6

Suppose that $A=\prod _1^n A_i$ and $B=\prod _1^n B_i$, where $A_i\subset B_i$. Then ${\mathcal I}(A,B)=\prod {\mathcal I}(A_i,B_i)$, $N{\mathcal I}(A,B)=\prod N{\mathcal I}(A_i,B_i)$ and $L{\mathcal I}(A,B)=\prod L{\mathcal I}(A_i,B_i)$.

Proof

Every A-submodule of B has the form $M=\prod M_i$, where each $M_i$ is an $A_i$-submodule of $B_i$. If M is invertible with inverse $N=\prod N_i$, then there are $m_j=(m_{ij})\in M$, $n_j=(n_{ij})\in N$ so that $\sum _j m_{ij}n_{ij}=1$ for all i. This shows that each $M_i$ is an invertible $A_i$-submodule of $B_i$, and hence that the natural map from ${\mathcal I}(A,B)$ to $\prod {\mathcal I}(A_i,B_i)$ is an injection. To see that it is a surjection, suppose that $M_i$ are invertible $A_i$-submodules of $B_i$ with inverses $N_i$. Then for each i there are $m_{ij}$ and $n_{ij}$ so that $\sum _j m_{ij}n_{ij}=1$. Thus $\prod M_i$ is an invertible A-submodule of B.

Since $(\prod _1^n A_i)[t]=\prod _1^n(A_i[t])$, the assertions about $N{\mathcal I}$ and $L{\mathcal I}$ follow by replacing $A_i$ with $A_i[t]$ and $A_i[t,1/t]$, and similarly for $B_i$. $\square$

5 Main theorem

The goal of this section is to show that ${\mathcal I}$ is a contracted functor, whose contraction is an étale cohomology group. We refer the reader to [3] for basic properties of étale sheaves and étale cohomology.

Theorem 5.1

${\mathcal I}$ is a contracted functor on ring extensions, and its contraction is

$$\begin{aligned} L{\mathcal I}(A,B)= H_{\mathrm {et}}^0({\text {Spec}}A,(f_*{\mathbb Z})/{\mathbb Z})=H_{\mathrm {nis}}^0({\text {Spec}}A,(f_*{\mathbb Z})/{\mathbb Z}). \end{aligned}$$

Here $(f_*{\mathbb Z})/{\mathbb Z}$ denotes the quotient sheaf in the étale topology. Theorem 5.1 is just the special case $S={\text {Spec}}(A)$ and $X={\text {Spec}}(B)$ of the following result.

Theorem 5.2

${\mathcal I}$ is a contracted functor on faithful affine maps, with contraction

$$\begin{aligned} L{\mathcal I}(f)= H_{\mathrm {et}}^0(S,(f_*{\mathbb Z})/{\mathbb Z}) = H_{\mathrm {nis}}^0(S,(f_*{\mathbb Z})/{\mathbb Z}). \end{aligned}$$

Corollary 5.3

$L{\mathcal I}(f)\cong L{\mathcal I}(f_{\mathrm {red}})$.

We begin the proof of Theorem 5.2 by generalizing (3.2) to the étale and Nisnevich topologies on S. Recall that if ${\mathcal F}$ is an étale sheaf on S (a sheaf on $S_{\mathrm {et}}$) then it is also a Nisnevich and a Zariski sheaf, and $H_{\mathrm {et}}^0(S,{\mathcal F})=H_{\mathrm {nis}}^0(S,{\mathcal F})=H_{\mathrm {zar}}^0(S,{\mathcal F})={\mathcal F}(S)$. This remark applies for example to the sheaves of units. To avoid confusion, it will be convenient to write ${\mathcal O}^\times _S$ and $f_*{\mathcal O}^\times _X$ for the étale sheaves $U\mapsto \Gamma (U,{\mathcal O}_U)^\times$ and $U\mapsto \Gamma (f^{-1}U,{\mathcal O}_{f^{-1}U})^\times$, instead of the traditional $\mathbb {G}_m$ and $f_*(\mathbb {G}_m|_{X_{\mathrm {et}}})$. Of course, they are also sheaves for the Nisnevich topology on S.

Lemma 5.4

The Zariski quotient sheaf $f_*{\mathcal O}^\times _X/{\mathcal O}^\times _S$ is an étale sheaf. Consequently, ${\mathcal I}(f)\cong H_{\mathrm {et}}^0(S,f_*{\mathcal O}^\times _X/{\mathcal O}^\times _S) \cong H_{\mathrm {nis}}^0(S,f_*{\mathcal O}^\times _X/{\mathcal O}^\times _S)$.

Proof

Since $H_{\mathrm {zar}}^1(S,{\mathcal O}^\times _S)=H_{\mathrm {et}}^1(S,{\mathcal O}^\times _S)$ and $H_{\mathrm {et}}^1(S,f_*{\mathcal O}^\times _X)$ is a subgroup of $H_{\mathrm {et}}^1(X,{\mathcal O}^\times _X)$, we have a commutative diagram:

From the 5-lemma, we see that the middle vertical map is an isomorphism, i.e., that $f_*{\mathcal O}^\times _X/{\mathcal O}^\times _S$ is an étale sheaf, and hence a Nisnevich sheaf. The final assertion follows from (3.2). $\square$

Notation 5.5

Given a scheme S, we write S[t] for $S\times {\text {Spec}}({\mathbb Z}[t])$; there is a natural map $p^{S,t}:S[t]\rightarrow S$. When the base S is clear we simply write $p^t$, so that $p^t_*{\mathcal O}^\times _{S[t]}$ denotes the direct image sheaf on S; it is both a Zariski and an étale sheaf on S. Similarly, we write S[t, 1 / t] for $S\times {\text {Spec}}({\mathbb Z}[t,1/t])$, with projection $p:S[t,1/t]\rightarrow S$, and also write $p_*{\mathcal O}^\times _{S[t,1/t]}$ for the direct image sheaf on S. Given $f:X\rightarrow S$ then, by abuse of notation, we will also write $f_*p^t_*{\mathcal O}^\times _{X[t]}$ for the direct image sheaf on S associated to the composition $X[t]\rightarrow X\rightarrow S$, etc.

For notational simplicity, we shall write ${\mathcal O}^\times$ and $f^T_*{\mathcal O}^\times$ for the étale sheaves ${\mathcal O}^\times _{S[t,1/t]}$ and $f[t,1/t]_*{\mathcal O}^\times _{X[t,1/t]}$ on S[t, 1 / t]. Thus Lemma 5.4 yields the formula

$$\begin{aligned} {\mathcal I}(f[t,1/t])\cong H_{\mathrm {et}}^0\left( S[t,1/t],f^T_*{\mathcal O}^\times /{\mathcal O}^\times \right) \cong H_{\mathrm {et}}^0\left( S,p_*\left( f^T_*{\mathcal O}^\times /{\mathcal O}^\times \right) \right) . \end{aligned}$$

Replacing $H_{\mathrm {et}}^0$ with $H_{\mathrm {nis}}^0$ yields an analogous formula.

Example 5.5.1

The analogue of the formulas ${\mathcal I}(f[t,1/t]) \cong H_{\mathrm {et}}^0(S,p_*(f^T_*{\mathcal O}^\times /{\mathcal O}^\times ))$ and $L{\mathcal I}(A,B)=H_{\mathrm {et}}^0({\text {Spec}}A,(f_*{\mathbb Z}/{\mathbb Z}))$ fail for the Zariski cohomology. To see this, consider the subring A of $B=k[x]$ defining the node. It is not hard to see that

$$\begin{aligned} {\mathcal I}(f[t,1/t])\cong {\text {Pic}}(A[t,1/t]) = {\text {Pic}}(A)\oplus {\mathbb Z}\quad \text {and} \quad L{\mathcal I}(f)\cong L\!{\text {Pic}}(A)\cong {\mathbb Z}\end{aligned}$$

(see [11, 2.2]), yet $H_{\mathrm {zar}}^0(S,p_*(f^T_*{\mathcal O}^\times /{\mathcal O}^\times )_{\mathrm {zar}})={\text {Pic}}(A)$ and $H_{\mathrm {zar}}^0(S,(f_*{\mathbb Z}/{\mathbb Z})_{\mathrm {zar}})=0$.

A similar calculation for the local ring $A_\wp$ of the node and $B_\wp$ the (semilocal) normalization of $A_\wp$ shows that $L{\mathcal I}(A_\wp ,B_\wp )={\mathbb Z}$.

Recall that a local ring A is hensel if every finite A-algebra B is a direct product of local rings. A Nisnevich sheaf on ${\text {Spec}}(A)$ is zero if and only if it is zero on ${\text {Spec}}(A^h_\wp )$ for every prime ideal $\wp$, where $A^h_\wp$ is the henselization of the local ring $A_\wp$.

Lemma 5.6

If A is a hensel local ring then $L{\mathcal I}(A,B)=H^0({\text {Spec}}B,{\mathbb Z})/{\mathbb Z}$.

Proof

By [11, 2.5], $L\!{\text {Pic}}(A)=0$. Since $LU(A)={\mathbb Z}$ and $LU(B)=H^0({\text {Spec}}B,{\mathbb Z})$, the sequence of Proposition 3.4 yields the result. $\square$

Remark 5.6.1

As noted in [11, 1.2.1], $H^0({\text {Spec}}B,{\mathbb Z})/{\mathbb Z}$ is a free abelian group for every B. We saw in Corollary 3.6 that $L{\mathcal I}(f)$ is always torsionfree.

If we fix $f:X\rightarrow S$ and view $L{\mathcal I}$ as the presheaf $U\mapsto L{\mathcal I}(U,f^{-1}U)$ on the étale site of S, Lemma 5.6 says that the associated étale sheaf is $f_*{\mathbb Z}/{\mathbb Z}$. Therefore we have a canonical map $a_f:L{\mathcal I}(f)\rightarrow H_{\mathrm {et}}^0(S,f_*{\mathbb Z}/{\mathbb Z})$.

Theorem 5.7

The canonical map $a_f:L{\mathcal I}(f)\rightarrow H_{\mathrm {et}}^0(S,f_*{\mathbb Z}/{\mathbb Z})$ is an isomorphism.

Proof

We claim there is a commutative diagram whose rows are the exact sequence of Proposition 3.4 and the cohomology sequence associated to ${\mathbb Z}\rightarrow f_*{\mathbb Z}\rightarrow f_*{\mathbb Z}/{\mathbb Z}$:

Given this claim, the theorem follows from the 5-lemma.

The left three vertical maps are the canonical maps from the evident presheaves to the global sections of the associated sheaves, so the left two squares commute. Since the right two vertical maps are the natural isomorphisms of [11, 5.5], the right square also commutes. Thus we only need to show that the remaining square commutes.

Recall from 5.5 that ${\mathcal O}^\times$ and $f^T_*{\mathcal O}^\times$ are the étale sheaves ${\mathcal O}^\times _{S[t,1/t]}$ and $f[t,1/t]_*{\mathcal O}^\times _{X[t,1/t]}$ on S[t, 1 / t]. The sheafification of $A[t,1/t]^\times \rightarrow H^0(S,{\mathbb Z})$ on S is a map $\partial _S:p_*{\mathcal O}^\times \rightarrow {\mathbb Z}$; it induces a map $Rp_*{\mathcal O}^\times \rightarrow {\mathbb Z}$ in the derived category of étale sheaves. Similarly, the sheafification of $B[t,1/t]^\times \rightarrow H^0(S,f_*{\mathbb Z})$ on S induces a map $f_*\partial _X:Rp_*(f^T_*{\mathcal O}^\times )\rightarrow f_*{\mathbb Z}$. Thus we have a morphism of triangles in the derived category.

(5.7.1)

Note that $H_{\mathrm {et}}^0(S,Rp_*({\mathcal O}^\times )[1])=H_{\mathrm {et}}^1(S[t,1/t],{\mathcal O}^\times ) ={\text {Pic}}(S[t,1/t])$ and, by Lemma 5.4,

$$\begin{aligned} H_{\mathrm {et}}^0\left( S,Rp_*\left( f^T_*{\mathcal O}^\times /{\mathcal O}^\times \right) \right) = H_{\mathrm {et}}^0\left( S[t,1/t],f^T_*{\mathcal O}^\times /{\mathcal O}^\times \right) ={\mathcal I}(f[t,1/t]). \end{aligned}$$

(5.7.2)

Thus applying $H_{\mathrm {et}}^0$ to the right-hand square in (5.7.1) yields the commutative square

The left map factors as ${\mathcal I}(f[t,1/t])\rightarrow L{\mathcal I}(f){\;\buildrel a_f \over \longrightarrow }\; H_{\mathrm {et}}^0(S,f_*{\mathbb Z}/{\mathbb Z})$, and the right map factors as ${\text {Pic}}(S[t,1/t])\rightarrow L{\text {Pic}}(S)\cong H_{\mathrm {et}}^1(S,{\mathbb Z})$. The top map is the map in (3.3), fitting into the commutative square with surjective vertical maps

which is implicit in Proposition 3.4. The claim follows. $\square$

Corollary 5.8

The Nisnevich quotient sheaf $f_*{\mathbb Z}/{\mathbb Z}$ is an étale sheaf.

Proof

By Lemma 5.6, it suffices to observe that if S is hensel local we have $L{\mathcal I}(f)=H^0_{\mathrm {et}}(S,f_*{\mathbb Z}/{\mathbb Z})$. $\square$

It remains to show that ${\mathcal I}$ is a contracted functor. Sheafifying the sequence $Seq(U,1_S)$ yields the sequence of sheaves on S:

$$\begin{aligned} 1\rightarrow {\mathcal O}^\times _S {\;\buildrel \Delta \over \longrightarrow }\; p^t_*\left( {\mathcal O}^\times _{S[t]}\right) \times p^{1/t}_*\left( {\mathcal O}^\times _{S[1/t]}\right) {\;\buildrel \pm \over \longrightarrow }\; p_*\left( {\mathcal O}^\times _{S[t,1/t]}\right) {\;\buildrel \partial _S \over \longrightarrow }\; {\mathbb Z}\rightarrow 1. \end{aligned}$$

(5.9)

In addition, Bass’ contraction $t_A:H^0({\text {Spec}}A,{\mathbb Z})\rightarrow A[t,1/t]^\times$ is natural in A, so we may sheafify it to obtain a morphism $t_S:{\mathbb Z}\rightarrow p_*({\mathcal O}^\times _{S[t,1/t]})$ of (Zariski or étale) sheaves on S.

Lemma 5.10

The sequence (5.9) of sheaves on S is split exact, with splitting $t_S$.

Proof

On an affine open ${\text {Spec}}(R)$ of S, this is just the sequence

$$\begin{aligned} 1 \rightarrow R^\times {\;\buildrel \Delta \over \longrightarrow }\; R[t]^\times \times R[1/t]^\times {\;\buildrel \pm \over \longrightarrow }\; R[t,1/t]^\times {\;\buildrel \partial _R \over \longrightarrow }\;{\mathbb Z}\rightarrow 1. \end{aligned}$$

The fact that it is exact, and naturally split by $t_S$ is proven in [1, XII.7.8]; see [11, 7.2]. $\square$

Corollary 5.11

Given a faithful affine map $f:X\rightarrow S$, the direct image of the sequence (5.9) on X is a split exact sequence of (Zariski or étale) sheaves on S, with splitting $f_*t_X$:

$$\begin{aligned} 1\rightarrow f_*{\mathcal O}^\times _X {\;\buildrel \Delta \over \longrightarrow }\; f_*p^t_*\left( {\mathcal O}^\times _{X[t]}\right) \times f_*p^{1/t}_*\left( {\mathcal O}^\times _{X[1/t]}\right) {\;\buildrel \pm \over \longrightarrow }\; f_*p_*\left( {\mathcal O}^\times _{X[t,1/t]}\right) {\;\buildrel f_*\partial _X \over \longrightarrow }\; f_*{\mathbb Z}\rightarrow 1. \end{aligned}$$

The global sections of the sequences in 5.10 and are of course Bass’ sequences Seq(U, S) and Seq(U, X). By 5.5, the sheafification of ${\mathcal I}(f[t,1/t])$ on S is $p_*(f^T_*{\mathcal O}^\times /{\mathcal O}^\times )$. Theorem 5.7 says that the sheafification of ${\mathcal I}(f[t,1/t])\rightarrow L{\mathcal I}(f)$ on S is a canonical map $\partial _f: p_*(f^T_*{\mathcal O}^\times /{\mathcal O}^\times ) \rightarrow f_*{\mathbb Z}/{\mathbb Z}$.

Proposition 5.12

The map $\partial _f:p_*(f^T_*{\mathcal O}^\times /{\mathcal O}^\times ) \rightarrow f_*{\mathbb Z}/{\mathbb Z}$ is split by a natural map of sheaves on S:

$$\begin{aligned} t_f:f_*{\mathbb Z}/{\mathbb Z}\rightarrow p_*\left( f^T_*{\mathcal O}^\times /{\mathcal O}^\times \right) . \end{aligned}$$

Proof

Applying the left exact functor $p_*$ to $1\rightarrow {\mathcal O}^\times \rightarrow f^T_*{\mathcal O}^\times \rightarrow f^T_*{\mathcal O}^\times /{\mathcal O}^\times \rightarrow 1$, we get exactness of the middle row in the following commutative diagram of sheaves on S.

(The top and bottom rows are tautologically exact.) The maps $t_S$, $f_*t_X$ induce the map $t_f$; since $\partial _S\, t_S$ and $\partial _X\, t_X$ are the identity, so are $(f_*\partial _X)(f_*t_X)$ and $\partial _f\, t_f$. $\square$

Proof of Theorem 5.2

By Theorem 5.7, we have $L{\mathcal I}(f)\cong H_{\mathrm {et}}^0(S,f_*{\mathbb Z}/{\mathbb Z})$. By Proposition 5.12, we have a natural section $t_f$ of the sheaf map $\partial _f$. By 5.5, the global sections of the map $\partial _f$ is the map ${\mathcal I}(f[t,1/t])\rightarrow L{\mathcal I}(f)$ in $Seq({\mathcal I},f)$. Hence the global sections of $t_f$ provide the required natural splitting. $\square$

Here is an easy consequence of Theorem 5.2, which is related to Lemma 4.3.

Corollary 5.13

Suppose that $f:A\hookrightarrow B$ and $g:B\hookrightarrow C$ are extensions. Then there is a short exact sequence

$$\begin{aligned} 1 \rightarrow L{\mathcal I}(A,B) \rightarrow L{\mathcal I}(A,C) \rightarrow L{\mathcal I}(B,C). \end{aligned}$$

More generally, given faithful affine maps $X{\;\buildrel g \over \longrightarrow }\;T{\;\buildrel f \over \longrightarrow }\;S$, there is an exact sequence

$$\begin{aligned} 1 \rightarrow L{\mathcal I}(f) \rightarrow L{\mathcal I}(fg) \rightarrow L{\mathcal I}(g). \end{aligned}$$

Proof

Applying $f_*$ to the exact sequence $0\rightarrow {\mathbb Z}\rightarrow g_*{\mathbb Z}\rightarrow (g_*{\mathbb Z})/{\mathbb Z}\rightarrow 0$ on T (or ${\text {Spec}}(B)$) yields the exact sequence of Nisnevich sheaves on S (or ${\text {Spec}}(A)$):

$$\begin{aligned} 1\rightarrow (f_*{\mathbb Z})/{\mathbb Z}\rightarrow (fg)_*{\mathbb Z}/{\mathbb Z}\rightarrow f_*(g_*{\mathbb Z}/{\mathbb Z}). \end{aligned}$$

Now apply the left exact global sections functor and use Theorem 5.1. $\square$

6 The vanishing of $L{\mathcal I}(A,B)$

In this section, we discuss some conditions on $A\subset B$ under which $L{\mathcal I}(A,B)=0$. We begin by noting two elementary consequences of the sheaf property of $f_*{\mathbb Z}/{\mathbb Z}$: (i) if $s,t\in A$ are comaximal then $L{\mathcal I}(A,B)\subset L{\mathcal I}(A[\frac{1}{s}],B[\frac{1}{s}])\oplus L{\mathcal I}(A[\frac{1}{t}],B[\frac{1}{t}])$, and (ii) if $L{\mathcal I}(A_\wp ,B_\wp )=0$ for every prime $\wp$ of A then $L{\mathcal I}(A,B)=0$. The converse does not hold:

Example 6.1

If $A={\mathbb C}[x]$ and $B={\mathbb C}[x,y]/(y^2-x^2)$ then $L{\mathcal I}(A,B)=0$, but if $\wp \ne xA$ we have $L{\mathcal I}(A_\wp ,B_\wp )={\mathbb Z}$ (use Proposition 3.4).

If $A=k[s,s^{-1}]$ and $B=k[x,x^{-1}]$ with $s=x^2$ then $L{\mathcal I}(f)=0$ but $L{\mathcal I}(A_\wp ,B_\wp )={\mathbb Z}$ for every nonzero prime $\wp$ of A. Thus $(f_*{\mathbb Z})/{\mathbb Z}$ is nonzero; its stalk is ${\mathbb Z}$ at any closed point, but is 0 at the generic point.

A simple necessary condition for $L{\mathcal I}(f)$ to vanish is for the Nisnevich sheaf $f_*{\mathbb Z}/{\mathbb Z}$ to vanish. It is not enough for the Zariski sheaf $f_*{\mathbb Z}/{\mathbb Z}$ to vanish; Example 5.5.1 shows that even if A is a local ring we can have $(f_*{\mathbb Z}/{\mathbb Z})_{\mathrm {zar}}=0$ and $H^0_{\mathrm {zar}}(A,f_*{\mathbb Z}/{\mathbb Z})=0$ but $L{\mathcal I}(f)=H^0_{\mathrm {nis}}(A,f_*{\mathbb Z}/{\mathbb Z})\ne 0$.

For finite morphisms, we have a simple criterion. We say that a map $f:X\rightarrow S$ is connected if it for every point s of S, the fiber $X_s=f^{-1}(s)$ is connected or empty. For a map ${\text {Spec}}(B)\rightarrow {\text {Spec}}(A)$, this means that for each prime ideal $\wp$ of A either there is no prime of B over $\wp$ or else the fiber ring $B\otimes _A k(\wp )$ is connected.

Proposition 6.2

Suppose that $f:X\rightarrow S$ is finite. Then

(a) the Nisnevich sheaf $(f_*{\mathbb Z})/{\mathbb Z}$ is zero if and only if f is connected.

(b) If f is connected then $L{\mathcal I}(f)=0$.

Proof

Since the problem is local in S, we may suppose that $S={\text {Spec}}(A)$ and $X={\text {Spec}}(B)$, with A a local ring. Let $A^h$ be the henselization of $A_\wp$, and set $B'=B\otimes _A A^h$. Since f is finite, $B'$ is a product of $n\ge 1$ hensel local rings $B_i$, each finite over $A^h$; see [3, 1.4.2]. Since $B/\wp B = B'/\wp B' = \prod B_i/\wp B_i$, the fiber of f at $\wp$ has n components, and is connected iff $n=1$, i.e., iff $B'$ is hensel local. Since the stalk of $f_*{\mathbb Z}/{\mathbb Z}$ at $\wp$ is zero iff $B'$ is hensel local, the result follows. $\square$

Examples 6.3

The hypothesis in 6.2 that f be finite is necessary.

(a) If $A={\mathbb Z}$ and $B={\mathbb Z}[\frac{1}{p}]\times {\mathbb Z}/p$ then $f:{\text {Spec}}(B)\rightarrow {\text {Spec}}(A)$ is quasi-finite and connected, but $L{\mathcal I}(f)={\mathbb Z}$ by Corollary 3.5. Here $f_*{\mathbb Z}/{\mathbb Z}$ is a skyscraper sheaf (at p).

(b) If A is the coordinate ring $k[x,y]/(y^2-x^3-x^2)$ of the node, and $B=A[1/x]$ then $A\subset B$ is étale and connected, yet $L{\mathcal I}(A,B)\cong L\!{\text {Pic}}(A)\ne 0$ by Corollary 3.5. In this case, $f_*{\mathbb Z}/{\mathbb Z}$ is the skyscraper sheaf ${\mathbb Z}$ at the nodal point.

(c) If $A=k[x]$ and $B=A[b,e]/(e^2-e-bx)$ then f is not connected, as $B/xB\cong k[b]\times k[b]$. On the other hand, $f_*{\mathbb Z}/{\mathbb Z}=0$ and hence $L{\mathcal I}(f)=0$. In fact, if $\wp \ne xA$ then $B\otimes A^h\cong A^h[e]$. In this case, f has relative dimension 1.

Examples 6.4

Even if f is finite but not connected we may still have $L{\mathcal I}(f)=0$.

a) $L{\mathcal I}({\mathbb R}[x],{\mathbb C}[x])=0$, but ${\mathbb R}[x]\subset {\mathbb C}[x]$ is not connected. In fact, $f_*{\mathbb Z}/{\mathbb Z}$ is nonzero exactly at those primes $\wp$ with ${\mathbb R}[x]/\wp \cong {\mathbb C}$. This example shows that the rank of $f_*{\mathbb Z}/{\mathbb Z}$ is not semicontinuous.

b) If $A=k[x]$ and $B=k[x,y]/(y^2=x^3+x^2)$ is the coordinate ring of the node then $L{\mathcal I}(f)=H^0({\text {Spec}}(A),f_*{\mathbb Z}/{\mathbb Z})=0$ but $(f_*{\mathbb Z})/{\mathbb Z}$ is nonzero because the stalk is ${\mathbb Z}$ at every point except at $x=0,-1$ and at the generic point (where the stalks are 0).

We now turn to the connection between $L{\mathcal I}$ and seminormalization.

Definition 6.5

(Swan [10, §2]) An extension $A\subset B$ is subintegral if B is integral over A, and ${\text {Spec}}(B)\rightarrow {\text {Spec}}(A)$ is a bijection inducing isomorphisms on all residue fields.

We say that A is seminormal in B if whenever $b\in B$ and $b^{2}, b^{3}\in A$ then $b\in A$. The seminormalization of A in B is the largest subring ${}^{^+}\!\!\!A_B$ of B which is subintegral over A. By [10, 2.5], ${}^{^+}\!\!\!A_B$ is seminormal in B.

These notions extend to faithful affine maps of schemes in the evident way; the seminormalization of S in X may be constructed by gluing together the seminormalizations on each affine open. We omit the details.

Remark 6.5.1

The condition that $N{\mathcal I}(A,B)=0$ is equivalent to A being seminormal in B, and implies that $N^i{\mathcal I}(A,B)=0$ for all $i>0$; this was proven in [8, 1.5,1.7]. More generally, a faithful affine map $f:X\rightarrow S$ is seminormal if and only if $N{\mathcal I}(f)=0$. This follows from the affine case, since both $N{\mathcal I}$ and seminormality are Zariski-local on S.

Proposition 6.6

$L{\mathcal I}(A,{}^{^+}\!\!\!A_B)=0$ and $L{\mathcal I}(A,B)\cong L{\mathcal I}({}^{^+}\!\!\!A_B,B)$.

Proof

The first assertion follows from Proposition 3.4, since $LU(A)=LU({}^{^+}\!\!\!A_B)$ (because ${\text {Spec}}(A)\rightarrow {\text {Spec}}({}^{^+}\!\!\!A_B)$ is a bijection) and $L\!{\text {Pic}}(A)=L\!{\text {Pic}}({}^{^+}\!\!\!A_B)$, by [11, 5.4]. (The hypothesis in [11, 5.4] that A be reduced is not needed in its proof.)

Now ${\mathcal I}(A[t,1/t],B[t,1/t])\rightarrow {\mathcal I}({}^{^+}\!\!\!A_B[t,1/t],B[t,1/t])$ is onto by [7, 4.1]. Hence the map $L{\mathcal I}(A,B)\rightarrow L{\mathcal I}({}^{^+}\!\!\!A_B,B)$ is onto. By Corollary 5.13, the kernel is $L{\mathcal I}(A,{}^{^+}\!\!\!A_B)=0$.   $\square$

Definition 6.7

(Asanuma) A ring extension $A\subset B$ is called anodal if every $b\in B$ such that $(b^{2}- b)\in A$ and $(b^{3}- b^{2})\in A$ belongs to A.

If $A\subset B$ is anodal then every idempotent of B belongs to A, so $H^0(A,{\mathbb Z})=H^0(B,{\mathbb Z})$. In particular, every anodal extension of a domain is connected. If A is a field then $A\subset B$ is anodal if and only if B is connected. The first author proved that the composition of anodal extensions is anodal; see [7, 3.1].

The following result generalizes a result of Asanuma (see [11, 3.4]), who considered the case $B=\text {frac}(A)$, as well as several results of the first author in [7].

Theorem 6.8

Let $A\subset B$ be an extension.

• (1) If $L{\mathcal I}(A, B)=0$ then $A\subset B$ is anodal.

• (2) If A is a 1-dimensional domain, and $A\subset B$ is an integral, birational and anodal extension, then $L{\mathcal I}(A,B)=0$.

Example 6.3(b) shows that the integral hypothesis is necessary in Theorem 6.8(2). Example 6.9 shows that not all integral, birational anodal extensions have $L{\mathcal I}(f)\!=\!0$.

Proof

(cf. Onoda-Yoshida [4, 1.10]) Let $b\in B$ be such that $b^2-b, b^3-b^2$ are in A; we need to show that $b\in A$. Consider the finite subring $C=A[b]$ of B. If $L{\mathcal I}(A,B)=0$, then $L{\mathcal I}(A,C)=0$ by Corollary 5.13. Let ${\mathfrak a}$ denote the ideal $(b^2-b)C$ of C; it is also an ideal of A, so $L{\mathcal I}(A/{\mathfrak a},C/{\mathfrak a})=0$ by Lemma 4.4. By Proposition 3.4, this implies that $H^0(A/{\mathfrak a},{\mathbb Z})\cong H^0(C/{\mathfrak a},{\mathbb Z})$. Since the image $\bar{b}$ of b is idempotent in $C/{\mathfrak a}$, this forces $\bar{b}\in A/{\mathfrak a}$ and hence $b\in A$, proving (1).

(2) Now suppose that A is a 1-dimensional domain, and that $A\subset B$ is an integral, birational and anodal extension. Since B is the union of finite A-algebras $B_\lambda$, all of which are anodal over A, we may assume that B is a finite A-algebra by Lemma 4.5. Since $A\subset B$ is finite and birational, the conductor ideal ${\mathfrak c}$ is nonzero, so $\dim A/{\mathfrak c}=0$. By [11, 3.6], the extension $\bar{f}:A/{\mathfrak c}\subset B/{\mathfrak c}$ is anodal because $A\subset B$ is. In particular, $\bar{f}$ is connected. Since $\dim (A/{\mathfrak c})=0$, $L\!{\text {Pic}}(A/{\mathfrak c})=0$ (by [11, 1.6.1]) and hence $L{\mathcal I}(\bar{f})=0$ by Proposition 3.4. By Lemma 4.4, $L{\mathcal I}(A,B)=L{\mathcal I}(A/{\mathfrak c},B/{\mathfrak c})=0$. $\square$

Here is an example of a 2-dimensional integral, birational and anodal extension with $L{\mathcal I}(A,B)\ne 0$, Thus Theorem 6.8(2) does not extend to $\dim (A)>1$.

Example 6.9

Let X be the coordinate axes in the plane, and $f:X\rightarrow S$ the quotient identifying each axis with the normalization of the node S. Consider the pushout $S\rightarrow S'={\text {Spec}}(A)$ of the tautological inclusion of X in $\mathbb {A}^2={\text {Spec}}(B)$ along f.

The map $\mathbb {A}^2\rightarrow S'$ is the map ${\text {Spec}}(B)\rightarrow {\text {Spec}}(A)$ of Example 3.5 in [11]. By construction, A is a 2-dimensional domain whose integral closure is $B=k[x,y]$, so $A\subset B$ is an integral, birational extension. It is shown in [11, 3.5.2] that $A\subset B$ is anodal and $L\!{\text {Pic}}A={\mathbb Z}$. Since B is normal, $L\!{\text {Pic}}B=0$. Since A, B are domains, $LU(A)=LU(B)={\mathbb Z}$ (by Example 2.4). By Proposition 3.4, $L{\mathcal I}(A,B)\cong L\!{\text {Pic}}(A)\cong {\mathbb Z}$.