aa r X i v : . [ m a t h . AG ] D ec COMPUTING p -ADIC INTEGRALS USING MOTIVIC INTEGRATIONKARL RÖKAEUSAbstra
t. We use the theory of motivi
integration in order to give a geometri
explanation of thebehavior of some p -adi
integrals. 1. Introdu
tionIn [Sko09℄ the author
omputes the integral I p = R Z np | Q ≤ i 7→ | X ( F q ) | .It extends by
ontinuity to a homomorphism C q : K ( Var F p ) → R . Now, when X is a variety over a
omplete dis
rete valuation ring, we may de(cid:28)ne a measure µ X on
ertain subsets of its ar
spa
e, X ∞ .This measure takes values in K ( Var k ) , where k is the residue (cid:28)eld of the dis
rete valuation ring. (Insteadof taking values in d M k , whi
h is the standard
hoi
e.) In
ase X = A n Z p we will have X ∞ = W ( F p ) n ,and C q µ X ( A ) = µ Haar ( A ∩ W ( F q ) n ) for every measurable subset A ⊂ X ∞ .Using this theory of motivi
integration, we prove that if X = A n Z p then I = Z X ∞ | Y ≤ i 7→ | X ( F q ) | , are well-de(cid:28)ned and
ontinuous. In this paper the questions of
onvergen
e will mainly boil down to the following result,whi
h is Example 2.14 in [Rök08℄:Proposition 2.1. Let k be any (cid:28)eld. If { e i } i ∈ N is a sequen
e of integers su
h that e i → ∞ , then P i ∈ N L − e i is
onvergent in K ( Var k ) .It follows for example that − L − is invertible. On the other hand, P n ≥ n L − n is not
onvergent; − L − is not invertible (
ontrary to with respe
t to the dimension (cid:28)ltration).We have a partial ordering on K ( Var k ) , de(cid:28)ned on K ( Var k ) as x ≤ y if x + [ X ] = y for some s
heme X , whi
h we may often use to redu
e questions about
onvergen
e to Proposition 2.1.The
onstru
tion of the motivi
measure theory in [Rök08℄ is the standard one, used in the appendixof [DL02℄, in [Loo02℄ and in [Seb04℄. We will not go through these de(cid:28)nitions here, we just tell what theresults are in the
ase of interest in this paper.From now on, we assume that X = A d O where O is a
omplete dis
rete valuation ring with residue(cid:28)eld k , of one of the following types: • k is a perfe
t (cid:28)eld, of prime
hara
teristi p , and O = W ( k ) , where W is the ring s
heme of Wittve
tors
onstru
ted with respe
t to the prime p . • k is a (cid:28)eld, and O = k [[ t ]] .OMPUTING p -ADIC INTEGRALS USING MOTIVIC INTEGRATION 3(We are mainly interested in the
ases when O is either Z p or Q [[ t ]] so that k = F p and Q respe
tively.)Then the spa
e of ar
s X ∞ of X = A d O is the set X ∞ = ( W ( k ) d , O = W ( k ) k [[ t ]] d , O = k [[ t ]] , where k is an algebrai
losure of k . We use the standard terminology in
onne
tion with the Witt ve
tors,see the appendix. Moreover, in order to get a uniform notation, we will write ( x i ) i ≥ for P i ≥ x i t i ∈ k [[ t ]] .For every n , if O = W ( k ) then X n is the s
heme whose R -points is W dn ( R ) for every k -algebra R ,where W n is the Witt ve
tors of length n . If instead O = k [[ t ]] then X n ( R ) = ( R [ t ] / ( t n )) d for every k -algebra R . We use π n to denote the proje
tion X ∞ → X n ( k ) .The de(cid:28)nitions of stable and measurable subsets of X are the standard ones. Moreover, the measureof a subset A ⊂ X ∞ whi
h is stable of level n is µ X ( A ) := [ π n ( A )] L − nd ∈ K ( Var k ) . The measure ofa general measureable set, and of a measurable fun
tion, is de(cid:28)ned in the standard way. The following(Propositions . and . in [Rök08℄) are the properties that we are after:Proposition 2.2. Let X = A d Z p . When A ⊂ X ∞ is measurable we have, for every power q of p , µ Haar ( A ∩ W ( F q ) d ) = C q ( µ X ( A )) . Moreover, if f : X ∞ → K ( Var k ) is measurable, then C q R X ∞ f dµ X = R W ( F q ) d C q ◦ f dµ Haar .2.2. Motivi
integrals of absolute values of polynomials. We
ontinue to use O to denote a
om-plete dis
rete valuation ring with residue (cid:28)eld k .To show that the integrals we are interested in exists we need the following simple approximation:Lemma 2.3. Let X = A d O and let A be a measurable subset of X ∞ . Then ≤ µ X ( A ) ≤ and dim µ X ( A ) ≤ .Proof. Suppose (cid:28)rst that A is stable of level n . We have π n ( A ) ⊂ X n = A dnk , giving immediatelythat dim A = dim π n ( A ) − nd ≤ nd − nd = 0 , i.e., dim µ X ( A ) ≤ . Also ≤ [ π n ( A )] ≤ L dn , hen
e ≤ µ X ( A ) ≤ .In the general
ase there are, by de(cid:28)nition, stable subsets A i su
h that µ X ( A ) = lim i →∞ µ X ( A i ) , i.e., µ X ( A ) = ( µ X ( A i )) i ∈ N . Sin
e the equalities holds for stable sets it follows by de(cid:28)nition that i ∈ N ≤ µ X ( A ) ≤ (1) i ∈ N = 1 . The same reasoning goes for the statement about dimension. (cid:3) Lemma 2.4. Let X = A d O and let A i be measurable subsets of X ∞ . If e i → ∞ as i → ∞ , then the sum P i ∈ N µ X ( A i ) L − e i is
onvergent.Proof. By Lemma 2.3, ≤ µ X ( A i ) L − e i ≤ L − e i . By Proposition 2.1 the sum P i ∈ N L − e i is
onvergentif and only if dim L − e i → −∞ , i.e., e i → ∞ . In this
ase it follows from Lemma 2.17 of [Rök08℄ that P i ∈ N µ X ( A i ) L − e i is
onvergent. (cid:3) Let X = A O and let X ∞ be its ar
spa
e. We have a fun
tion ord : X ∞ → N ∪ {∞} , mapping x to the biggest power of the uniformizer dividing x , with the properties that ord ab = ord a + ord b and ord( a + b ) ≥ min { ord a, ord b } . We
ontinue to write X for A d O . If f ∈ O [ X , . . . , X d ] it de(cid:28)nes afun
tion X ∞ → X ∞ . So for n ∈ N we may
onsider the subset { ord f ≥ n } := { ( a , . . . , a d ) ∈ X ∞ :ord f ( a , . . . , a d ) ≥ n } . When O = Z p this set has the property that { ord f ≥ n } ∩ Z dp = { ( a , . . . , a d ) ∈ Z dp : ord p f ( a , . . . , a d ) ≥ n } .Lemma 2.5. Let X = A d O . The subset { ord f ≥ n } ⊂ X ∞ is stable of level n and µ X ( { ord f ≥ n } ) =[ π n ( { ord f ≥ n } )] L − dn .Proof. To simplify the notation, we prove this for the spe
ial
ase when O = W ( k ) ; the general
ase issimilar. When we view X ∞ as the k -points on the ring s
heme W d , we see that { ord f ≥ n } is a
tually the k -points on a
losed subs
heme. For let k [ X i , X i , . . . , X iN , . . . ] di =1 represent W d . Let f , f . . . be theuniversal polynomials de(cid:28)ning f in the Witt ve
tors, i.e., f n ∈ k [ X i , . . . , X in ] di =1 and if R is any k -algebraand r i = ( r i , r i , . . . ) ∈ W ( R ) for i = 1 , . . . , d , then f ( r , . . . , r d ) = ( f ( r , . . . , r d ) , f ( r , . . . , r d ) , . . . ) ∈ KARL RÖKAEUS W ( R ) . (We write f n ( r , . . . , r d ) for f n ( r , . . . , r d ; . . . ; r n , . . . , r dn ) ∈ R .) We then see that { ord f ≥ n } is identi(cid:28)ed with { ( x , . . . , x d ) ∈ W ( k ) d : f ( x , . . . , x d ) ≡ n ) } ⊂ X ∞ , i.e., { ( x , . . . , x d ) ∈ W ( k ) d : f i ( x , . . . , x d ) = 0 for i = 0 , . . . , n − } ⊂ X ∞ , This in turn is the k -points on the
lose subs
heme Spec k [ X i , X i , . . . , X iN , . . . ] di =1 ( f , . . . , f n − ) ⊂ W d . Now π m ( X ∞ ) = W dm ( k ) . Hen
e, for m ≥ n , we see that π m ( { ord f ≥ n } ) is the k -points on the k -s
heme Spec k [ X i , . . . , X i,m − ] di =1 ( f , . . . , f n − ) . In what follows we identify π m ( { ord f ≥ n } ) with its underlying s
heme. We then see that π m ( { ord f ≥ n } ) = π n ( { ord f ≥ n } ) × k A d ( m − n ) k . The result follows. (cid:3) Next
onsider the fun
tion a L − ord f ( a ) : X ∞ → K ( Var k ) . For a ∈ W ( k ) we write | a | := L − ord a and we want to
ompute the integral R X ∞ | f | dµ X . The following proposition shows that the integralexists.Proposition 2.6. Let X = A d O . Let A be a measurable subset of X ∞ , and f ∈ O [ X , . . . , X d ] . Theintegral R A | f | dµ X = R A L − ord f dµ X exists. Moreover, when X = A d Z p we have, for q any power of p , C q R A | f | dµ X = R A ∩ W ( F q ) d | f | p dµ Haar .Proof. By de(cid:28)nition the integral equals µ X ( f = 0) · P i ∈ N µ X (cid:0) A ∩ { ord f = i } (cid:1) L − i . By Lemma 2.5, { ord f = i } is stable, hen
e A ∩ { ord f = i } is measurable. The integral therefore exists by Lemma 2.4.Next, by Proposition 2.2, C q R A L − ord f dµ X = R A ∩ W ( F q ) d | f | p dµ Haar . (cid:3) The primary purpose of this paper is to show that the integral of the absolute value of a
ertainpolynomial in many variables is a rational fun
tion in L , with
oe(cid:30)
ients in Z . For this we begin withsome lemmas about general integrals of this kind of fun
tions.Lemma 2.7. Let A = S i ∈ N A i be a disjoint union of stable subsets and suppose that P i ∈ N µ X ( A i ) is
on-vergent (so that A is measurable). Then for any f ∈ O [ X , . . . , X d ] , we have R A | f | dµ X = P i ∈ N R A i | f | dµ X .Proof. By Proposition 2.6 the integral exists. Sin
e A i ∩{ ord f = m } is stable, and A i ∩{ ord f = m } ⊂ A i ,it follows from Lemma 3.8 of [Rök08℄ that the sum P i ∈ N µ X (cid:0) A i ∩ { ord f = m } (cid:1) is
onvergent. Hen
e,sin
e the union A ∩ { ord f = m } = S i ∈ N A i ∩ { ord f = m } is disjoint it follows from Proposition 3.7 oflo
.
it. that µ X (cid:0) A ∩ { ord f = m } (cid:1) = P i ∈ N µ X (cid:0) A i ∩ { ord f = m } (cid:1) . We may therefore write Z A | f | dµ X = X m ∈ N µ X (cid:0) A ∩ { ord f = m } (cid:1) L − m = X m ∈ N X i ∈ N µ X (cid:0) A i ∩ { ord f = m } (cid:1) L − m . Be
ause of Lemma 2.4, if we do the above summation over an enumeration of N , it is
onvergent. Hen
eby Lemma 2.19 of lo
.
it it equals X i ∈ N X m ∈ N µ X (cid:0) A i ∩ { ord f = m } (cid:1) L − m = X i ∈ N Z A i | f | dµ X . (cid:3) Let f , . . . , f r ∈ O [ X , . . . , X d ] . For α = ( α , . . . , α r ) ∈ N r we write { ord f i = α i } ri =1 for the subset { a ∈ X ∞ : ord f i ( a ) = α i } ri =1 ⊂ X ∞ .Lemma 2.8. Let X = A d O . For I ⊂ ( N ∪ {∞} ) d ((cid:28)nite or in(cid:28)nite), let U I := S α ∈ I { ord X i = α i } di =1 .Let f ∈ O [ X , . . . , X d ] . Then R U I | f | dµ X = P α ∈ I R { ord X i = α i } | f | dµ X . (In parti
ular the integral exists.)OMPUTING p -ADIC INTEGRALS USING MOTIVIC INTEGRATION 5Proof. We show that U I is measurable, the result then follows from Lemma 2.7. Sin
e the union U I = S α ∈ I { ord X i = α i } di =1 is disjoint it su(cid:30)
es, by Proposition 3.7 of [Rök08℄, to prove
onvergen
e of thesum X α ∈ I µ X ( { ord X i = α i } di =1 ) . Let N be a large integer. We have π N ( { ord X i = α i } di =1 ) ⊂ π N ( { ord X i ≥ α i } di =1 ) . The underlyings
heme of this latter set is Spec k [ X i , . . . , X iN ] di =1 ( X i , . . . , X i,α i − ) di =1 = Spec k [ X iα i , . . . , X iN ] di =1 . Hen
e [ π N ( { ord X i = α i } di =1 )] ≤ L Nd − P di =1 α i and
onsequently µ X ( { ord X i = α i } di =1 ) ≤ L − P di =1 α i .Now, when I is in(cid:28)nite we see that as α varies over I , max { α i } di =1 → ∞ , hen
e that − P di =1 α i → −∞ .So by Proposition 2.1, P α ∈ I L − P di =1 α i is
onvergent, hen
e by Lemma 2.17 of lo
.
it, P α ∈ I µ X ( { ord X i = α i } di =1 ) is
onvergent. (cid:3) 3. The motivi
integral of a polynomial in one variableLet O be a
omplete dis
rete valuation ring with residue (cid:28)eld k ,
onstru
ted as in Subse
tion 2.1. Inthe pre
eding se
tion we proved the existen
e of the integral R X ∞ | f | dµ X ∈ K ( Var k ) , where X = A d O ,and f is a polynomial in d variables with
oe(cid:30)
ients in O . In this se
tion we
ompute this integralmore expli
itly, in the
ase when d = 1 . This type of
omputation is standard in the theory of motivi
integration. We in
lude this se
tion anyway, in order to (cid:28)x notation for the later se
tions, and also togive a simple example.In this se
tion we let X = A O . We use the notation of the pre
eding se
tion. In parti
ular, let f i ∈ k [ X , . . . , X i ] be the universal polynomials that de(cid:28)ne the fun
tion x f ( x ) : X ∞ → X ∞ . Morepre
isely, the f i are su
h that if O = W ( k ) and X := ( X , X , . . . , X i , . . . ) ∈ W ( k [ X , X , . . . , X i , . . . ]) then f ( X ) = (cid:0) ( f ( X ) , f ( X , X ) , . . . , f i ( X , . . . , X i ) , . . . (cid:1) ∈ W ( k [ X , X , . . . , X i , . . . ]) If O = k [[ t ]] and X := P i ≥ X i t i ∈ k [ X , . . . , X i , . . . ][[ t ]] then f ( X ) = P i ≥ f i ( X , . . . , X i ) t i .Let x = ( x , x , . . . , x i , . . . ) ∈ X ∞ (re
all that we write ( x , x , . . . ) for P i ≥ x i t i in order to get auniform notation). Write f i ( x ) for f i ( x , . . . , x i ) . That the inequality ord f ( x ) ≥ n holds is then equivalentto f ( x ) = f ( x ) = · · · = f n − ( x ) = 0 . If m ≥ n we then see that π m ( x ) = ( x , . . . , x m − ) ∈ X m ( k ) belongs to π m (cid:0) { ord f ≥ n } (cid:1) if and only if f ( x ) = f ( x ) = · · · = f n − ( x ) = 0 . Therefore, π m (cid:0) { ord f ≥ n } (cid:1) equals the k -points on the k -s
heme Spec k [ X ,...,X m − ]( f ,...,f n − ) , and π − m π m (cid:0) { ord f ≥ n } (cid:1) = { ord f ≥ n } . In whatfollows we will identify π m (cid:0) { ord f ≥ n } (cid:1) with its underlying s
heme.Proposition 3.1 (Motivi
Newton's Lemma). Let f ∈ O [ X ] and assume that f is non-
onstant andseparable. Consider the subset { ord f ≥ n } ⊂ X ∞ , where n ≥ . We have an isomorphism of k -s
hemes π n (cid:0) { ord f ≥ n } (cid:1) → π (cid:0) { ord f ≥ n } (cid:1) . In parti
ular, µ X (cid:0) { ord f ≥ n } (cid:1) = h Spec k [ X ]( f ) i L − n .Proof. Let R i := k [ X , . . . , X i − ]( f , . . . , f i − ) i ≥ . Then π i (cid:0) { ord f ≥ n } (cid:1) = Spec R i for i = 1 , . . . , n and we want to prove that R n ≃ R . We do this byproving that the
anoni
al homomorphism R i → R i +1 is an isomorphism for every i ≥ .To simplify notation, we do this only in the mixed
hara
teristi
ase. Let x := ( X , . . . , X i ) ∈ W i +1 (cid:0) k [ X , . . . , X i ] (cid:1) and ˜ x := ( X , . . . , X i − , . From (A.1) it follows that x = ˜ x + V i r( X i ) , so we may KARL RÖKAEUSTaylor expand to get f ( x ) = f (cid:0) ˜ x + V i r( X i ) (cid:1) = f (˜ x ) + ∂f∂X (˜ x ) · V i r( X i ) + O (cid:0) V i r( X i ) (cid:1) ∈ W i +1 (cid:0) k [ X , . . . , X i ] (cid:1) . (3.2)Here f (˜ x ) = ( f , . . . , f i − , q ) , where q is a polynomial in k [ X , . . . , X i − ] . Moreover, sin
e π f = f , itfollows that if ∂f∂X ( x ) = ( f ∗ , . . . , f ∗ i ) then f ∗ = ∂f ∂X . Hen
e ∂f∂X (˜ x ) = ( ∂f ∂X , . . . ) . Finally by PropositionA.2 (cid:0) V i r( X i ) (cid:1) = F i V i (cid:0) r( X i ) (cid:1) = 0 ∈ W i +1 (cid:0) k [ X , . . . , X i ] (cid:1) . Hen
e if we write expli
itly we see thatthe right hand side of (3.2) is ( f , . . . , f i − , q ) + ( ∂f ∂X , . . . ) · (0 , . . . , , X i ) =( f , . . . , f i − , q ) + (0 , . . . , , ( ∂f ∂X ) p i X i )=( f , . . . , f i − , q + ( ∂f ∂X ) p i X i ) . On the other hand, by de(cid:28)nition, f ( x ) = ( f , . . . , f i ) ∈ W i +1 (cid:0) k [ X , . . . , X i ] (cid:1) hen
e we get the identity(3.3) f i ( X , . . . , X i ) = q ( X , . . . , X i − ) + ( ∂f ∂X ) p i · X i in k [ X , . . . , X i ] .We shall also use the hypothesis that f is separable modulo V . This means that ∂f ∂X is invertible in R . Let ( ∂f ∂X ) − and h ∈ k [ X ] be su
h that ∂f ∂X · ( ∂f ∂X ) − = 1 + hf in k [ X ] .We now prove that R i → R i +1 is inje
tive. Let g ∈ k [ X , . . . , X i − ] . We have to prove that if g = 0 ∈ R i +1 then g = 0 ∈ R i . So suppose that g = h f + · · · + h i f i ∈ k [ X , . . . , X i ] , where h j ∈ k [ X , . . . , X i ] . By (3.3) this gives g = h f + · · · + h i − f i − + h i · (cid:0) q + ∂f ∂X p i · X i (cid:1) ∈ k [ X , . . . , X i ] . Substituting − q · ( ∂f ∂X ) − p i for X i then gives g = h ∗ f + · · · + h ∗ i − f i − + h ∗ i · ( q − q + h ∗ f ) ∈ k [ X , . . . , X i − ] , where the h ∗ j and h ∗ are polynomials in k [ X , . . . , X i − ] . Hen
e g = 0 ∈ R i and
onsequently R i → R i +1 is inje
tive.Finally we prove that R i → R i +1 is surje
tive. It su(cid:30)
es to show that X i is in the image of R i .Working in R i +1 , (3.3) be
omes(3.4) q + ∂f ∂X p i · X i . Now ∂f ∂X ∈ R is invertible, and R = 0 by assumption. Sin
e R → R i +1 is inje
tive it hen
e followsthat the image of ∂f ∂X in R i +1 is invertible,
onsequently we
an write (3.4) as X i = − q · ∂f ∂X − p i ∈ R i +1 . The right hand side involves only the variables X , . . . , X i − , hen
e is in the image of R i . (cid:3) Proposition 3.5. Let f ∈ O [ X ] and assume that f ∈ k [ X ] is separable and non-
onstant. Then Z | f | dµ X = 1 − [Spec k [ X ] / ( f )] 1 L + 1 ∈ K ( Var k ) . Proof. By de(cid:28)nition we have Z X ∞ | f | dµ X = X m ≥ L − m µ X { ord f = m } . Sin
e { ord f = m } = { ord f ≥ m } \ { ord f ≥ m + 1 } we have µ X { ord f = m } = h Spec k [ X ]( f ) i · ( L − m − L − ( m +1) ) OMPUTING p -ADIC INTEGRALS USING MOTIVIC INTEGRATION 7for m ≥ . For m = 0 we have µ X (ord f = 0) = µ X ( X ∞ \ { ord f ≥ } ) = 1 − h Spec k [ X ]( f ) i L − . Therefore,using Proposition 2.1, Z X ∞ | f | dµ X =1 + h Spec k [ X ]( f ) i · (cid:18) − L − + X m ≥ L − m (cid:0) L − m − L − ( m +1) (cid:1)(cid:19) =1 − h Spec k [ X ]( f ) i · L + 1 . (cid:3) Example 3.6. We look at the
ase when O = Z p : If f = aX + b , where a ∈ Z × p , then k [ X ]( f ) = k .Sin
e [Spec k ] = 1 we have R X ∞ | aX + b | dµ X = LL +1 , showing in parti
ular that if q is a power of p then R W ( F q ) | aX + b | dX = qq +1 More generally, assume that f is su
h that f is irredu
ible of degree d . Then Spec k [ X ] / ( f ) ≃ F p d ,hen
e R X ∞ | f | dµ X = 1 − [Spec F pd ] L +1 . Applying C q for di(cid:27)erent powers of p shows that Z W ( F q ) | f | p dµ Haar = ( − d/ ( q + 1) q = p i where d | i q = p i where d ∤ i . 4. Change of variablesWe prove three theorems about manipulation of these kind of integrals. Re
all that we use O to denotea
omplete dis
rete valuation ring with perfe
t residue (cid:28)eld k .4.1. Linear
hange of variables. A linear
hange of variables is easy to do also in the motivi
ase:Proposition 4.1. Let X = A d O and let a ij ∈ O be su
h that the determinant of M = ( a ij ) is in O × .Given f ∈ Z p [ X , . . . , X d ] , de(cid:28)ne g ( X , . . . , X d ) := f (cid:0) ( X , . . . , X d ) M (cid:1) . Then R X ∞ | f | dµ X = R X ∞ | g | dµ X .Proof. We (cid:28)rst prove that µ X { ord f ≥ n } = µ X { ord g ≥ n } for every n . We have a map { ord f ≥ n } → { ord g ≥ n } , given by ( x , . . . , x d ) ( x , . . . , x d ) M − . This is a bije
tion, for it is well de(cid:28)ned sin
e g ( xM − ) = f ( xM − M ) = f ( x ) ≡ n ) , and it has a well de(cid:28)ned inverse x xM . Therefore π n { ord f ≥ n } and π n { ord g ≥ n } are isomorphi
(viewed as subs
hemes of X n ),
onsequently [ π n { ord g ≥ n } ] =[ π n { ord f ≥ n } ] .It follows that µ X { ord f = n } = µ X { ord g = n } for every n , hen
e that the integrals are equal. (cid:3) X := A d O and Y := A e O . Moreover, let Z := A d + e O = X × O Y . We may then identify ( Z ) ∞ with X ∞ × Y ∞ . Our aim is to show the separation ofvariables result, Theorem 4.4. We do this using two partial results, whi
h we state as the following twolemmas:Lemma 4.2. If A ⊂ X ∞ and B ⊂ Y ∞ are stable, then A × B ⊂ Z ∞ is stable, and µ Z ( A × B ) = µ X ( A ) µ Y ( B ) .Proof. Sin
e A and B are stable there is an integer n with the property that there are a (cid:28)nite number of k -varieties V i su
h that π n ( A ) = S i V i ( k ) , and a (cid:28)nite number of k -varieties U i su
h that π n ( A ) = S i U i ( k ) .We have π n ( A × B ) = π n ( A ) × π n ( B )= [ i V i ( k ) × [ j U j ( k )= [ i,j V i ( k ) × U j ( k )= [ i,j ( V i × k U j )( k ) , KARL RÖKAEUShen
e [ π n ( A × B )] = P i,j [ V i × k U j ] = P i,j [ V i ][ U j ] = ( P i [ V i ])( P j [ U j ]) = [ π n ( A )][ π n ( B )] . Therefore µ X ×Y ( A × B ) = [ π n ( A × B )] L − n ( d + e ) = ([ π n ( A )] L − nd )([ π n ( B )] L − ne ) = µ X ( A ) µ Y ( B ) . (cid:3) Lemma 4.3. If A ⊂ X ∞ and B ⊂ Y ∞ are measurable, then A × B ⊂ Z ∞ is measurable, and µ Z ( A × B ) = µ X ( A ) µ Y ( B ) .Proof. Let A m and C mi be stable subsets of X ∞ su
h that A ∆ A m ⊂ S i ∈ N C mi . Let u m := P i ∈ N µ X ( C mi ) be
onvergent and lim m →∞ u m = 0 . Let B m and D mi be stable subsets of Y ∞ su
h that B ∆ B m ⊂ S i ∈ N D mi , where v m := P i ∈ N µ X ( D mi ) is
onvergent and lim m →∞ v m = 0 . Then ( A × B ) ∆( A m × B m ) =( A ∆ A m ) × ( B ∆ B m ) ⊂ S i ∈ N C mi × D mi . By Lemma 4.2, µ Z ( C mi × D mi ) = µ X ( C mi ) µ Y ( D mi ) , hen
e byLemma 2.21 of [Rök08℄, the sum s m := P i ∈ N µ Z ( C mi × D mi ) is
onvergent, and s m ≤ u m v m . Sin
e u m and v m tends to zero the same holds for u m v m and
onsequently also for s m . Hen
e, sin
e A m and B m are stable, µ Z ( A × B ) = lim m →∞ µ Z ( A m × B m ) = lim m →∞ µ X ( A m ) µ Y ( B m ) = µ X ( A ) µ Y ( B ) . (cid:3) Theorem 4.4 (Separation of variables). Let X := A d O , Y := A e O and Z = A d + e O . If A ⊂ X ∞ and B ⊂ Y ∞ are measurable, and f ∈ O [ X , . . . , X d ] , g ∈ O [ Y , . . . , Y e ] , then R A × B | f g | dµ Z = R A | f | dµ X · R B | g | dµ Y .Proof. By Proposition 2.6 the integral is
onvergent: Z A × B | f g | dµ Z = X ξ ∈ N µ Z (( A × B ) ∩ { ord f g = ξ } ) L − ξ Sin
e { ord f g = ξ } = S µ + ν = ξ { ord f = µ } × { ord g = ν } we have ( A × B ) ∩ { ord f g = ξ } = [ µ + ν = ξ (cid:0) A ∩ { ord f = µ } (cid:1) × (cid:0) B ∩ { ord g = ν } (cid:1) and sin
e this is a disjoint union of measurable sets it follows from the previous lemma that Z A × B | f g | dµ Z = X ξ ∈ N (cid:18) X µ + ν = ξ µ X (cid:0) A ∩ { ord f = µ } (cid:1) · µ Y (cid:0) B ∩ { ord g = ν } (cid:1)(cid:19) L − ξ Sin
e this sum is
onvergent, Lemma 2.20 of [Rök08℄ says that we may rearrange it to obtain X µ ∈ N µ X (cid:0) A ∩ { ord f = µ } (cid:1) L − µ ! X ν ∈ N µ Y (cid:0) B ∩ { ord f = ν } (cid:1) L − ν ! = Z A | f | dµ X · Z B | g | dµ Y (cid:3) k [ X • , . . . , X • N ] for the polynomial ring k [ X i , . . . , X iN ] ni =1 . As usual,we use Q N to denote the universal polynomials de(cid:28)ning Q ∈ O [ X , . . . , X n ] , so that Q N ∈ k [ X • , . . . , X • N ] .(See the dis
ussion in the proof of Lemma 2.5, or in the beginning of Se
tion 3.)Lemma 4.5. Let X = A n O . Let Q ∈ O [ X , . . . , X n ] be a form of degree s . If ord x i > for i = 1 , . . . , n ,then ord Q ≥ s . Moreover for every ξ ∈ N µ X ( { ord Q > ξ + s, ord x i > } ni =1 ) = L − n µ X ( { ord Q > ξ } ) . Proof. For N su(cid:30)
iently large, π N +1 (ord ∆ > ξ + s, ord x i > is the spe
trum of the algebra k [ X • , . . . , X • N ] (cid:0) Q , . . . , Q ξ + s , X • (cid:1) . In the mixed
hara
teristi
ase, it follows from Corollary A.4 that the
lass of this in K ( Var k ) equals(sin
e [ X ] = [ X red ] for any s
heme X ) the
lass of the spe
trum of k [ X • , . . . , X • N ] (cid:0) Q ( X • ) , . . . , Q ξ ( X • , . . . , X • ξ +1 ) (cid:1) . In the equal
hara
teristi
ase, this is straight forward to prove. Now, using the
hange of variables X • i X • i − , the spe
trum of this algebra is π N (ord Q > ξ ) . The result follows. (cid:3) OMPUTING p -ADIC INTEGRALS USING MOTIVIC INTEGRATION 9Theorem 4.6. Let X = A n O , let Q ∈ O [ X , . . . , X n ] be a form of degree s and let A = { ord x i > } ni =1 ⊂X ∞ . Then R A | Q | dµ X = L − s − n R X ∞ | Q | dµ X .Proof. By the (cid:28)rst part of Lemma 4.5, { ord Q = ξ, ord x i > } ni =1 = ∅ for ξ < s , hen
e Z A | Q | dµ X = X ξ ≥ µ X { ord Q = ξ + s, ord x i > } L − ( ξ + s ) . Using the se
ond part of the lemma it follows that this equals X ξ ≥ L − n µ X (ord Q = ξ ) L − ( ξ + s ) = L − s − n Z X ∞ | Q | dµ X . (cid:3) 5. The motivi
integral of the absolute value of a produ
t of linear formsAs before, we let O be a
omplete dis
rete valuation ring with residue (cid:28)eld k , of one of the typesdes
ribed in Subse
tion 2.1. We de(cid:28)ne, for any n ∈ N , X n = A n O , and we let X n ∞ be its spa
e of ar
s.The main result of this se
tion is Theorem 5.10, and also Theorem 5.15. In Theorem 5.10 we givea re
ursive method to
ompute I = R X n ∞ | Q ℓ i | dµ X n , where ℓ i ∈ Z [ X , . . . , X n ] are linear forms. When O = k [[ t ]] , these forms are arbitrary; when O = W ( k ) we need the forms to be of a rather spe
ial type.The restri
tion in the se
ond
ase is taken
are of in Theorem 5.15, when we give a re
ursion that worksfor general forms in
ase O = W ( k ) , provided that the
hara
teristi
of k is su(cid:30)
iently large.When appli
able, these theorems also give a fun
tion f ∈ Z ( T ) su
h that I = f ( L ) . If we let O = Z p this also gives a motivi
explanation to the phenomenon dis
ussed in the introdu
tion. For by applying C q to I , for di(cid:27)erent powers q of p , we get R W ( F q ) | Q ℓ i | dµ Haar = f ( q ) .Remark 5.1. As mentioned in the introdu
tion, it is not true in general that the motivi
integral ofthe absolute value of a polynomial is equal to f ( L ) , with f ∈ Z ( T ) . This
an be seen from the integral R X ∞ | x + 1 | dµ X : When O = Z p with p ≡ , then by Example 3.6 this integral is equal to − [Spec F p ] / ( L + 1) , and by applying the point
ounting homomorphism for di(cid:27)erent powers of p wesee that this
annot be equal to f ( L ) for f ∈ Z ( T ) .Let us mention one remaining question about these integrals: From the
omputations performedin Theorem 5.10 and 5.15 it is
lear that f ∈ Z ( T ) , the rational fun
tion with the property that R X n ∞ | Q ℓ i | dµ X n = f ( L ) , is independent of O , provided that we
hoose O among the rings { W ( k ) : p su(cid:30)
iently large } ∪ { k [[ t ]] } . In parti
ular, we have R Z np | Q ℓ i | dµ Haar = f ( p ) for p big enough. It wouldbe desirable to have a motivi
explanation also for this fa
t. This
an probably be a
hieved using thetheory of motivi
integration developed in [CL08℄. Alternatively, we indi
ate in the following remark howthe problem
ould be handled using geometri
motivi
integration.Remark 5.2. By Theorem 6.1 of [DL01℄, if O = Q [[ t ]] , if Y = { P = 0 } ⊂ X n where P is a polyno-mial, and if J ( T ) = P i ≥ [ Y i +1 ] T i ∈ M Q [[ T ]] , then the following holds: Firstly, J ( T ) is rational, withdenominator
onsisting of fa
tors of the form − L a T b . Moreover, if we
hoose representatives for the
oe(cid:30)
ients of J ( T ) , de(cid:28)ned over Z , and then
ount F p -points on them, then for p su(cid:30)
iently large weget the power series J p ( T ) = P i ≥ |{ x ∈ ( Z / ( p i +1 )) n : P ( x ) = 0 }| T i .The pro
ess of
hoosing representatives for elements of M Q , and then
ounting F p -points on them forall p , de(cid:28)nes a homomorphism C : M Q → Q p Q / ∼ , where ( a p ) and ( b p ) are equivalent if a p = b p foralmost all p . (The (cid:28)lter produ
t with respe
t to the Fré
het (cid:28)lter.) Sin
e R Z np | P | dµ Haar = 1 + p − n − (1 − p ) J p ( p − − n ) , one
ould de(cid:28)ne the motivi
integral of P by (cid:28)rst
omputing J as a rational fun
tion,and then de(cid:28)ne the integral to be I = 1 + L − n − (1 − L ) J ( L − − n ) ∈ M Q [(1 − L i ) − ] i ≥ . This integralthen has the property that C I = (cid:0)R Z np | P | dµ Haar (cid:1) p ∈ Q p Q / ∼ . For example, let P = X + 1 , and let m = [Spec Q [ X ] / ( P ( X ))] . Using Theorem 3.1 one sees that J ( T ) = m/ (1 − T ) , so the integral of P is − m/ ( L + 1) . Hen
e, for p su(cid:30)
iently large the value of R Z p | P | dµ Haar is if p ≡ and ( p − / ( p + 1) if p ≡ (a result that of
ourse is true for all p ). Probably the method used to0 KARL RÖKAEUSprove Theorem 5.10
an be used also to
ompute J ( T ) when P is a produ
t of linear forms, showing thatthe integral equals f ( L ) , hen
e that R Z np | Q ℓ i | dµ Haar = f ( p ) for p big enough.We now give an example showing that f is not independent of p for all p , only for p su(cid:30)
iently large.For (cid:28)x a prime l . Let p be any prime di(cid:27)erent from l , and let X = A Z p . Then, using Proposition 4.1,Proposition 4.4, and Proposition 3.5, we see that Z X ∞ | ( x + x )( x − ( l − x ) | dµ X = Z X ∞ | y y | dµ X = (cid:16)Z X ∞ | y | dµ X (cid:17) = L ( L +1) . If this formula were true for p = l , then it would follow that R Z p | ( x + x )( x − ( p − x ) | dx dx = p ( p +1) ,
ontradi
ting the following example.Example 5.3. Consider the linear mapping ( x , x ) ( x + x , x − ( p − x ) : Z p → Z p . It is easy to
he
k that it is inje
tive, that its image is S p − a =0 ( a + p Z p ) , and that its Ja
obian is
onstant of absolutevalue /p . Hen
e Z Z p | ( x + x )( x − ( p − x ) | dx dx = p p − X a =0 Z ( a + p Z p ) | y y | dy dy . Now if a = 0 , R ( a + p Z p ) | y y | dy dy = ( R a + p Z p dy ) = 1 /p , whereas R ( p Z p ) | y y | dy dy = ( R p Z p | y | dy ) =( R Z p | y | dy − R Z × p dy ) = ( p/ ( p + 1) − ( p − /p ) = 1 / ( p ( p + 1)) , hen
e Z Z p | ( x + x )( x − ( p − x ) | dx dx = p + p − p +1) . p -adi
omputation, but requires a di(cid:27)erent, more
ompli
ated,method, we begin by brie(cid:29)y dis
uss the p -adi
method, and why it does not translate to the motivi
ase.This is done in the following example:Example 5.4. Suppose we want to
ompute I = R A | α − α | dα , where we integrate over the set A ⊂ Z p of tuples su
h that α ≡ α mod p and α α mod p . One way to do this is to express the integralin terms of an integral of the same fun
tion, but now integrated over all of Z p . This is what is done forgeneral integrals, and in the motivi
setting in Theorem 5.9. We illustrate this method on the integral I : Choosing a set of representatives of the
osets of p Z p , x ˜ x : F p → Z p , we may write ea
h su
h α i uniquely as α i = ˜ x i + pβ i . Moreover, α i ≡ α j mod p if and only if x i = x j in F p . Hen
e I = X x ∈ F p : x = x = x Z α i =˜ x i + pβ i | α i − α j | dα. On ea
h of these integrals we perform the
hange of variables α i = ˜ x i + pβ i (the x i are (cid:28)xed). TheJa
obian of this has absolute value p − , hen
e ea
h of the integrals in the sum equal p − R Z p | pβ − pβ | dβ .(This step do not generalize to arbitrary linear forms, sin
e then the ˜ x i will not
an
el.) Sin
e there are p ( p − terms in the sum, we (cid:28)nd that I = p ( p − p − − R Z p | β − β | dβ .The main problem in translating this to the motivi
setting is that it is not possible to partition theintegration set in this manner. However, this partition is not visibly in the result of the
omputation, sothe result might still be possibly to translate into the motivi
setting. This is done in Theorem 5.9. Also,note that in that theorem we work in the most general
ase possibly,
ontrary to in this example.To prove Theorem 5.9 we (cid:28)rst needs two lemmas about the Witt ve
tors. See Appendix A for notationused in
onne
tion with the Witt ve
tors.Lemma 5.5. Let k be a perfe
t (cid:28)eld of
hara
teristi p , where p is a prime di(cid:27)erent from , andlet ℓ = aX + bY ∈ W ( k )[ X, Y ] be a linear form in two variables whose
oe(cid:30)
ients are multipli
ativeOMPUTING p -ADIC INTEGRALS USING MOTIVIC INTEGRATION 11representatives. For x = ( x , x , . . . ) ∈ W ( A ) , where A is a k -algebra, de(cid:28)ne ˜ x = ( x , x , . . . ) ∈ W ( A ) .Let y ∈ W ( A ) and de(cid:28)ne ˜ y similarly. Suppose that ℓ ( x, y ) ≡ . Then ℓ ( x, y ) = V ℓ (˜ x, ˜ y ) . If p = 2 the result holds provided ℓ = X − Y .Proof. When p = 2 we have for every ring A that − − r(1) = r( − ∈ W ( A ) . This follows in thestandard way by (cid:28)rst proving it when p is invertible in A . In this
ase W ∗ is an isomorphism, and sin
e W ∗ r( x ) = ( x, x p , x p , . . . ) we have W ∗ r(1) + W ∗ r( − 1) = 0 so the result follows. But if it this result istrue for a ring A , it holds for every sub- and quotient ring, hen
e for every ring.Let a = r a and b = r b . The
ondition ℓ ( x, y ) ≡ then means that a x + b y = 0 . Fromwhat was said above it follows that a r x + b r y = r( a x ) + r( b y ) = r( a x ) + r( − a x ) = 0 . We thenhave ℓ ( x, y ) = ℓ (r x + V ˜ x, r y + V ˜ y )=( a r x + b r y ) + ( a V ˜ x + b V ˜ y )= V(F a · ˜ x ) + V(F b · ˜ y )= V ℓ (˜ x, ˜ y ) . When p = 2 we may still prove the result when ℓ = X − Y , for if ℓ ( x, y ) ≡ then x = y , hen
e r( x ) = r( y ) . (cid:3) Throughout this se
tion, we
ontinue to use the notation of Subse
tion 4.3: We write k [ X • , . . . , X • N ] for the polynomial ring k [ X i , . . . , X iN ] ni =1 , and use Q N to denote the universal polynomials de(cid:28)ning Q ∈ O [ X , . . . , X n ] .Lemma 5.6. Fix a perfe
t (cid:28)eld k of
hara
teristi p . Let P = Q di =1 ℓ i ∈ W ( k )[ X , . . . , X n ] , where the ℓ i are linear forms, all of whose
oe(cid:30)
ients are multipli
ative representatives, at most two of whi
h arenon-zero. Moreover, if p = 2 , assume that all the forms are of the type X i − X j . Let x , . . . , x n ∈ W ( A ) ,where A is a k -algebra, be su
h that ℓ i ( x , . . . , x n ) ≡ for every i , and de(cid:28)ne ˜ x i as in Lemma5.5. Then P ( x , . . . , x n ) = F d − V d P ( ˜ x , . . . , ˜ x n ) . In parti
ular, working in W ( k [ X • n ] n ∈ N ) , we have P ξ = 0 for ξ < d , and P ξ + d = P ξ ( X • , . . . , X • ξ +1 ) p d − for ξ ∈ N .Proof. Using the pre
eding lemma and Corollary A.3 we get P ( x , . . . , x n ) = d Y i =1 ℓ i ( x , . . . , x n )= d Y i =1 V ℓ i ( ˜ x , . . . , ˜ x n )= F d − V d d Y i =1 ℓ i ( ˜ x , . . . , ˜ x n )= F d − V d P ( ˜ x , . . . , ˜ x n ) (cid:3) In the
ase of equal
hara
teristi
, this lemma also holds, but for any set of linear forms. Its proof isstraight forward, we just state the result:Lemma 5.7. Let k be a (cid:28)eld, and let O = k [[ t ]] . Let P = Q di =1 ℓ i ∈ O [ X , . . . , X n ] , where the ℓ i arelinear forms. Let x , . . . , x n ∈ A [[ t ]] , where A is a k -algebra, be su
h that ℓ i ( x , . . . , x n ) ≡ t forevery i , and de(cid:28)ne ˜ x i as in Lemma 5.5 (re
all that we write elements of power series rings as tuples).Then P ( x , . . . , x n ) = t d P ( ˜ x , . . . , ˜ x n ) . In parti
ular, working in k [ X • n ] n ∈ N [[ t ]] , we have P ξ = 0 for ξ < d , and P ξ + d = P ξ ( X • , . . . , X • ξ +1 ) for ξ ∈ N .2 KARL RÖKAEUSLemma 5.8. Let S be a (cid:28)nite set. Let Q = Q i ∈ S ℓ i , where the ℓ i ∈ O [ X , . . . , X n ] are linear formssatisfying the following
onditions: • If O = k [[ t ]] then the ℓ i are arbitrary • If O = W ( k ) , where k is a (cid:28)eld of prime
hara
teristi
di(cid:27)erent from , then ea
h ℓ i is linearform in at most two variables, and its
oe(cid:30)
ients are multipli
ative representatives. • If O = W ( k ) , where k is a (cid:28)eld of
hara
teristi , assume that all the forms are of the type X i − X j Let T be a subset of S . If ord ℓ i > for i ∈ T , then ord Q ≥ | T | . Moreover, for every integer ξ ≥ − wehave µ X n (ord Q > ξ + | T | , ord ℓ i > , i ∈ T, ord ℓ i = 0 , i ∈ S \ T ) = [ H T ] L n µ X n (ord Q > ξ ) , where H T is the subvariety of X n = A nk given by ( ℓ i ) = 0 for i ∈ T and ( ℓ i ) = 0 for i ∈ S \ T .Proof. For N su(cid:30)
iently large, π N +1 (ord Q > ξ + | T | , ord ℓ i > , i ∈ T, ord ℓ i = 0 , i / ∈ T ) is the spe
trumof the algebra k [ X • , . . . , X • N ][( ℓ i ) − ] i/ ∈ T (cid:0) Q , . . . , Q ξ + | T | , ( ℓ i ) (cid:1) i ∈ T . By Lemma 5.6 or 5.7 the
lass of this equals the
lass the spe
trum of k [ X • , . . . , X • N ][( ℓ i ) − ] i/ ∈ T (cid:0) Q ( X • ) , . . . , Q ξ ( X • , . . . , X • ξ +1 ) , ( ℓ i ) (cid:1) i ∈ T . Now, sin
e the ( ℓ i ) only involves the variables X • , we may write this as k [ X • ][( ℓ i ) − ] i/ ∈ T (( ℓ i ) ) i ∈ T ⊗ k k [ X • , . . . , X • N ]( Q ( X • ) , . . . , Q ξ ( X • , . . . , X • ξ +1 )) The spe
trum of the (cid:28)rst fa
tor is H T whereas, using the
hange of variables X • i X • i − , the spe
trumof the se
ond fa
tor is π N (ord Q > ξ ) . The result follows. (cid:3) Theorem 5.9. Let S be a (cid:28)nite set. For i ∈ S , let ℓ i ∈ O [ X , . . . , X n ] be a linear forms, satisfying the
onditions of Lemma 5.8. Let Q = Q i ∈ S ℓ i . For T ⊂ S , de(cid:28)ne Q T = Q i ∈ T ℓ i . Let H T be the subvarietyof X n = A nk given by ( ℓ i ) = 0 for i ∈ T and ( ℓ i ) = 0 for i ∈ S \ T . Then Z ord ℓ i =0 ,i ∈ T ord ℓ i =0 ,i/ ∈ T | Q | dµ X n = [ H T ] L −| T |− n Z X n ∞ | Q T | dµ X n . Proof. By the (cid:28)rst part of the lemma, { ord Q = ξ, ord ℓ i > , i ∈ T, ord ℓ i = 0 , i / ∈ T } = ∅ for ξ < | T | ,hen
e Z ord ℓ i =0 ,i ∈ T ord ℓ i =0 ,i/ ∈ T | Q | dµ X n = Z ord ℓ i =0 ,i ∈ T ord ℓ i =0 ,i/ ∈ T | Q T | dµ X n = X ξ ≥ µ X n (ord Q T = ξ + | T | , ord ℓ i > , i ∈ T, ord ℓ i = 0 , i / ∈ T ) L − ( ξ + | T | ) . Using the se
ond part of the lemma it follows that this equals X ξ ≥ [ H T ] L − n µ X n (ord Q T = ξ ) L − ( ξ + | T | ) = [ H T ] L −| T |− n Z X n ∞ | Q T | dµ X n . (cid:3) O be a
omplete dis
rete valuation ring. Let Q = Q i ∈ S ℓ i , where ℓ i ∈ O [ X , . . . , X n ] are linear forms, satisfying the
onditions of Lemma 5.8. Then R X n ∞ | Q | dµ X n is a rational fun
tion in L ,i.e., there is an f ∈ Z ( T ) su
h that the integral equals f ( L ) . Moreover, f may be
omputed expli
itly byre
ursion.OMPUTING p -ADIC INTEGRALS USING MOTIVIC INTEGRATION 13Proof. This re
ursion is immediate, using Theorem 5.9: Write, for T ⊂ S , Q T := Q i ∈ T ℓ i . We have Z X n ∞ | Q | dµ X = Z ℓ i =0 , i ∈ S | Q | dµ X + X T ( S Z ℓ i =0 , i ∈ Tℓ i =0 , i ∈ S \ T | Q T | dµ X . Theorem 5.9 now shows that(5.11) (1 − [ H S ] L −| S |− n ) Z X n ∞ | Q | dµ X = X T ( S [ H T ] L −| T |− n Z X n ∞ | Q T | dµ X . The right hand side is known indu
tively. Moreover, sin
e dim H S ≤ n it follows that [ H S ] L −| S |− n ∈ F ≤−| S | M k , hen
e it is not equal to . Also, by the (cid:28)rst part of the proof of Theorem 5.12, H S is ana(cid:30)ne spa
e, hen
e equal to L m for some m . Therefore, by Proposition 2.1, − [ H S ] L −| S |− n is invertible.Hen
e R X n ∞ | Q | dµ X is as a rational fun
tion in L and the
lasses of various hyper plane arrangements.Be
ause of the next theorem, Theorem 5.12, it follows that the integral is a rational fun
tion in L . (cid:3) The following theorem is of
ourse already well known. We provide a proof for
ompleteness.Theorem 5.12. Let V be a (cid:28)nite dimensional k -spa
e, I a (cid:28)nite set and for every i ∈ I , let ℓ i : V → k be a linear fun
tion. For S ⊂ I , let H S = { ℓ i = 0 , i ∈ S, ℓ i = 0 , i / ∈ S } ⊂ A V , i.e, H S ( R ) = { x ∈ R ⊗ k V : ℓ i ( x ) = 0 , i ∈ S, ℓ i ( x ) ∈ R × , i / ∈ S } for every k -algebra R . Then there is a polynomial p ∈ Z [ X ] su
h that [ H S ] = p ( L ) ∈ K ( Var k ) .Proof. First, let U = ∩ i ∈ S ker ℓ i . Then H S = { ℓ i | U = 0 , i / ∈ S } ⊂ A U . We may therefore assume that S = ∅ . We now prove the
laim by indu
tion on the number of hyperplanes: First, let ℓ : V → k benon-zero. We may then
hoose a basis of V , { e , . . . , e d } su
h that ℓ ( e ) = 1 and ℓ ( e i ) = 0 for i > .Hen
e { ℓ = 0 } = Spec k [ X , . . . , X d ][ X − ] , and
onsequently [ { ℓ = 0 } ] = L d − ( L − ∈ K ( Var k ) .Assume now that the
laim holds for any k -spa
e V and for any
olle
tion of less than n hyperplanes.Let H = { ℓ i = 0 } ni =1 ⊂ A V . De(cid:28)ne U = ker ℓ n . Then { ℓ i = 0 } n − i =1 ⊂ A V is the disjoint union of H and { ℓ i | U = 0 } n − i =1 ⊂ A U ⊂ A V , hen
e [ H ] = [ { ℓ i = 0 } n − i =1 ] − [ { ℓ i | U = 0 } n − i =1 ] ∈ K ( Var k ) , and we are done byindu
tion. (cid:3) We illustrate this with an example.Example 5.13. We apply this to the example that motivated this
omputations, the integral V n := R X n ∞ (cid:12)(cid:12)Q ≤ i