Examples on a Conjecture about Makar-Limanov Invariants of Affine Unique Factorization Domains
aa r X i v : . [ m a t h . A C ] O c t EXAMPLES ON A CONJECTURE ABOUT MAKAR-LIMANOVINVARIANTS OF AFFINE UNIQUE FACTORIZATION DOMAINS
ZIQI LIU
Abstract.
The author introduces a conjecture about Makar-Limanov invariantsof affine unique factorization domains over a field of characteristic zero. Then theauthor finds that the conjecture does not always hold when k is not algebraicallyclosed and gives some examples where the conjecture holds. Contents
1. Introduction 22. Preliminaries 32.1. Affine unique factorization domains 32.2. Nilpotent derivations and Makar-Limanov invariants 53. On the Makar-Limanov’s Conjecture 73.1. Several existed results 73.2. On Danielewski domains 83.3. On Koras-Russell threefolds 103.4. On Finston-Maubach domains 124. Comments 14References 15
This paper is an undergraduate research work. I thank professor Xiaosong Sun in Jilin Universityfor his help and suggestions when I worked on this paper. Introduction
Throughout the work, A [ n ] denotes the polynomial ring over a given ring A with n variables; k denotes a field of characteristic zero; A n denotes the affine space over k of dimension n . An affine k -variety is an irreducible algebraic set over k ; given anaffine variety V ⊆ A n , define I ( V ) := { f ∈ k [ n ] : f ( a ) = 0 , ∀ a ∈ V } ; given a subring I of A [ n ] , define Z ( I ) := { a ∈ A n : f ( a ) = 0 , ∀ f ∈ I } .In paper [14], Makar-Limanov provides several conjunctures about Makar-Limanovinvariant of an affine unique factorization domain. One of his conjectures is that Conjecture 1. If A is an affine UFD over C , then ML ( A ) = ML ( A [1] ).Here we are going to discuss a generalized version of this conjecture which originallyput forward on complex field C . Conjecture 2. If A is an affine UFD over k , then ML ( A ) = ML ( A [1] ).Based on following parts, one can see that ML ( A [1] ) ⊆ ML ( A ) and k ⊆ ML ( A ).Therefore, we only need to focus on ML ( A [1] ) ⊇ ML ( A ) when ML ( A ) = k .In Section 2, the author introduces the definition of an affine unique factorizationdomain and some methods to identify affine unique factorization domains. Also, theauthor reviews the definition of a locally nilpotent derivation over a k -algebra andthen introduce the Makar-Limanov invariant of a k -algebra.In Section 3, the author lists some important results about Conjecture 2. Basedon those results, we only need to consider affine UFD A over k with ML ( A ) = A and ML ( A ) = k . Then, the author proves that Conjecture 2 holds for Danielewskidomains when k is algebraically closed and is false when k = R . Also, the authorproves that Conjecture 2 is always true for Koras-Russell threefolds. Moreover, theauthor checks this conjecture on affine UFDs constructed by Finston and Maubach.In this paper, one of such affine UFDs is called a Finston-Maubach domain.In section 4, the author makes several comments on this paper as well as theresearch topic. Preliminaries
Affine unique factorization domains.Definition 2.1.1.
If a nonzero ring (with multiplicative identity) R has no nonzerozero divisors, then it is a domain . If the domain R is also commutative, then it isan integral domain . Example 2.1.2.
The ring of integers Z is an integral domain while the Hurwitzinteger H is a domain but not an integral domain. Definition 2.1.3.
Given a k -vector space A and a binary operation · : A × A → A ,denote the product of x and y in k by xy here and the addition in A by +, if(1) ( a + b ) · c = a · c + b · c ;(2) c · ( a + b ) = c · a + c · b ;(3) ( x a )( y b ) = ( xy )( a · b );holds for all a , b , c ∈ A and x, y ∈ k , then the pair ( A, · ) is an algebra over k , orsimply a k -algebra , also denoted by A . Definition 2.1.4.
If an integral domain A containing k is finitely generated as a k -algebra, then A is an affine domain over k . Theorem 2.1.5. (1) Given an affine k -variety V , its coordinate ring k [ V ] := A [ n ] / I ( V ) is an affinedomain over k .(2) If k is algebraically closed, and let A be an affine domain over k . Then thereexist an affine k -variety, such that A ∼ = k [ V ] .Proof. (1) We know that I := I ( V ) is prime since V is irreducible, therefore k [ V ] := A [ n ] /I is an integral domain. Note that k [ V ] is clearly a finitely generated k -algebra,it is an affine domain over k .(2) Choose generators a , . . . , a n of A , the surjective homomorphism F : k [ n ] → A sending f to f ( a , . . . , a n ) yields A ∼ = k [ n ] /I where I = Ker( F ). Since A is an integraldomain, I is a prime ideal. Let V := Z ( I ), we know I = I ( V ) from the Hilbert’sNullstellensatz, so A ∼ = k [ V ]. (cid:3) Remark 2.1.6.
Some very details of this proof can be found in book [9]. One cansee that if k is algebraically closed, then there is a bijection between the set of allaffine domains over k and the set of all affine varieties over k . Definition 2.1.7.
If each element x = 0 of an integral domain A is a product of aunit in A and prime elements p i in A , that is x = u · p · · · · · p m then A is a unique factorization domain , abbreviated to a UFD. If a UFD is alsoan affine domain over k , then it is a affine unique factorization domain over k , abbreviated to an affine UFD over k . In addition, an affine k -variety X is called factorial if its coordinate ring k [ X ] is an affine UFD. Remark 2.1.8.
Since proving an integral domain to be a UFD is hard in general, it isnatural to find special methods to tell an affine UFD over k from an affine k -domain.That topic is also interesting but not what we would mainly discuss in this note. Inthis case, I will list several important results without proofs. Proofs of those resultsand relevant definitions can be found in [16], [9] and [8]. ZIQI LIU
Lemma 2.1.9.
Given a noetherian domain A , following statements are equivalent:(1) A is a UFD;(2) every height one prime ideal of A is principal;(3) A is integrally closed and the Weil divisor class group Cl(Spec( A )) is trivial. Remark 2.1.10.
Since each affine k -domain is noetherian, so Lemma 2.1.9 givesus a way to test affine UFDs. For example, if V is a smooth k -variety, then fromProposition 6.15 in [9] we know that Pic( k [ V ]) ∼ = Cl( k [ V ]). In this case, k [ V ] is anaffine UFD implies that the Picard group Pic( k [ V ]) should be trivial. Example 2.1.11.
This two examples show some power of this lemma.(1) A = R [ x, y ] / ( x + y −
1) is not an affine UFD over R ;(2) A = C [ x , . . . , x n ] / ( x + · · · + x n −
1) is not an affine UFD over C for each n > Proof.
In fact, we could obtain that Pic( A ) = Z / Z and Pic( A ) = Z . (cid:3) Proposition 2.1.12.
Given a smooth affine curve C over k , then its coordinate ring k [ C ] is integrally closed. Example 2.1.13.
Note that x + y = 1 is a smooth affine curve in A , the affine k -domain A = C [ x, y ] / ( x + y −
1) is an affine UFD over C since Pic(A) is trivial. Proposition 2.1.14.
Given an affine k -domain A and a multiplicatively closed set S ⊆ A generated by a set of prime elements, then the localization S − A is a UFDimplies that A is an affine UFD over k . Example 2.1.15.
Given n > R [ x , . . . , x n ] / ( x + · · · + x n −
1) is an affine UFDover R . Proof.
Denote this affine R -domain by A , one can see that since A/ ( x n − ∼ = R [ x , . . . , x n ] / ( x + · · · + x n − + x n − , x n − ∼ = R [ x , . . . , x n ] / ( x + · · · + x n − , x n − ∼ = R [ x , . . . , x n − ] / ( x + · · · + x n − )is an integral domain, x n − A . Let t = ( x n − − , one has x + · · · + x n − ⇒ x + · · · + x n − + 1 t + 2 t = 0which implies that t ∈ R [ tx , . . . , tx n − ]. Therefore A x n − = A [ s ] / (1 − s ( z − A [ t ] = R [ tx , . . . , tx n − , t ]is a UFD and then A is an affine UFD by Proposition 2.1.14. (cid:3) Remark 2.1.16.
What’s more, Proposition 2.1.14 is actually a corollary of a famoustheorem of Masayoshi Nagata (see Theorem 6.3 in [16]). One can then prove thefollowing well-known theorem given by Masayoshi Nagata and Abraham A. Klein. Ifail to find direct sources and give an adapted one according to Theorem 8.2 in [16].
Theorem 2.1.17.
Given n ≥ , if f ( x , . . . , x n ) is a non-degenerate quadratic form,then affine k -domain A = k [ x , . . . , x n ] / ( f ) is an affine UFD. Nilpotent derivations and Makar-Limanov invariants.Definition 2.2.1.
Let A be an algebra over k , if a homomorphism of k -algebras D : A → A satisfies the Leibniz rule D ( ab ) = D ( a ) b + aD ( b ) for all a, b ∈ A then it is a k -derivation of A . Der k ( A ) denotes the set of all k -derivations of A . Proposition 2.2.2.
Given an algebra A over k and a k -derivation of A , then(1) D ( a ) = 0 for all a ∈ k ;(2) Ker( D ) := { a ∈ A : D ( a ) = 0 } is a subalgebra of A ;(3) Nil( D ) := { a ∈ A : ∃ n a ∈ N + , D n a ( a ) = 0 } is a subalgebra of A .Proof. (1) Let a = b = 1, one has D (1) = 1 D (1) + D (1)1 = 2 D (1)which implies that D (1) = 0, then D ( a ) = D ( a ·
1) = aD (1) = 0 for all a ∈ k .(2) From (1), we know that k ⊆ Ker( D ), so Ker( D ) can succeed the algebra structureof A . Then it is sufficient to prove that Ker( D ) is a vector space over k , which is nothard by the definition of vector spaces.(3) From (1), we know that k ⊆ Nil( D ), so Nil( D ) can succeed the algebra structureof A . Then it is sufficient to prove that Nil( D ) is a vector space over k , which is nothard by the definition of vector spaces. (cid:3) Remark 2.2.3.
In (2), one can also prove Ker( D ) to be a vector space over k by theisomorphism theorem for vector spaces (modules over a field). One can also see thatKer( D ) ⊆ Nil( D ) as a subalgebra. Definition 2.2.4.
Given an algebra A over k and a k -derivation of A , if Nil( D ) = A ,then D is locally nilpotent or is a locally nilpotent k -derivation . The set ofall locally nilpotent k -derivations of A is denoted by LND k ( A ) or just LND( A ) whenwith no hazards to rise confusion. Definition 2.2.5.
Given a locally nilpotent k -derivation over k -algebra A and a nonzero element a ∈ A , we definedeg D ( a ) := max n ∈ N { D n ( a ) = 0 } as the degree of a by D . Specifically, we define deg D (0) = −∞ . Example 2.2.6.
For each k -algebra A , the zero endomorphism θ A : A → A whichsends all elements to 0 is a locally nilpotent k -derivation, called zero derivation . Example 2.2.7.
All partial derivations are locally nilpotent k -derivation of k [ n ] . Definition 2.2.8.
Given an algebra A over k , we define set ML ( A ) := \ D ∈ LND k ( A ) Ker( D )as Makar-Limanov invariant of k -algebra A . Example 2.2.9.
The partial derivation ∂∂x : R [ x, y ] → R [ x, y ] is a locally derivationover the field of real numbers R and one can see deg ∂∂x ( x ) = 3. Example 2.2.10.
Given a positive integer n , ML ( k [ n ] ) = k . Proof.
First, we know that k ⊆ ML ( k [ m ] ) from Proposition 2.2.2. Then since allpartial derivations are locally nilpotent derivations over k , so k ⊇ ML ( k [ m ] ). (cid:3) ZIQI LIU
Example 2.2.11.
Given an affine UFD A := C [ x, y ] / ( x + y −
1) over the complexfield C , one has ML ( A ) = A . Proof.
It is sufficient to prove that the only element in LND C ( A ) is zero derivation.Given a nonzero D ∈ LND( A ), one has0 = D (1) = D ( x + y ) = 2 xD ( x ) + 2 yD ( y )from which we could get that D ( x ) = yp ( x, y ) and D ( y ) = − xp ( x, y )for a nonzero p ( x, y ) ∈ A . Then one hasdeg D ( x ) − D ( yp ( x, y )) = deg D ( y ) + deg D ( p ( x, y ))but also deg D ( y ) − D ( − xp ( x, y )) = deg D ( x ) + deg D ( p ( x, y ))which implies that deg D ( p ( x, y )) = −
1, impossible. (cid:3)
Example 2.2.12. ([12], Example 11) Given affine domain A := C [ x, y ] / ( x − y )over the complex field C (by the way, A is not a UFD), then ML ( A ) = A . Proof.
It is sufficient to prove that the only element in LND C ( A ) is zero derivation.If D ∈ LND( A ) is not zero, then one has D ( x ) = c for all c ∈ C − { } ; otherwise2 yD ( y ) = D ( y ) = D ( x ) = 3 x D ( x ) = 3 cx which leads to D ( y ) = cx y / ∈ A , a contradiction.Therefore, D ( x ) is nonconstant in A . Since D ( y ) = x y D ( x ) ∈ A , x | D ( x ) and thenone can set D ( x ) = yp ( x ) + xq ( x ) ∈ A . Let m := deg D ( x ) , n := deg D ( y ), one has n = 12 deg D ( y ) = 12 deg D ( x ) = 32 m so n > m . Notice that0 = D m +1 ( x ) = D m ( yp ( x ) + xq ( x )) = D m ( yp ( x )) + D m ( xq ( x ))and D m +1 ( xq ( x )) = d m +1 xq ( x )d x k D m +1 ( x ) = 0we could obtain D m +1 ( yp ( x )) = 0. In this case, one can see n + deg D ( p ( x )) = deg D ( y ) + deg D ( p ( x )) = deg D ( yp ( x )) ≤ m This is impossible, hence the only element in LND C ( A ) is zero derivation. (cid:3) Remark 2.2.13.
Personally, I do not really agree with Makar-Limanov’s way tocompute ML ( A ) in this example because he uses the undefined deg D ( t ) where t is aparameter for A ∼ = C [ t , t ]. On the Makar-Limanov’s Conjecture
Several existed results.
Here we list some results about Conjecture 2. Fromnow on, transcendence degree of an affine k -domain A is denoted by tr.deg k A . Theorem 3.1.1. ([12], Lemma 21)
Given an affine domain A over k . If ML ( A ) = A , then ML ( A [1] ) = A . Remark 3.1.2.
In this case, one can see ML ( A ) = ML ( A [1] ) = A for the affinedomain A in Example 2.2.11. Theorem 3.1.3. ([3], Lemma 2.3)
Given an affine domain A over k with tr.deg k A = 1 . Then(1) A ∼ = k [1] if and only if ML ( A ) = k ;(2) A = k [1] if and only if ML ( A ) = A . Remark 3.1.4.
Here we could get that ML ( A ) = ML ( A [1] ) holds for each affinedomain A over k with tr.deg k A = 1. Theorem 3.1.5. ([7], Theorem 9.12)
Given an affine unique factorization domain A over an algebraically closed field k with tr.deg k A = 2 . Then ML ( A ) = k is equivalent to A = k [2] . Remark 3.1.6.
Example 3.1.7 shows that tr.deg k A = 2 is necessary for this theorem. Example 3.1.7.
Consider the affine C -domain A = C [ x, y, z, w ] / ( xz − yw − A is an affine UFD over C and ML ( A ) = C . Proof. (1) First, since A/ ( w ) ∼ = C [ x, y, z ] / ( xy −
1) = ( C [ x, y ] / ( xy − z ] ∼ = ( C [ x ][ 1 x ])[ z ] ∼ = ( C [ x ]) x [ z ]is an integral domain (actually an affine UFD over C ), w is prime in A .Let x ′ = xw , y ′ = y, z ′ = z , then x ′ , y ′ , z ′ are algebraically independent and A w ∼ = A [ 1 w ] = C [ x ′ , y ′ , z ′ ][ 1 w ] = C [ x ′ , y ′ , z ′ ]is a UFD over C . Therefore, by Proposition 2.1.14, A is an affine UFD over C .(2) One can check that the derivations D , D , D and D on A generated by D ( x ) = 0 , D ( y ) = z, D ( z ) = 0 , D ( w ) = xD ( x ) = 0 , D ( y ) = w, D ( z ) = x, D ( w ) = 0 D ( x ) = z, D ( y ) = 0 , D ( z ) = 0 , D ( w ) = yD ( x ) = w, D ( y ) = 0 , D ( z ) = y, D ( w ) = 0are locally nilpotent. Here one can see Ker( D ) and C [ x, z ] are both algebraicallyclosed in A . What’s more, one has C [ x, z ] ⊆ Ker( D ) ⊂ A andtr.deg C Ker( D ) = tr.deg C C [ x, z ] = 2Hence Ker( D ) = C [ x, z ]. Similarly, we could obtain Ker( D ) = C [ x, w ] , Ker( D ) = C [ y, z ] and Ker( D ) = C [ y, w ]. In this case, C = \ i =1 Ker( D i ) ⊇ ML ( A ) ⊇ C and thus ML ( A ) = C . Moreover, since C ⊆ ML ( A [1] ) ⊆ ML ( A ) = C , it is clearthat ML ( A [1] ) = ML ( A ) = C . (cid:3) ZIQI LIU
On Danielewski domains.Definition 3.2.1.
Given integer n ≥ p ∈ S where S := [ d ≥ { p ∈ k [ x, y ] | d = deg y p = deg y p (0 , y ) } we define the subring D ( n,p ) of k [ x, x − , y ] by D ( n,p ) := k [ x, y, z ] / ( x n z − p ( x, y ))to be a Danielewski domain over k . Remark 3.2.2.
Here we should notice that we could assume deg x p ( x, y ) < n . If amonomial x s y t appears in p with s ≥ n , then one has x n ( z − x s − n y t ) = p ′ ( x, y ) for p ′ ( x, y ) = p ( x, y ) − x s y t . Therefore, one has D ( n,p ) ∼ = D ( n,p ′ ) . Repeating this, we couldeventually obtain a p ( x, y ) ∈ S with deg x p ( x, y ) < n and D ( n,p ) ∼ = D ( n,p ) .Danielewski domains over k are clearly affine k -domains and we want to knowwhen one such domain would become an affine UFD over k . Firstly, we introducethe following simple and well-known result. Proposition 3.2.3.
The Danielewski domain D (1 ,p ) = k [ x, y, z ] / ( xz − p ( y )) is an affine UFD if and only if p ∈ k [ y ] − k is irreducible.Proof. If nonconstant polynomial p ( y ) ∈ k [ y ] is irreducible, then the ideal ( p ( y )) isprime in both k [ y ] and k [ y, z ]. Therefore, the quotient ring D (1 ,p ) / ( x ) = k [ y, z ] / ( p ( y ))is an integral domain which implies that x is prime in D (1 ,p ) . Note that( D (1 ,p ) ) x ∼ = D (1 ,p ) [ 1 x ] = k [ x, y, x ]is a UFD over k , D (1 ,p ) is an affine UFD over k according to Proposition 2.1.14.On the other hand, a reducible p ∈ k [ y ] − k can be decomposed as a product ofnonconstant polynomials in k . In this case, D (1 ,p ) cannot be UFD or each irreducibleelement in D (1 ,p ) is prime, which implies a contradiction form xz = p ( y ). (cid:3) Example 3.2.4.
Consider affine k -domain A = k [ x, y, z ] / ( xz − y − A is an affine UFD when k = R and is not when k = C . Proposition 3.2.5.
If Danielewski domain D (1 ,p ) is a UFD, then ML ( D (1 ,p ) ) = k .Proof. One can check that the derivations D and D on D (1 ,p ) generated by D ( x ) = 0 , D ( y ) = x, D ( z ) = p ′ ( y ) D ( x ) = p ′ ( y ) , D ( y ) = z, D ( z ) = 0are locally nilpotent. Here one can see Ker( D ) and k [ x ] are both algebraically closedin D (1 ,p ) . What’s more, one has k [ x ] ⊆ Ker( D ) ⊂ D (1 ,p ) andtr.deg k Ker( D ) = tr.deg kk [ x ] = 1Hence Ker( D ) = k [ x ]. Also, we could obtain that Ker( D ) = k [ z ]. In this case, k = Ker( D ) ∩ Ker( D ) ⊇ ML ( D (1 ,p ) ) ⊇ k and thus ML ( D (1 ,p ) ) = k , which implies that ML (( D (1 ,p ) ) [1] ) = k . (cid:3) This proposition implies that Makar-Limanov’s conjecture holds for D (1 ,p ) and wewant to look into other cases. Furthermore, Alhajjar proved the following theorem inhis Ph.D. thesis [1] based on some works of Makar-Limanov and Freudenburg listedthis result in his book [7] (Theorem 9.2). Theorem 3.2.6. ([1], Proposition 6.16) If n, d ≥ and deg y p ( x, y ) = d for p ∈ S ,then ML ( D ( n,p ) ) = k [ x ] . Note that if D ( n,p ) is an affine UFD with n, d ≥ y p ( x, y ) = d , then x should be prime and the quotient ring D ( n,p ) / ( x ) = k [ y, z ] / ( p y (0 , y ))is an integral domain. Therefore, p (0 , y ) ∈ k [ y ] is irreducible. However, it is impossi-ble when k is algebraically closed because deg y p (0 , y ) = deg y p ( x, y ) = d >
1. So herewe only need to consider potential counterexamples to Makar-Limanov’s conjecturewith a non-algebraically closed field.
Example 3.2.7.
Consider the Danielewski domain A = R [ x, y, z ] / ( x z − ( y + 1)),we could prove the following statements.(1) A is an affine UFD;(2) ML ( A [1] ) = R . Proof. (1) First, since A/ ( x ) = R [ y, z ] / ( y + 1) is an integral domain, x is prime inaffine R -domain A . Then from z = y +1 x , one can see A x ∼ = A [ 1 x ] = R [ x, y, x ]is a UFD. Therefore, A is an affine UFD by Proposition 2.1.14.(2) Consider the Danielewski domain B = R [ x, y, z ] / ( xz − ( y + 1)), we know that itis an affine UFD by Example 3.2.4 and then ML ( B [1] ) = ML ( B ) = R from Proposition 3.2.5. Moreover, according to Theorem 10.1 in [7], one has ML ( A [1] ) = ML ( B [1] ) = R so we are done. (cid:3) Remark 3.2.8.
This example implies that when k is not algebraically closed, it ispossible that ML ( A [1] ) = ML ( A ) for certain affine UFD over k . Therefore, we willonly consider the conjecture on algebraically closed fields in the following part.Now we are moving to Danielewski domains D ( n,p ) with deg y p ( x, y ) = 1 and n ≥ p ( x, y ) = a ( x ) y + b ( x ) where a, b ∈ k [ x ] and a (0) = 0.Note that gcd ( x n , a ( x )) = 1, then one can see k [ x, y, z ] / ( x n z − a ( x ) y − b ( x )) ∼ = k [ x, y, z ] / ( z ) ∼ = k [ x, y ]from k [ x, y, z ] ∼ = k [ x, y, x n z − a ( x ) y − b ( x )]. Therefore, the Makar-Limanov invariantof one such Danielewski domain is k .Based on the previous discussions in this part, we clearly have the following resultas a conclusion. Theorem 3.2.9. If k is algebraically closed, then ML ( D ( n,p ) ) = ML (( D ( n,p ) ) [1] ) foreach Danielewski domain D ( n,p ) over k . On Koras-Russell threefolds.Definition 3.3.1. A Koras-Russell threefold of the first kind is defined by R = k [ x, y, z, w ] / ( x + x d y + z u + w v )where d, u, v ≥ u, v ) = 1.In this part, we are going to investigate the conjecture on Koras-Russell threefoldswhich are clearly affine domains. First of all, we want to know when a Koras-Russellthreefold R is an affine UFD. Note that x + x d y + z u + w v = 0 ⇒ y = − x d − − z u x d − w v x d ∈ k [ x, z, w ][ 1 x ]then one can see R x ∼ = R [ 1 x ] = k [ x, z, w ][ 1 x ]Hence R x is a localization of a polynomial ring which is a UFD. It is not hard tocheck that the element x is prime in R . Therefore, we are able to see the followingresult according to Proposition 2.1.14. Proposition 3.3.2.
Each Koras-Russell threefold R is an affine UFD. Then we wonder if ML ( R ) = ML ( R [1] ) also always holds. In order to figure outthis problem, we need to introduce a fundamental result at first.As a concise version of Makar-Limanov’s original proof in [13], Daigle, Freudenburgand Moser-Jauslin proved a feature of Koras-Russell threefolds in [5]. Based on theirresult, Freudenburg computed Makar-Limanov invariants of those threefolds in newedition of his book [7]. Theorem 3.3.3. ([7], Theorem 9.9)
Let R be a Koras-Russell threefold R = k [ x, y, z, w ] / ( x + x d y + z u + w v ) where d, u, v ≥ and gcd( u, v ) = 1 . Then ML ( R ) = k [ x ] . In their arguments and computations, some propositions and lemmas are impor-tant. We are going to introduce some less-known definitions relevant to those propo-sitions and lemmas at first and then give the propositions and lemmas.
Definition 3.3.4.
Let G be an abelian group, and let B be a ring. A G -grading of B is a family { B a } a ∈ G of subgroups of ( B, +) such that B = ⊕ a ∈ G B a and B a B b ⊆ B a + b for all a, b ∈ G and a ring (or domain/integral domain/affine k -domain) B with a G -grading is calleda G -grading ring (or domain/integral domain/affine k -domain). Definition 3.3.5.
Given B = ⊕ a ∈ G B a a G -grading ring, a nonzero f ∈ B is called G -homogeneous if f ∈ B a for a unique a ∈ G . Here we say that f is of degree a and write deg G f = a . Definition 3.3.6.
Given B = ⊕ a ∈ G B a a G -grading ring and a nonzero f ∈ B . f denotes the highest-degree homogeneous summand of f . In case f = 0, wedefine 0 = 0. Here one can see x = x if x is G -homogeneous. Definition 3.3.7.
Given B = ⊕ a ∈ G B a a G -grading k -domain and a k -algebra R ⊆ B , we define the associated G -graded domain R to be the k -subalgebra of B generated by the set { r | r ∈ R, r = 0 } . Definition 3.3.8.
Given B = ⊕ a ∈ G B a a G -grading k -domain and D ∈ Der k B , ifdeg G D := max { deg G ( Df ) − deg G ( f ) : f = 0 , f ∈ B } exists, then we define D : B → B as Dx = ( Dx, deg G ( Df ) − deg G ( f ) = deg G D , deg G ( Df ) − deg G ( f ) < deg G D Moreover, one can see deg G ( D ) = deg G D and Ker( D ) ⊆ Ker( D ). Lemma 3.3.9. ([5], Lemma 3.7)
Let ( G, ≤ , + , be a totally ordered abelian groupand B = ⊕ a ∈ G B a a G -graded integral domain. Let A = ⊕ a ≤ B a , x ∈ B , and R = A [ x ] . Then R = A [ x ] . Proposition 3.3.10. ([5], Theorem 3.8)
Let ( G, ≤ , + , be a totally ordered abeliangroup, B a G -graded k -affine domain, and R ⊂ B a k -subalgebra such that B is alocalization of R . Let deg G : R → G ∪ {−∞} be the restriction of the degree functionon B determined by the grading. Then deg( D ) is defined for every D ∈ Der k ( R ) . Proposition 3.3.11. ([5], Corollary 6.3)
Let ( G, ≤ , + , be a totally ordered abeliangroup, B = ⊕ a ∈ G B a a G -graded integral domain containing Z , where B is finitelygenerated as a A -algebra. Then one can write B = A [ x , . . . , x n ] where x i = 0 ishomogeneous of degree d i = 0 for each i . Let H i = h d , . . . , ˆ d i , . . . , d n i for each i ∈ [ n ] . Then for every G -homogeneous D ∈ LND k ( B ) the following conditions hold.(1) For each i ∈ [ n ] such that H i = G ( B ) , D x i = 0 .(2) For every choice of distinct i, j ∈ [ n ] such that H i = G ( B ) and H j = G ( B ) , onehas Dx i or Dx j = 0 . In line with their works, we are able to make the following computation.
Theorem 3.3.12.
Let R be a Koras-Russell threefold R = k [ x, y, z, w ] / ( x + x d y + z u + w v ) where d, u, v ≥ and gcd( u, v ) = 1 . Then ML ( R [ t ]) = k [ x ] .Proof. Let group G = Z and define a total order (cid:22) on G by lexicographical ordering.Consider a G -grading on B = k [ x, x − , z, w, t ] with x, z, w, t homogeneous anddeg G ( x, z, w, t ) = (( − , , , (0 , − v, , (0 , − u, , (0 , , − u, v ) = 1, one has { f ∈ B | deg( f ) (cid:22) (0 , , } = k [ x, z, w, t ] ⊂ R [ t ]We set A = { f ∈ B | deg( f ) (cid:22) (0 , , } = k [ x, z, w, t ]. The degree function deg G on B restricts to affine k -domain R [ t ], where deg G y = ( d, − uv, R [ t ] = A [ y ] = k [ x, z, w, t, y ]. Since y = − x − d ( x + z u + w v ) in B , onecan see that y = − x − d ( z u + w v ) and then x d y + z u + w v = 0.Given a nonzero D ∈ LND k ( R [ t ]). By Proposition 3.3.10, the induced G -homogeneousderivation D of R [ t ] is nonzero and locally nilpotent. Since h deg G y, deg G z, deg G w, deg G t i = d Z × Z × Z h deg G x, deg G y, deg G w, deg G t i = Z × u Z × Z h deg G x, deg G y, deg G z, deg G t i = Z × v Z × Z are proper subgroups of G ( R [ t ]) = Z , we know that at least two of Dx, Dz, Dw mustbe zero according to Proposition 3.3.11.When Dz = Dw = 0, one has D ( x d y ) = 0 and then Dx = Dy = 0. In this case, theonly possible is Dt = 0.When Dx = Dz = 0 or Dx = Dz = 0, one can see that either Dy = 0 or Dx = Dy = Dz = Dw = 0 , Dt = 0Therefore, either Ker D ⊂ A or Ker D = k [ x, y, z, w ].Now suppose that Dx = 0. Choose f, g ∈ Ker D which are algebraically independent.Let f , g ∈ k [ x, y, z, w, t ] and f , g ∈ k [ y, z, w, t ] be such that f = xf + f and g = xg + g Here f and g should be algebraically independent in R . Otherwise, there exists P ∈ k [2] with P ( f , g ) = 0. But then P ( f, g ) ∈ xR [ t ], which implies that Dx = 0,a contradiction. In addition, since deg G ( xf ) ≺ deg G f , one has deg G f = deg G f .Similarity, deg G g = deg G g . Now one has f , g ∈ Ker D where f = f = f ( y, z, w, t ) and g = g = g ( y, z, w, t )If Ker D ⊂ A , then Dy = 0 and f , g ∈ k [ z, w, t ]. In this case, k [ z, w, t ] is thealgebraic closure of k [ f , g ] and thus k [ z, w, t ] ⊂ Ker D . However, we could obtain0 = D ( x d y ) and then D = 0, a contradiction.If Ker D = k [ x, y, z, w ], then D restricts to be a zero derivation in R . So D restrictsto a zero derivation in R , a contradiction.Therefore, we always get a contradiction with Dx = 0, which implies that Dx = 0.So one has k [ x ] ⊆ ML ( R [ t ]). From Theorem 3.3.3, we know that ML ( R [ t ]) ⊆ ML ( R ) = k [ x ]and thus ML ( R [ t ]) = k [ x ]. (cid:3) Here one has proved that the Conjecture 2 holds for all Koras-Russell threefolds.However, methods applied in this proof rely on specific structures of Koras-Russellthreefolds and are hard to be generalized.3.4.
On Finston-Maubach domains.
In paper [6], Finston and Maubach con-struct a series of affine UFDs (called Finston-Maubach domains in this paper) withnon-trivial Makar-Limanov invariant based on Brieskorn-Catalan-Fermat rings. Weare going to check Conjecture 2 on Finston-Maubach domains in the following part.
Definition 3.4.1.
Given n ≥
3, we define F ∈ C [ x , . . . , x n , y , . . . , y n ] as F := x d + · · · + x d n n + ( x y − x y ) e + · · · + ( x n y − x y n ) e n where 1 d + · · · + 1 d n + 1 e + · · · + 1 e n ≤ n − R given by R := C [ x , . . . , x n , y , . . . , y n ] / ( F )is called a Finston-Maubach domain of order n . Proposition 3.4.2.
Each Finston-Maubach domain R is an affine UFD. Proof.
It is clear that R is an affine domain. Moreover, Lemma 3.5 in paper [6] tellsus that R is a UFD. One can also try to prove it independently by Proposition 2.1.14and the factoriality of Brieskorn-Catalan-Fermat rings with n ≥ (cid:3) Then we are going to compute the Makar-Limanov invariant of Finston-Maubachdomains based on Finston and Maubach’s results. For that, one has to introduce alemma at first.
Lemma 3.4.3. ([6], Lemma 2.7)
Let A be an affine Q -domain. Consider a subset F = { f , f , . . . , f n } of A and positive integers d , . . . , d n satisfying: • f := f d + f d + · · · + f d n n is a prime element of A . • Non nontrivial sub-sum of F d , F d , . . . , F d n n lies in ( f ) .Additionally, assume that d + 1 d + · · · + 1 d n ≤ n − Set R := A/ ( f ) and let D ∈ LND( R ) , one has D ( f i ) = 0 for all ≤ i ≤ n . Proposition 3.4.4. ML ( R ) = C [ n ] for a Finston-Maubach domain R of order n .Proof. Firstly, it is not hard to check that D ∈ LND( R ) where D is generated by D ( x i ) = 0 and D ( y i ) = x i . Given D ∈ LND( R ), by Lemma 3.4.3, one has D ( x i ) = 0and D ( l i ) = 0 where l i = x i y − x y i . Therefore, one has x D ( y i ) = y D ( x i ) , ∀ i ∈ { , , . . . , n } Since R is a UFD, D ( y i ) = αx i for some α ∈ R . Then one can see D = αD . Noticethat D is nonzero if and only if α is nonzero, one can see Ker( D ) = C [ x , . . . , x n ] foreach nonzero D ∈ LND( R ). Hence one has ML ( R ) = C [ x , . . . , x n ] = C [ n ] . (cid:3) Now we are going to compute ML ( R [1] ) for a Finston-Maubach domain R of order n and we also need to introduce several lemmas in [2] and [6]. Lemma 3.4.5. ([2], Theorem 3.1)
Let f , f , . . . , f n ∈ K [ s ] , where K is an alge-braically closed field containing Q , Assume f d + f d + · · · + f d n n = 0 . Additionally,assume that for every ≤ i < i < · · · < i s ≤ n , f d i i + f d i i + · · · + f d is i s = 0 = ⇒ gcd { f i , f i , · · · , f i s } = 1 Then n X i =1 d i ≤ n − implies that all f i are constant. Lemma 3.4.6. ([6], Lemma 2.2)
Let D be a locally nilpotent derivation on a domain A containing Q . Then A embeds into K [ s ] , where K is some algebraically closed fieldof characteristic zero, in such way D = ∂ s on K [ s ] . Proposition 3.4.7. ML ( R [1] ) = C [ n ] for a Finston-Maubach domain R of order n .Proof. Let Finston-Maubach domain R = C [ x , . . . , x n , y , . . . , y n ] / ( F ) where F := x d + · · · + x d n n + ( x y − x y ) e + · · · + ( x n y − x y n ) e n Suppose D ∈ LND( R [ t ]), where D = 0. By Lemma 3.4.6 with K an algebraic closureof the quotient field of ( R [ t ]) D , we realize D as partial derivation ∂ s on K [ s ] ⊇ R [ t ]. Notice that F ( s ) = 0 and there cannot be a subsum of x d + · · · + x d n n + l e + · · · + l e n n tobe zero, where l i = x i y − x y i . Then by Lemma 3.4.5, one has x i , l i to be constant in K [ s ]. In this case, one has D ( x i ) = ∂ s x i = 0 and D ( y i ) = ∂ s l i = 0 in R [ t ]. Therefore,one has ML ( R [ t ]) ⊇ C [ x , . . . , x n ]. In other hands, we already know that ML ( R [ t ]) ⊆ ML ( R ) = C [ x , . . . , x n ]So ML ( R [1] ) = ML ( R ) = C [ n ] . (cid:3) Here we know that Conjecture 2 is also valid as to Finston-Maubach domains. Theproof in Proposition 3.4.7 is nothing new but to imitate the proof of Finston andMaubach for the almost rigidity of Finston-Maubach domains.4.
Comments
This work was originally started as an extension of my Bachelor thesis. At thattime, I planned to study Conjecture 2 by observing examples. However, I found thatmost of existing methods of computing Makar-Limanov invariants rely on specificstructure of an affine domain. So it may be hard to obtain a promising way to solvethe conjecture by them. Additionally, my first semester for a master degree will beginsoon and I will hardly have time for this work. In this case, I decide to stop it here.In my opinion, the core of this conjecture is to build a connection between theUFD property and Makar-Limanov invariant. It is difficult because the UFD itself isa really profound area, which is too fundamental to be given a description by locallynilpotent derivations. If we want to use current ways to compute Makar-Limanovinvariants, we need to focus on the structure of ( f ) in an affine UFD k [ n ] / ( f ). Orone can try to interpret Makar-Limanov invariants from the perspective of geometry,which may bring us some new ideas.These three examples in this paper are computed by three different methods. Thefirst one is simple and direct. The second one uses a famous technique in this area,called homo-generalization. Readers can find more information about this techniquein [7], [11] and [12]. The third example has some connection with the famous ABC-theorem and one can find more details in [2] and [7]. References [1] Bachar Alhajjar.
Locally Nilpotent Derivations of Integral Domains . Ph.D. thesis, Univer-site de Bourgogne, 2017.[2] Michiel de Bondt. Another generalization of Mason’s ABC-theorem. Preprint,arXiv:0707.0434, 2009.[3] Anthony Crachiola and Leonid Makar-Limanov. On the rigidity of small domains.
Jounralof Algebra , 284(1):1-12, 2005.[4] Anthony Crachiola and Leonid Makar-Limanov. An algebraic proof of a cancellation the-orem for surfaces.
Jounral of Algebra , 320(8):3113-3119, 2008.[5] Daniel Daigle, Gene Freudenburg and Lucy Moser-Jauslin. Locally nilpotent derivations ofrings graded by an abelian group.
Advanced Studies in Pure Mathematics , 75:29-48, 2017.[6] David Finston and Stefan Maubach. Constructing (almost) rigid rings and a UFD havinginfinitely generated Derksen and Makar-Limanov invariant.
Canadian Mathematics Bul-letin , 53(1):77-86, 2010.[7] Gene Freudenburg.
Algebraic Theory of Locally Nilpotent Derivations (Second Edition) .Springer, New York, The United States, 2017.[8] Robin Hartshorne.
Algebraic Geometry . Springer, New York, The United States, 1977.[9] Gregor Kemper.
A Course in Commutative Algebra . Springer, Berlin, Germany, 2009.[10] Shulim Kaliman and Leonid Makar-Limanov. On the Russell-Koras contractible threefolds.
Jounral of Algebraic Geometry , 6(2):247-268, 1997.[11] Shulim Kaliman and Leonid Makar-Limanov. AK-invariant of affine domains. In
Affinealgebraic geometry , pages 231-255. Osaka University Press, Osaka, Japan, 2007.[12] Leonid Makar-Limanov.
Locally nilpotent derivations, a new ring invariant and applica-tions . Lecture notes, ResearchGate:265356937, 1998.[13] Leonid Makar-Limanov. On the hypersurface x + x y + z + t in C or a C -like threefoldwhich is not C . Israel Journal of Mathematics , 96:419-429, 1996.[14] Leonid Makar-Limanov. Some conjectures, examples and counterexamples.
AnnalesPolonici Mathematici , 76(1-2):139-145, 2001.[15] Leonid Makar-Limanov. On the group of automorphisms of a surface x n y = p ( z ). IsraelJournal of Mathematics , 121:113-123, 2001.[16] Pierre Samuel.
On unique factorization domains . Lecture notes, AlgebraNote 30, 1998.
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