F -thresholds and test ideals of Thom-Sebastiani type polynomials
Manuel González Villa, Delio Jaramillo-Velez, Luis Núñez-Betancourt
aa r X i v : . [ m a t h . A C ] M a y . F -THRESHOLDS AND TEST IDEALS OF THOM-SEBASTIANI TYPEPOLYNOMIALS MANUEL GONZ ´ALEZ VILLA , DELIO JARAMILLO-VELEZ , AND LUIS N ´U ˜NEZ-BETANCOURT Abstract.
We provide a formula for F -thresholds of a Thom-Sebastiani type polynomial overa perfect field of prime characteristic. This result extends the formula for the F -pure thresholdof a diagonal hypersurface. We also compute the first test ideal of Thom-Sebastiani type poly-nomials. Finally, we apply our result to find hypersurfaces where the log canonical thresholdsequals the F -pure thresholds for infinitely many prime numbers. Introduction
The F -threshold of a polynomial f , with respect to an ideal I , c I ( f ), is a numerical invariantgiven by the asymptotic Frobenius order of f in I [MTW05, HMTW08, DSNnBP18]. In partic-ular, if m is a maximal ideal, c m ( f ) equals to the F -pure threshold at m . This invariant measuresthe severity of the singularities of f at the point corresponding to m [TW04]. This F -threshold isthe analogue in prime characteristic of the log-canonical threshold [HY03, TW04, MTW05]. Thisnumber is usually very difficult to compute. However, there are formulas for diagonal [Her15],binomial [Her14], Calabi-Yau [BS15], Elliptic Curves [BS15, Pag18], and quasi-homogeneous onedimensional hypersurfaces [HNnBWZ16].An important aspect of a numerical invariant is its computability in concrete examples. Find-ing formulas for Thom-Sebstiani type polynomials is a classic strategy to enlarge the class ofconcrete hypersurfaces for which a invariant can be computed. This type of polynomial can bewritten as the sum of two polynomials in a different set of variables. Examples of this typeof polynomials include, for instance, diagonal hypersurfaces, certain binomials, and suspensionsby n ≥ F -thresholds and the first test ideal of a Thom-Sebastiani typepolynomial. Our methods are inspired in the work of Hern´andez [Her15] on diagonal hypersur-faces.In our first main result, we provide a formula for the F -threshold of a Thom-Sebastiani typepolynomial with respect to the sum of two ideals. Theorem A (see Theorem 3.4) . Let K be a perfect field of prime characteristic p . Let R = K [ x , . . . , x n ] and R = K [ y , . . . , y m ] with maximal homogeneous ideals m = ( x , . . . , x n ) and Mathematics Subject Classification.
Primary 13A35; Secondary 14B05.
Key words and phrases. F -thresholds, test ideals, Thom-Sebastiani type polynomials, log canonical threshold. The first author was partially supported by Spanish national grant MTM2016-76868-C2-1-P. The second author was partially supported by CONACyT Fellowship 862006. The third author was partially supported by CONACyT Grant 284598 and C´atedras Marcos Moshinsky. m = ( y , . . . , y m ) respectively. Let g ∈ √ I ⊆ m and g ∈ √ I ⊆ m , where I and I areideals. Let f = g + g ∈ R ⊗ K R , a = c I ( g ), and a = c I ( g ). If a + a ≤
1, then c I + I ( f ) = a + a , if L = ∞ , h a i L + h a i L + p L , if L < ∞ , where L = sup n N ∈ N | a ( e )1 + a ( e )2 ≤ p − ≤ e ≤ N o . Unlike previous work on computation of F -thresholds, we do not assume any shape on the F -pure thresholds of either g or g . Furthermore, we are able to compute F -thresholds withrespect to ideals other than the maximal ideal. This is useful to compute lower bound of thenumber of F -jumping numbers for a Thom-Sebastiani type polynomial (see Remark 3.6).An important conjecture regarding the F -pure threshold states that lct ( f ) = c m ( f p ) forinfinitely many prime numbers p , where f ∈ Q [ x , . . . , x n ] and f p ∈ F p [ x , . . . , x n ] is its modelmodule p (see [MTW05, Conjecture 3.6] and references therein). Using Theorem A , we are ableto show several families of polynomials where this conjecture holds (see Section 4).We are able to compute the first test ideal of a Thom-Sebastiani type polynomial at the F -pure threshold. We point out that, unlike previous work, we do not assume that the test idealsof g or g are monomial. In fact, we do not assume any shape of the ideals. Furthermore, wedo not make any assumption on the characteristic. Theorem B (see Theorem 5.5) . Let K be a perfect field of prime characteristic p . Let R = K [ x , . . . , x n ] and R = K [ y , . . . , y m ] with maximal homogeneous ideals m = ( x , . . . , x n )and m = ( y , . . . , y m ) respectively. Let f = g + g ∈ R = R ⊗ K R = K [ x , . . . , x n , y , . . . , y m ],where g ∈ m and g ∈ m . Let a = c m ( g ) = r s , a = c m ( g ) = r s , and m =( x , . . . , x n , y , . . . , y m ), where r i , s i ∈ N . Suppose that a + a ≤
1, and set L = sup n N ∈ N | a ( e )1 + a ( e )2 ≤ p − ≤ e ≤ N o , and d = sup n e ≤ L | a ( e )1 + a ( e )2 ≤ p − o . Then, τ (cid:16) f c m ( f ) (cid:17) = ( f ) if c m ( f ) = 1 ,τ ( g a ) + τ ( g a ) if c m ( f ) p − e · N , (cid:16) g ⌈ p d a ⌉ (cid:17) [1 /p d ] + (cid:16) g ⌈ p d a ⌉ (cid:17) [1 /p d ] if c m ( f ) ∈ Z [ p ] & c m ( f ) = 1 . We point out that Theorem B includes several cases that were not computed before, even fordiagonal hypersurfaces (see Example 5.6).
Convention.
In this manuscript, p denotes a prime number and K denotes a perfect field ofcharacteristic p . 2. Background
Expasions in base p . In this section we recall the notion of the non-terminating base p expansion of a number in (0 , Definition 2.1.
Let α ∈ (0 , p be a prime number. The non-terminating base p expansionof α is the expression α = X e ≥ α ( e ) p e , with 0 ≤ α ( e ) ≤ p −
1, such that for all n >
0, there exists -THRESHOLDS AND TEST IDEALS OF THOM-SEBASTIANI TYPE POLYNOMIALS 3 e ≥ n with α ( e ) = 0. The number α ( e ) is unique and it is called the e th digit of the non-terminating base p expansion of α . Example 2.2.
Let α = ab be a rational number in (0 , p ≡ b , then p = bw + 1for some w ≥
1. The non-terminating base p expansion of α is periodic, and it is given by α = X e ≥ awp e .The elementary notion of adding without carrying for integers (base p ) extends to non-terminating base p expansions. Definition 2.3.
Let α ∈ (0 , e ≥
1, the e th truncation of the non-terminating base p expansion of α is defined by h α i e = α (1) p + · · · + α ( e ) p e . We use the conventions h α i ∞ = α and h i e = 0.(2) Let p < n , and α , α ∈ (0 , α and α add without carrying in base p if α ( e )1 + α ( e )2 ≤ p − e ≥ . Remark 2.4.
The non-terminating base p expansions of α and α n add without carrying if andonly if the base p expansions of the integers p e h α i e and p e h α n i e add without carrying for all e > Lemma 2.5 ([Her11, Lemma 4.6]) . Let α ∈ (0 , . Then the following statements hold. (1) If α p e · N , then ⌊ p e α ⌋ = p e h α i e . (2) ⌈ p e α ⌉ = p e h α i e + 1 . (3) Let α , α ∈ (0 , . If α and α add without carrying in base p and α := α + α ≤ ,then α ( e ) = α ( e )1 + α ( e )2 . To end this section we include some useful results on divisibility for integers.
Lemma 2.6 ([Luc78, L.E02]) . Let k , k ∈ N and set N = k + k . Then (cid:18) Nk i (cid:19) = N ! k ! k ! p if and only if k , k add without carrying in base p . Theorem 2.7 (Dirichlet) . Given n and m non-zero natural numbers with gcd( n, m ) = 1 , thereexists infinitely many prime numbers p such that p ≡ n mod m . Theorem 2.8 (Dirichlet) . For any collection α , . . . , α n of rational numbers, there exists infin-itely many prime numbers p such that ( p − · α i ∈ N for ≤ i ≤ n . Test ideals and F-thresholds.Notation 2.9.
Throughout this subsection, R denotes the polynomial ring K [ x , . . . , x n ] over K , and m denotes its homogeneous maximal ideal ( x , . . . , x n ) . The Frobenius map F : R → R is the p th power morphism given by r r p . For everyideal I ⊆ R , the e -th Frobenius power of I , denoted by I [ p e ] , is the ideal generated by the set (cid:8) g p e | g ∈ I (cid:9) . Since R is a regular ring, then the Frobenius map F : R → R is flat. Therefore,we have that ( I [ p e ] : R J [ p e ] ) = ( F e ( I ) : R F e ( J )) = F e (( I : R J )) = ( I : R J ) [ p e ] . M. GONZ ´ALEZ VILLA, D. JARAMILLO-VELEZ, AND L. N ´U ˜NEZ-BETANCOURT
Remark 2.10.
The ring R is finitely generated and free over R p e = K [ x p e , . . . , x p e n ], with basisthe set of monomials (cid:8) µ | µ m [ p e ] (cid:9) . Definition 2.11.
For an ideal I ⊆ R and a positive integer e , let I [1 /p e ] denote the uniquesmallest ideal J of R with respect to the inclusion such that I ⊆ J [ p e ] . Proposition 2.12 ([BMS09, Proposition 2.5]) . Let e , . . . , e s be a basis of R over R p e . Let f , . . . , f r be generators of the ideal I of R . For each i ∈ { , . . . , s } let f i = P sj =1 a p e i,j e j with a i,j ∈ R . Then, I [1 /p e ] = ( a i,j | ≤ i ≤ r, ≤ j ≤ s ) . Definition 2.13 ([HY03, BMS08]) . Let I ⊆ R be an ideal, c a positive real number, and e apositive integer. Then, the test ideal of I with exponent c is τ ( I c ) = S e ≥ (cid:0) I ⌈ cp e ⌉ (cid:1) [1 /p e ] . Remark 2.14.
Let I be an ideal of R . Then, f p e ∈ I [ p e ] if and only if f ∈ I . This followsbecause R is finitely generated and free over R p e .If α is a rational number with a power of p in the denominators, the test ideal can be obtainedfrom the following formula. Proposition 2.15 ([BMS08, Lemma 2.1]) . Let r, e ∈ N . Then, τ ( f r/p e ) = ( f r ) [1 /p e ] . Definition 2.16.
The F -threshold of a non-zero proper ideal a ⊆ √ J ⊆ m with respect to anideal J , denoted by c J ( a ), is defined as lim e →∞ ν J a ( p e ) p e , where ν J a ( p e ) = max (cid:8) l ∈ N | a l J [ p e ] (cid:9) . It is known that the previous limit exits [MTW05, Lemma 1.1], and c J ( a ) is a rational number[BMS08, Theorem 3.1]. If a = ( f ), we simply write ν Jf ( p e ), c J ( f ) and τ ( f c ). Proposition 2.17 ([MTW05, Proposition 1.9]) . Let J ⊆ m be an ideal whose radical contains f = 0 . For every non-negative integer e , we have (cid:10) c J ( f ) (cid:11) e = ν Jf ( p e ) p e F-thresholds of a Thom-Sebastiani type polynomial
In this section we focus in stuying the F -threshold of a Thom-Sebastiani type polynomialwith respect to the sum of two ideals in different variables. In particular, we prove Theorem A.We start fixing notation for this section. Notation 3.1.
Let { x , . . . , x n } and { y , . . . , y m } be two disjoint sets of variables. Considerthe rings of polynomials R = K [ x , . . . , x n ] and R = K [ y , . . . , y m ] . Let m = ( x , . . . , x n ) and m = ( y , . . . , y m ) denote their maximal homogeneous ideals. Let I ⊆ m and I ⊆ m be two ideals. Let a = c I ( g ) and a = c I ( g ) . Let f = g + g ∈ R = R ⊗ K R = K [ x , . . . , x n , y , . . . , y m ] , where g ∈ R and g ∈ R . We start with a lemma that may be know to the experts. We add it for the sake of complete-ness. -THRESHOLDS AND TEST IDEALS OF THOM-SEBASTIANI TYPE POLYNOMIALS 5
Lemma 3.2.
We consider Notation 3.1. Let θ, e ∈ N . Then, f θ ∈ I [ p e ]1 R + I [ p e ]2 R if and only if (cid:0) θj (cid:1) = 0 , g j ∈ I [ p e ]1 , or g θ − j ∈ I [ p e ]2 for every ≤ j ≤ θ .Proof. We first assume that (cid:0) θj (cid:1) = 0 , g j ∈ I [ p e ]1 , or g θ − j ∈ I [ p e ]2 for every 0 ≤ j ≤ θ . Then, f θ = P θj =0 (cid:0) θj (cid:1) g j g θ − j ∈ I [ p e ]1 R + I [ p e ]2 R .We now assume that f θ ∈ I [ p e ]1 R + I [ p e ]2 R . We proceed by contradiction. We assume that theset A := (cid:26) j ∈ N | (cid:18) θj (cid:19) = 0 , g j I [ p e ]1 , g θ − j I [ p e ]2 , and j ≤ θ (cid:27) is not empty. Set t := min A .By our choice of t , we have that P t − j =0 (cid:0) θj (cid:1) g j g θ − j ∈ I [ p e ]1 R + I [ p e ]2 R , and therefore h = f − t − X j =0 (cid:18) θj (cid:19) g j g θ − j = θ X j = t (cid:18) θj (cid:19) g j g θ − j ∈ I [ p e ]1 R + I [ p e ]2 R. Let β = ν I g ( p e ). We have that t ≤ β and g β − t h = θ X j = t (cid:18) θj (cid:19) g β + j − t g θ − j ∈ I [ p e ]1 R + I [ p e ]2 R. Moreover, (cid:0) θj (cid:1) g β + j − t g θ − j ∈ I [ p e ]1 R + I [ p e ]2 R for every j > t . Then, (cid:0) θt (cid:1) g β g θ − t ∈ I [ p e ]1 R + I [ p e ]2 R .Since (cid:0) θt (cid:1) = 0, we deduce that g β g θ − t ∈ I [ p e ]1 R + I [ p e ]2 R. Since g β I [ p e ]1 and g θ − t I [ p e ]2 , we have that g β = 0 in R /I [ p e ]1 and g θ − t = 0 in R /I [ p e ]2 .Then, 0 = g β ⊗ g θ − t in R /I [ p e ]1 ⊗ K R /I [ p e ]2 = R/ ( I [ p e ]1 R + I [ p e ]2 R ) . Hence, g β g θ − t I [ p e ]1 R + I [ p e ]2 R, a contradiction. (cid:3) We start proving Theorem A by finding a lower bound for F -thresholds of a Thom-Sebastianitype polynomial with respect to an ideal that has also Thom-Sebastiani type shape. Lemma 3.3.
We consider Notation 3.1. Then, c I + I ( g + g ) ≥ h a i L + h a i L + 1 p L , where L = sup n N ∈ N | a ( e )1 + a ( e )2 ≤ p − for all ≤ e ≤ N o . Proof.
We note that a ( L +1)1 + a ( L +1)2 ≥ p , by our definition of L . There exists α , α ∈ N suchthat α + α = p − ≤ α < a ( L +1)1 and 0 ≤ α ≤ a ( L +1)2 . For any integer e ≥ L + 2, weset η ( e ) = ( η ( e ) , η ( e )) := ( h a i L , h a i L ) + (cid:18) α p L +1 + p − p L +2 + · · · + p − p e , α p L +1 (cid:19) , = ( h a i L , h a i L ) + p L +1 · α + p e − ( L +1) − p e − ( L +1) ! , α p L +1 ! . M. GONZ ´ALEZ VILLA, D. JARAMILLO-VELEZ, AND L. N ´U ˜NEZ-BETANCOURT
Thus, η ( e ) ≤ h a i L +1 and η ( e ) < h a i L +1 . The base p expansions of the integers p e η ( e ) and p e η ( e ) are p e η ( e ) = p e − ( L +1) α + p e − ( L +2) ( p −
1) + · · · + p − , and p e η ( e ) = p e − ( L +1) α . Since α + α = p −
1, the integers p e η ( e ) and p e η ( e ) add without carrying in base p . Thus(1) (cid:18) p e ( η ( e ) + η ( e )) p e η ( e ) (cid:19) p by Lemma 2.6.Since η ( e ) ≤ h a i e and η ( e ) ≤ h a i e for every e ≥ L + 2, we have that p e η ( e ) ≤ p e h a i e = ν I g ( p e ) and p e η ( e ) ≤ p e h a i e = ν I g ( p e ). Therefore, g p e η ( e )1 I [ p e ]1 , g p e η ( e )2 I [ p e ]2 . Thus, f p e ( η ( e )+ η ( e )) ( I + I ) [ p e ] by Lemma 3.2. It follows that p e (cid:10) c I + I ( f ) (cid:11) e = ν I + I f ( p e ) ≥ p e ( η ( e ) + η ( e )) . Finally, we have (cid:10) c I + I ( f ) (cid:11) e ≥ η ( e ) + η ( e ) = h a i L + h a i L + 1 p L +1 ( α + α ) + p − p L +2 + · · · + p − p e = h a i L + h a i L + p − p L +1 + p − p L +2 + · · · + p − p e . Taking the limit when e → ∞ , we have that c I + I ( f ) ≥ h a i L + h a i L + p − p L +1 (cid:18) p + 1 p + . . . (cid:19) = h a i L + h a i L + 1 p L . (cid:3) Theorem 3.4.
Consider Notation 3.1. If a + a ≤ , then c I + I ( f ) = a + a , if L = ∞ , h a i L + h a i L + p L , if L < ∞ , where L = sup n N ∈ N | a ( e )1 + a ( e )2 ≤ p − for all ≤ e ≤ N o . Proof.
We split the proof into two cases.Case 1:
L < ∞ : By Lemma 3.3 we have that(2) c I + I ( f ) ≥ h a i L + h a i L + 1 p L We proceed by contradiction, and assume that the inequality in Equation 2 is strict. Then, ν I + I f ( p e ) p e > h a i L + h a i L + 1 p L for e ≫ . Multiplying by p e , we obtain that ν I + I f ( p e ) > p e ( h a i L + h a i L ) + p e − L = p e − L (cid:0) p L ( h a i L + h a i L ) + 1 (cid:1) for e ≥ L. -THRESHOLDS AND TEST IDEALS OF THOM-SEBASTIANI TYPE POLYNOMIALS 7 Therefore, f p e − L ( p L ( h a i L + h a i L )+1 ) = (cid:16) f p L ( h a i L + h a i L )+1 (cid:17) p e − L ( I + I ) [ p e ] = (cid:16) ( I + I ) [ p L ] (cid:17) [ p e − L ] . By Remark 2.14, it follows that f p L ( h a i L + h a i L )+1 ( I + I ) [ p L ] . This implies that there exists k = ( k , k ) ∈ N such that k + k = p L ( h a i L + h a i L ) + 1 and g k g k ( I + I ) [ p L ] . Inparticular, we have g k ( I ) [ p L ] , and g k ( I ) [ p L ] . By Proposition 2.17, we conclude that k ≤ ν I g ( p L ) = p L h a i L , and k ≤ ν I g ( p L ) = p L h a i L . Since k + k = p L ( h a i L + h a i L ) + 1, we get h a i L + h a i L + 1 p L = 1 p L ( k + k ) ≤ h a i L + h a i L , which is a contradiction. Therefore, it holds c I + I ( f ) = h a i L + h a i L + 1 p L . Case 2: L = ∞ : Since f ν I I f ( p e ) ( I + I ) [ p e ] , we conclude that there exists k , k ∈ N suchthat k + k = ν I + I f ( p e ) and g k g k ( I + I ) [ p e ] by Lemma 3.2. Since the polynomials g , g are in different sets of variables, we have g k I [ p e ]1 , g k I [ p e ]2 . It follows that k ≤ ν I g ( p e ) and k ≤ ν I g ( p e ). Since ν I + I f ( p e ) = k + k ≤ ν I g ( p e ) + ν I g ( p e ) , we have that ν I + I f ( p e ) p e ≤ ν I g ( p e ) p e + ν I g ( p e ) p e . Taking the limit when e → ∞ , we have that c I + I ( f ) = lim e →∞ ν I + I f ( p e ) p e ≤ lim e →∞ ν I g ( p e ) p e + ν I g ( p e ) p e ! = lim e →∞ ν I g ( p e ) p e + lim e →∞ ν I g ( p e ) p e = a + a . We now check that a + a ≤ c I + I ( f ). Since a and a add without carrying in base p ,Remark 2.4 and Lemma 2.6 imply that(3) (cid:18) p e ( h a i e + h a i e ) p e h a i e (cid:19) p. Moreover, ν I g ( p e ) = p e h a i e , and ν I g ( p e ) = p e h a i e by Proposition 2.17. Then,(4) g p e h a i e I [ p e ]1 , g p e h a i e I [ p e ]2 . Equations 3, 4, and Lemma 3.2 imply that f p e ( h a i e + h a i e ) ( I + I ) [ p e ] . Thus, p e ( h a i e + h a i e ) ≤ ν I + I f ( p e ) for all e ≫ . M. GONZ ´ALEZ VILLA, D. JARAMILLO-VELEZ, AND L. N ´U ˜NEZ-BETANCOURT
Dividing by p e , we have h a i e + h a i e ≤ ν I + I f ( p e ) p e for all e ≫ . Finally, taking the limit when e → ∞ , we obtain that a + a ≤ c I + I ( f ), and c I + I ( f ) = a + a . (cid:3) Example 3.5.
Let a, b, n ∈ Z > , and b ≥ a . Let g = x n ∈ R = K [ x ] , g = y a y b ∈ R = K [ y , y ], and f = g + g ∈ R ⊗ K R . Let I = m = ( x ), I = m = ( y , y ), and m = m + m = ( x, y , y ). Since a = c m ( g ) = n and c m ( g ) = b , we have that c m ( f ) = n + b , if L = ∞ , (cid:10) n (cid:11) L + (cid:10) b (cid:11) L + p L , if L < ∞ . where L = sup n N ∈ N | (cid:0) n (cid:1) ( e ) + (cid:0) b (cid:1) ( e ) ≤ p − ≤ e ≤ N o . It is worth mention-ing that c m ( f ) = c m ( x n + y b ). Remark 3.6. If g ∈ I and g ∈ I , then f ∈ I R + I R . Then, if we run over all the test ideal I i = τ ( g λ i i ) for λ i ∈ (0 , g i and whose F -thresholdsgive the jumping numbers in (0 , F -thresholds withrespect to τ ( g λ ) R + τ ( g λ ) R . This give F -jumping numbers of f , but not necessarily all.4. Relation with log canonical threshold
The behavior of the F -pure threshold for different reductions mod p and their connectionswith invariants over Q have attracted much interest. We refer the to recent surveys on thistopic [BFS13, TW18]. We are particularly interested in the relation between the log canonicalthreshold and the F -threshold. As consequence of Theorem 3.4, we give further evidence for theconjecture relating the F -pure and log-canonical threshold.Let f ∈ Q [ x , . . . , x n ], and let f p denote the polynomial over F p obtained by reducing eachcoefficient of f modulo p .A log resolution of f defined over Q is a proper birational map π Q : Y Q → A n Q , with Y Q smooth, such that f O Y Q is a principal and defines a simple normal crossing divisor D . The logcanonical threshold lct ( f ) of f at the origin 0 ∈ A n Q is the minimum number λ ∈ Q > suchthat the ideal ( π Q ) ∗ O Y Q ( K π Q − ⌊ λ · D ⌋ ) is contained in the maximal ideal ( x , x . . . , x n ) Q .Mustat¸ˇa, Tagaki, and Watanabe showed the following relation between the F-pure thresholdand the log canonical threshold. Theorem 4.1 ([MTW05, Theorem 3.4]) . Let f ∈ Q [ x , . . . , x n ] , and let f p denote the polynomialover F p obtained by reducing each coefficient of f modulo p . Then, lim p →∞ c m ( f p ) = lct ( f ) . Furthermore, they have formulated the following conjecture.
Conjecture 4.2 ([MTW05, Conjecture 3.6]) . Let f ∈ Q [ x , . . . , x n ] , and let f p denote thepolynomial over F p obtained by reducing each coefficient of f modulo p . There exist infinitelymany prime numbers p such that lct ( f ) = c m ( f p ) . We now use Theorem 3.4 to study Conjecture 4.2. -THRESHOLDS AND TEST IDEALS OF THOM-SEBASTIANI TYPE POLYNOMIALS 9
Theorem 4.3.
We consider Notation 3.1 with K = Q . Let f = g + g ∈ R ⊗ Q R be aThom-Sebastiani type polynomial as in the previous section. Assume that g ∈ R and g ∈ R satisfy Conjeture 4.2. Denote by B i , with i = 1 , the infinite set of primes p such that lct ( g i ) = c m (( g i ) p ) . Assume that a + a < and there exists an infinite set B ⊆ B ∩ B suchthat the non terminating base p extensions of a and a add without carrying for every p ∈ B .Then f = g + g satisfies the Conjeture 4.2.Proof. This is a consequence of the Thom-Sebastiani property for log canonical thresholds[Kol97b, Proposition 8.21] and Theorem 3.4. (cid:3)
We know provide a series of examples where the hypothesis of Theorem 4.3 are satisfied.
Example 4.4 (Thom-Sebastiani type sum of monomials) . If g ∈ R and g ∈ R monomials,then B = B is the set of all primes p . If a + a <
1, Theorem 2.8 implies the existenceof an infinite set B of primes p for which the non terminating base p expansions of a and a are periodic, and therefore add without carrying [Her11, Lemma 4.16 and Example 4.4]. Thenthe hypothesis of Theorem 4.3 are satisfied, and Conjecture 4.2 holds for f . Particular cases ofthis situation are diagonal hypersurfaces, and Thom-Sebastiani type binomials (or multinomials,since the argument extends to any finite number of summands).A case of the latteris Example 3.5. Notice that if gcd( a, b, n ) = 1 the hypersurfaces ofExample 3.5 are irreducible quasi-ordinary hypersurfaces with only one characteristic exponent.We point out that the log-canonical thresholds of irreducible quasi-ordinary hypersurfaces hasalready been computed [BGPGV12]. Example 4.5 (Suspensions or ramified cyclic coverings) . Let g ∈ R such that the set B contains and infinite number of primes p such that ( p − a ∈ N . An example of this is g = x b x b + x c x c ∈ R = K [ x , x ] a binomial hypersurface whose maximal splitting polytope P f contains a unique maximal point η = ( r s , r s ) ∈ Q , with r +1 s + r +1 s ≤ g = y d ∈ K [ y ]. Then a = d = lct ( g ) for all primes, and so, B is the set of allprimes. Finally, by Theorem 2.8, there is a infinite subset B of B = B ∩ B , consisting of allprimes p such that ( p − a , ( p − d ∈ N . We have that, if a + a <
1, then the non terminatingbase p expansions of a and a add without carrying [Her11, Lemma 4.16 and Example 4.4]. Weconclude that c m ( f ) = a + a = lct ( f ) for all p ∈ B . Hence, f = g + g satisfies Conjecture4.2. We note that it is possible to iterated the previous construction because f satisfies thehypothesis required on g .The following examples combine the previous cases. Example 4.6.
Let R = K [ x, y ], R = K [ w, z ], R = K [ t, u, v ], and g = x + y ∈ m = ( x, y ), g = z w + z w ∈ m = ( z, w ), and g = v u t ∈ m = ( t, u, v ). Set f = g + g + g ∈ R ⊗ K R ⊗ K R , and m = ( x, y, z, w, v, u, t ). We compute the F-pure threshold of f , and checkthat f satisfies Conjecture 4.2. • If p ≡ c m ( g ) = = lct ( g ). • If p ≡ c m ( g ) = [Her14, Theorem 4.1]. Moreover, Theorem 4.1implies that c m ( g ) = lct ( g ) = . • We have that c m ( g ) = = lct ( v u t ) for any prime p The set of primes p such that p ≡ c m ( f ) = = lct ( f ). Example 4.7 (Exploiting other properties of the log canonical threshold and the F -pure thresh-old) . Let g and g be polynomials satisfying the conditions in Theorem 4.3.We consider first f = ( g + g ) n . Then, for any prime p in the infinite set B , we have c m ( f ) = c m ( g ) + c m ( g ) n = lct ( g ) + lct ( g ) n = lct ( f ) . Thus, Conjecture 4.2 holds for f .We consider now f = g g . Then, we have c m ( f ) ≤ lct ( f ) ≤ min { lct ( g ) , lct ( g ) } = min { lct ( g ) , lct ( g ) } = c m ( f )for any integer p in the infinite set B [TW04, Proposition 3.2] [Kol97b, Theorem 8.20] [Her14,Lemma 3.3].Hence, c m ( f ) = lct ( f ) for any prime p in B , and so, Conjecture 4.2 holds for f . Example 4.8.
Consider the polynomial f = w x y z + 4 u v w x y z + 6 u v w x y z + 4 u v w x y z + u v , and p ≡ f = (cid:0) u v + w xy z (cid:1) = ( g + g ) for g = u v ∈ K [ u, v ] and g = w xy z ∈ K [ w, x, y, z ]. Then, c m ( f ) = c m ( g + g ) /
4, where m = ( u, v, w, x, y, z ). For the monomials g , g we get that c m ( g ) = 1 / lct ( g ) , and c m ( g ) = 1 /
11 = lct ( g ) , where m = ( u, v ), and m = ( w, x, y, z ). The number 1 / /
11 add without carryingwhenever p ≡ c m ( f ) = 1 / / /
11) = 9 /
154 by Theorem 3.4. There areinfinitely many prime numbers that satisfies p ≡ lct ( f ) = lct ( g + g ) / /
154 = c m ( f ) . Therefore Conjecture 4.2 holds for the polynomial f . Example 4.9.
Let f = w x t + uv w x + t y z + y zuv ∈ K [ t, u, v, w, x, y, z ], and p ≡ f = ( w x + y z )( t + uv ) = g g , for g = w x + y z ∈ K [ w, x, y, z ], and g = t + uv ∈ K [ t, u, v ]. We have that c m ( f ) =min { c m ( g ) , c m ( g ) } , where m = ( w, x, y, z ) and m = ( t, u, v ). Using Theorem 3.4, wecompute the F -threshold of g , and g . For p ≡ /
6, and 1 / c m ( g ) = 1 / / /
3. By Example 3.5, we obtain c m ( g ) = 5 / c m ( f ) = c m ( g ) = 2 /
3. There are infinitely many prime numbers that satisfies p ≡ c m ( f ) ≤ lct ( f ) ≤ min { lct ( g ) , lct ( g ) } = 2 / . Therefore Conjecture 4.2 holds for the polynomial f .5. Test ideal of a Thom-Sebastiani type polynomial
In this section we compute a formula for the first non-trivial test ideal of a Thom-Sebastianitype polynomial.
Notation 5.1.
Let { x , . . . , x n } and { y , . . . , y m } be two disjoint sets of variables. Consider therings of polynomials R = K [ x , . . . , x n ] and R = K [ y , . . . , y m ] . Denote by m = ( x , . . . , x n ) and m = ( y , . . . , y m ) their maximal homogeneous ideals. Let f = g + g ∈ R = R ⊗ K R = K [ x , . . . , x n , y , . . . , y m ] , where g ∈ R and g ∈ R . Let a = c m ( g ) , and a = c m ( g ) . -THRESHOLDS AND TEST IDEALS OF THOM-SEBASTIANI TYPE POLYNOMIALS 11 We start with some of preparation lemmas before proving Theorem B.
Lemma 5.2.
Let α , α ∈ R ≥ . Assume that α ( e )1 + α ( e )2 ≤ p − , and (cid:18) p e ( h α i e + h α i e ) p e h α i e (cid:19) p, then (cid:18) p e ( h α i e + h α i e ) + 1 p e h α i e + 1 (cid:19) p & (cid:18) p e ( h α i e + h α i e ) + 1 p e h α i e + 1 (cid:19) p. Proof.
From the definition of binomial coefficient we get the equalities (cid:18) p e ( h α i e + h α i e ) + 1 p e h α i i e + 1 (cid:19) = (cid:18) p e ( h α i e + h α i e ) p e h α i i e (cid:19) · p e ( h α i e + h α i e ) + 1 p e h α i i e + 1 , with i = 1 ,
2. The result follows from these equalities and the congruence p e ( h α i e + h α i e ) + 1 ≡ α ( e )1 + α ( e )2 + 1 p. The latter holds because α ( e ) i ≤ α ( e )1 + α ( e )2 ≤ p − (cid:3) Lemma 5.3.
Let α , α ∈ R ≥ , and assume that α + α < . Let L = sup n N ∈ N | α ( e )1 + α ( e )2 ≤ p − for all ≤ e ≤ N o and d = sup n e ≤ L | α ( e )1 + α ( e )2 ≤ p − o . Then, h α i d + h α i d + 1 p d = h α i L + h α i L + 1 p L . Proof.
We proceed by cases.If d = L = ∞ , then the result follows because h α i i ∞ = α i .If L = ∞ and d < ∞ , we have α ( e )1 + α ( e )2 = p − e ≥ d + 1. Hence α + α = h α i L + h α i L + 1 p L = h α i d + h α i d + ∞ X j = d +1 p − p j = h α i d + h α i d + 1 p d . If L < ∞ and d < ∞ , we have α ( e )1 + α ( e )2 = p − d + 1 ≤ e ≤ L . Hence h α i L + h α i L + 1 p L = h α i d + h α i d + p − p d +1 + · · · + p − p L + 1 p L = h α i d + h α i d + 1 p d . (cid:3) Lemma 5.4.
Consider Notation 5.1, and assume that a + a < . Let L = sup n N ∈ N | a ( e )1 + a ( e )2 ≤ p − for all ≤ e ≤ N o . If a ( e )1 + a ( e )2 ≤ p − for some e ≤ L , then (cid:16) f p e ( h a i e + h a i e )+1 (cid:17) [1 /p e ] = (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] + (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] . Proof.
Let θ i = p e h a i i e = ν m i g i ( p e ) for i = 1 ,
2, and θ = p e ( h a i e + h a i e ) = θ + θ . First, we check that (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] + (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] ⊆ (cid:16) f p e ( h a i e + h a i e )+1 (cid:17) [1 /p e ] for i = 1 ,
2. We choose elements w i,j ∈ R such that 1 , g i , . . . , g θ i i , w i, , . . . , w i,s i is a free basis of R i as R p e i -module. Then, B e = n g j g j , g j w ,k , w ,k g j , w ,k w ,k | ≤ j i ≤ θ i & 0 ≤ k i ≤ s i o is a free basis of R over R p e . We fix h j ∈ R such that g θ +11 = θ X j =0 h p e j g j + s X j =1 h p e θ + j w ,j . Then, (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] = ( g p e h a i e +11 ) [1 /p e ] = h h j | ≤ j ≤ θ + s i by Proposition 2.12. We consider f θ +1 in terms of the basis B e . We now show that each h p e j appears as a coefficient. Then, f θ +1 = θ +1 X k =0 (cid:18) θ + 1 k (cid:19) g k g θ +1 − k = g θ +12 + · · · + (cid:18) θ + 1 θ + 1 (cid:19) g θ +11 g θ + · · · + g θ +11 . By Lemma 5.2, we have (cid:18) θ + 1 θ i + 1 (cid:19) p. We expand g θ +11 g θ as g θ +11 g θ = ( θ X j =0 h p e j g j + s X j =1 h p e θ + j w ,j ) g θ = θ X j =0 h p e j g j g θ + s X j =1 h p e θ + j w ,j g θ . We note that the other summands are either multiples of w ,k g j , or g j w ,k , with 0 ≤ j ≤ θ ,0 ≤ j < θ , and 0 ≤ k i ≤ s i . Then, h j ∈ (cid:0) f θ +1 (cid:1) [1 /p e ] . We conclude that (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] ⊆ (cid:0) f p e ( h a i e + h a i e )+1 (cid:1) [1 /p e ] . Similarly, (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] ⊆ (cid:0) f p e ( h a i e + h a i e )+1 (cid:1) [1 /p e ] .We now show the other containment. From the expression f θ +1 = g θ +12 + (cid:18) θ + 11 (cid:19) g g θ + · · · + (cid:18) θ + 1 θ (cid:19) g θ g θ +12 + (cid:18) θ + 1 θ + 1 (cid:19) g θ +11 g θ + · · · + g θ +11 , we conclude that f θ +1 ∈ (cid:16) g θ +11 , g θ +12 (cid:17) . Then, (cid:16) f θ +1 (cid:17) [1 /p e ] ⊆ (cid:16) g θ +11 , g θ +12 (cid:17) [1 /p e ] = (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] + (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] . (cid:3) We are now ready to prove Theorem B. -THRESHOLDS AND TEST IDEALS OF THOM-SEBASTIANI TYPE POLYNOMIALS 13
Theorem 5.5.
Consider Notation 5.1. Let L = sup n N ∈ N | a ( e )1 + a ( e )2 ≤ p − for all ≤ e ≤ N o and d = sup n e ≤ L | a ( e )1 + a ( e )2 ≤ p − o . Then τ (cid:16) f c m ( f ) (cid:17) = ( f ) if c m ( f ) = 1 ,τ ( g a ) + τ ( g a ) if c m ( f ) p − e · N , (cid:16) g ⌈ p d a ⌉ (cid:17) [1 /p d ] + (cid:16) g ⌈ p d a ⌉ (cid:17) [1 /p d ] if c m ( f ) ∈ Z [ p ] & c m ( f ) = 1 . Proof.
It is well know that τ ( f ) = ( f ). This settles the first case. We assume that c m ( f ) = 1.If c m ( f ) Z [ p ], then a and a add without carrying by Theorem A. Thus, L = ∞ , and c m ( f ) = a + a . There exists infinity e ∈ N such that a ( e )1 + a ( e )2 ≤ p −
2, because c m ( f ) Z [ p ].The result follows from Lemma 5.4, by picking e ∈ N such that a ( e )1 + a ( e )2 ≤ p − τ ( g a ) = (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] , τ ( g a ) = (cid:16) g ⌈ p e a ⌉ (cid:17) [1 /p e ] , and τ ( f a + a ) = (cid:16) f ⌈ p e ( a + a ) ⌉ (cid:17) [1 /p e ] = (cid:16) f p e ( h a i e + h a i e )+1 (cid:17) [1 /p e ] . Finally, suppose that c m ( f ) ∈ p − e · N . Lemma 5.3 implies that c m ( f ) = h a i L + h a i L + 1 p L = h a i d + h a i d + 1 p d . Then, τ (cid:16) f c m ( f ) (cid:17) = τ (cid:18) f h a i d + h a i d + pd (cid:19) = (cid:16) f p d ( h a i d + h a i d )+1 (cid:17) [1 /p d ] = (cid:16) g ⌈ p d a ⌉ (cid:17) [1 /p d ] + (cid:16) g ⌈ p d a ⌉ (cid:17) [1 /p d ] by Proposition 2.15 and Lemma 5.4. (cid:3) Example 5.6 ([Her15, Example 4.9]) . Let p > a = p − , and b = p − .Consider the polynomials g = x b + · · · + x ba ∈ R = F p [ x , . . . , x a ], g = y pb + · · · + y pba ∈ R = F p [ y , . . . , y a ], and f = g + g ∈ R = R ⊗ F p R . Let m = h x , . . . , x a i ⊂ R , m = h y , . . . , y a i ⊂ R and m = h x , . . . , x a , y , . . . , y a i ⊂ R be the respective maximal homogeneousideals. Theorem 3.4 implies that c m ( f ) = a + a = p ∈ p − · N , where a = c m ( g ) = a/b , and a = c m ( g ) = a/pb . Since g = x b + · · · + x ba = ( x a ) p · x a + · · · + ( x aa ) p · x aa , and g = ( y b + · · · + y ba ) p · , we conclude from Theorem 5.5, that τ (cid:16) f c m ( f ) (cid:17) = (cid:16) g ⌈ pa ⌉ (cid:17) [1 /p ] + (cid:16) g ⌈ pa ⌉ (cid:17) [1 /p ] = h x a , . . . , x aa i + h y b + · · · + y ba i = h x a , . . . , x aa , y b + · · · + y ba i . Example 5.7.
Consider the following polynomial f = z w + z w + v u t ∈ K [ t, u, v, w, z ].For p = 97, we want to compute the test ideal of f at a = c m ( f ), where m = ( t, u, v, w, z ). Wesplit the polynomial f into g = z w + z w ∈ K [ w, z ], and g = v u t ∈ K [ t, u, v ]. Let a , a bethe F -threshold of g , and g , respectively. We know that a = 3 /
16, and a = 1 /
8, because g is a monomial [Her14, Example 4.3]. Since, a , and a add without carrying, by Theorem 3.4 wehave that a = a + a = 5 /
16. The test ideals τ ( g a ) = ( w, z ), and τ ( g a ) = ( t ) were computedusing Definition 2.13. Finally, Theorem 5.5 implies that τ ( f a ) = τ ( g a ) + τ ( g a ) = ( t, w, z ). Acknowledgments
The authors thank Daniel J. Hernndez for helpful comments and suggestions. The secondauthor started this work while pursuing a master’s degree at CIMAT. He thanks this institutionfor its support during his studies. We use Macaulay2 [GS] to compute several examples.
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