aa r X i v : . [ m a t h . A C ] D ec F -INVARIANTS OF DIAGONAL HYPERSURFACES DANIEL JES ´US HERN ´ANDEZ
Abstract.
In this note, we derive a formula for the F -pure threshold of diagonal hyper-surfaces over a perfect field of prime characteristic. We also calculate the associated testideal at the F -pure threshold, and give formulas for higher jumping numbers of Fermathypersurfaces. Introduction
Let R be a polynomial ring over a perfect field L of characteristic p >
0, and consider apolynomial f ∈ R . Using the Frobenius morphism R → R given by r r p , one may definea family of ideals { τ ( λ • f ) ⊆ R : λ > } called the test ideals of f . Test ideals (defined inthe context of tight closure ) were originally introduced in [HH90], and generalized to pairs in[HY03]. Test ideals vary with respect to λ in the following way: they shrink as λ increases,and are also stable to the right. We say that a parameter λ is an F -jumping number of f if τ ( λ • f ) = τ (( λ − ε ) • f ) for every 0 < ε < λ . We call the smallest F -jumping number the F -pure threshold of f and denote it by fpt ( f ). In this article, we consider these invariantswhen f is diagonal or Fermat . Recall that f is called diagonal if it is an L ∗ -linear combinationof x d , · · · , x d n n and Fermat if it is an L ∗ -linear combination of x d , · · · , x dd .In Theorem 3.1, we give a formula for the F -pure threshold of a diagonal hypersurface as afunction of the characteristic. In Theorem 3.3, we give a formula for the first non-trivial testideal of a diagonal hypersurface. Note that (classical) test ideals of diagonal hypersurfaceswere computed by McDermott in [McD01] and [McD03]. In Theorem 3.6, we give conditionsfor the existence of, and formulas for, higher jumping numbers of Fermat hypersurfaces. Fora detailed discussion of our main results, and for examples, see Section 3.0.1. Acknowledgements.
This work is part of the author’s Ph.D. thesis at the Universityof Michigan. I would like to thank Karen Smith for suggesting this problem, as well as EmilyWitt, whose observation led to the statement and proof of Theorem 3.1.1.
Test Ideals and F -pure thresholds Let L be a a perfect of characteristic p >
0, and let R = L [ x , · · · , x n ]. We will use m to denote the ideal ( x , · · · , x n ). As L is perfect, we have that R p e := L [ x p e , · · · , x p e n ] is thesubring of ( p e ) th powers of R . For every ideal I ⊆ R , let I [ p e ] denote the ideal generated bythe set (cid:8) g p e : g ∈ I (cid:9) . We call I [ p e ] the e th Frobenius power of I . Definition 1.1.
We will use B e to denote the set of monomials (cid:8) µ : µ / ∈ m [ p e ] (cid:9) . The readermay verify that B e is a free basis for R as an R p e -module. If f ∈ R is a non-zero polynomialand µ ∈ B e , we use Γ eµ ( f ) to denote the element of R such that f = P µ ∈ B e Γ eµ ( f ) p e µ . The author was partially supported by the National Science Foundation RTG grant number 0502170 atthe University of Michigan. emark 1.2. As R is finitely generated and free over R p e , it follows that f p e ∈ I [ p e ] if andonly if f ∈ I . Definition 1.3.
Let f ∈ R be a non-zero polynomial. We use ( f )[ pe ] to denote the idealgenerated by the set (cid:8) Γ eµ ( f ) : µ ∈ B e (cid:9) .Lemma 1.4 follows from [BMS08, Proposition 2.5], though we include a proof for the sakeof completeness. Lemma 1.4.
Let f ∈ R . If I ⊆ R is an ideal, then ( f )[ pe ] ⊆ I if and only if f ∈ I [ p e ] . Proof.
If ( f )[ pe ] ⊆ I , then f ∈ (cid:16) ( f )[ pe ] (cid:17) [ p e ] ⊆ I [ p e ] . Instead, suppose f ∈ I [ p e ] = ( a p e , · · · , a p e s ).Then, f = P si =1 g i · a p e i = P µ ∈ B e (cid:0)P si =1 a i Γ eµ ( g i ) (cid:1) p e µ . Thus, Γ eµ ( f ) = P si =1 a i Γ eµ ( g i ), andwe conclude that ( f )[ pe ] ⊆ I . (cid:3) Remark 1.5.
Lemma 1.4 shows that ( f )[ pe ] is the unique minimal ideal I such that f ∈ I [ p e ] .This shows that ( f )[ pe ] does not depend on the specific choice of basis B e for R over R p e . Definition 1.6.
For every λ ≥
0, the set (cid:26) (cid:0) f ⌈ p e λ ⌉ (cid:1) [ pe ] : e ≥ (cid:27) defines an increasing se-quence of ideals [BMS08, Lemma 2.8]. We call the stabilizing ideal the test ideal of f (withrespect to the parameter λ ), and denote it by τ ( λ • f ). In other words, τ ( λ • f ) = [ e ≥ (cid:0) f ⌈ p e λ ⌉ (cid:1) [ pe ] = (cid:0) f ⌈ p e λ ⌉ (cid:1) [ pe ] for all e ≫ . The following lemma, whose proof we omit, allows us to identify when the test idealstabilizes in an important special case.
Lemma 1.7. [BMS09, Lemma 2.1] If λ ∈ p e · N , then τ ( λ • f ) = (cid:0) f p e λ (cid:1) [ pe ].Test ideals form a decreasing sequence of ideals, and are stable to the right [BMS08,Proposition 2.11, Corollary 2.16] . That is, τ ( λ • f ) ⊆ τ ( λ ◦ • f ) if λ ≥ λ ◦ . Additionally,for every λ ≥ ε > τ ( λ • f ) = τ (( λ + δ ) • f ) whenever 0 ≤ δ < ε .This behavior motivates the following definition. Definition 1.8.
We say that λ > F -jumping number of f if τ ( λ • f ) = τ (( λ − ε ) • f ) for all 0 < ε < λ. By convention, we consider 0 an F -jumping number of f . Proposition 1.9. [BMS08, Proposition 2.25] A number γ > F -jumping number of f if and only if γ − F -jumping number of f .Let f be a non-zero, non-unit polynomial in R , and choose e ≫ p e > deg f . Itfollows that for every proper ideal I ( R , we have that f / ∈ I [ p e ] . This, combined with Lemma1.4 and Lemma 1.7, shows that ( f )[ pe ] = τ (cid:16) p e • f (cid:17) is not contained in any proper ideal of R , and thus must equal R . We see that τ ( λ • f ) = R for 0 < λ ≪
1, and so the smallestnon-zero F -jumping number of f is the minimal parameter λ such that τ ( λ • f ) = R . Thisjumping number is of particular interest, and is called the F -pure threshold of f . efinition 1.10. We call fpt ( f ) := sup { λ ∈ R ≥ : τ ( λ • f ) = R } the F -pure threshold of f , and we call fpt m ( f ) := sup { λ ∈ R ≥ : τ ( λ • f ) m = R m } the F -pure threshold of f at m .In our computations, we will use the following well known description of fpt m ( f ). Lemma 1.11. fpt m ( f ) = max (cid:8) λ > ∃ e λ ≥ f ⌈ p e λ ⌉ / ∈ m [ p e ] for all e ≥ e λ (cid:9) . Proof.
Comparing with Definition 1.10, we see it suffices to show τ ( λ • f ) m = R m if andonly if there exists e λ ≥ f ⌈ p e ⌉ / ∈ m [ p e ] for all e ≥ e λ . By definition, there exists e λ such that τ ( λ • f ) = (cid:0) f ⌈ p e λ ⌉ (cid:1) [ pe ] for all e ≥ e λ . For such an e , τ ( λ • f ) m = R m if and onlyif (cid:0) f ⌈ p e λ ⌉ (cid:1) [ pe ] m = R m . However, this occurs if and only if (cid:0) f ⌈ p e λ ⌉ (cid:1) [ pe ] m , which Lemma 1.4shows happens if and only if f ⌈ p e λ ⌉ / ∈ m [ p e ] . (cid:3) Some remarks on base p expansions Definition 2.1.
Let α ∈ (0 , p be a prime number. Let α ( d ) be the unique integerin [0 , p −
1] such that α = P d ≥ α ( d ) p d and such that α ( d ) = 0 is not eventually zero as afunction of d . We call α ( d ) the d th digit of the non-terminating base p expansion of α . Weadopt the convention that α (0) = 0 ( d ) = 0. Example 2.2. If α = p = p + P e ≥ , we see that α (1) = 0 and α ( e ) = p − e ≥ Definition 2.3. If λ = 0, we call h λ i e := P ed =1 λ ( d ) p d the e th truncation of λ (in base p ). Lemma 2.4. If λ ∈ [0 , ⌈ p e λ ⌉ = p e h λ i e + 1. Furthermore, if α ∈ [0 , ∩ p e · N and λ > α , then h λ i e ≥ α . Proof. As p e λ = p e h λ i e + p e · P d>e λ ( d ) p d , the first claim follows from the observation that0 < P d>e λ ( d ) p d ≤ p e . We also see that p e + h λ i e ≥ λ > α , so(1) 1 + p e h λ i e > p e α. By hypothesis, both sides of (1) are integers, and we conclude that p e h λ i e ≥ p e α . (cid:3) Definition 2.5.
Let ( λ , · · · , λ n ) ∈ [0 , n , and let p be a prime number. We say the e th digits of λ , · · · , λ n add without carrying (in base p ) if λ ( e )1 + · · · + λ ( e ) n ≤ p −
1, and wesay that λ , · · · , λ n add without carrying if all of their digits add without carrying. We saynatural numbers k , · · · , k n add without carrying (in base p ) if the obvious condition holds. Remark 2.6. If λ , · · · , λ n add without carrying (in base p ) and λ := P ni =1 λ i ≤
1, then λ ( e ) = λ ( e )1 + · · · + λ ( e ) n for all e ≥ Lemma 2.7. [Dic02, Luc78] Let k = ( k , · · · k n ) ∈ N n and set N = | k | = P k i . Then (cid:0) N k (cid:1) := N ! k ! ··· k n ! p if and only if k , · · · , k n add without carrying (in base p ). . Discussion of the main results F -pure theshholds of diagonal hypersurfaces. In our first result, we give a formulathe F -pure threshold of a diagonal hypersurface. Theorem 3.1.
Let L be the supremum over all N such that the e th digits of d , · · · , d n addwithout carrying for all 0 ≤ e ≤ N . If f is a L ∗ -linear combination of x d , · · · , x d n n , then fpt m ( f ) = d + · · · + d n if L = ∞ D d E L + · · · + D d n E L + p L if L < ∞ Formulas for the F -pure threshold of x + y and x + y are given in [MTW05, Example 4 . . Example 3.2.
We adopt decimal notation for base p expansions. For example, if a, b areintegers with 0 ≤ a, b ≤ p −
1, then .a b (base p ) will denote the unique number λ with theproperty that λ ( e ) = a for e odd and λ ( e ) = b for e even. Let f be a L ∗ -linear combinationof x and y . If p = 3, then12 = . . . . We see that carrying is required to add the second digits of and (but not the first),and Theorem 3.1 implies fpt m ( f ) = (cid:10) (cid:11) + (cid:10) (cid:11) + = 0 + + = . Similarly, one canshow that fpt m ( f ) = if p = 2. If p = 6 ω + 1 for some ω ≥
1, then12 = . ω (base p ) and 13 = . ω (base p ) . We notice that and add without carrying (in base p ), and Theorem 3.1 implies fpt m ( f ) = + = . Finally, if p = 6 ω + 5 for some ω ≥
0, then12 = . ω + 2 (base p ) and 13 = . ω + 1 4 ω + 3 (base p ) . Once more, we see that carrying is needed to add the second digits of and , (but not thefirst), and Theorem 3.1 implies(2) fpt m ( f ) = (cid:28) (cid:29) + (cid:28) (cid:29) + 1 p = 3 ω + 2 p + 2 ω + 1 p + 1 p = 5 ω + 4 p . The reader may verify that ω +4 p + p = , so we may rewrite (2) as fpt m ( f ) = − p . Thus,we recover the following formula from [MTW05, Example 4.3]: fpt m ( x + y ) = / p = 22 / p = 35 / p ≡ − p if p ≡ . .2. A computation of the first non-trivial test ideal.
Our second theorem computesthe value of the test ideal at the F -pure threshold. Theorem 3.3. If f is a L ∗ -linear combination of x d , · · · , x d n n , then τ ( fpt m ( f ) • f ) = ( f ) if fpt m ( f ) = 1 m if fpt m ( f ) = d + · · · + d n m if fpt m ( f ) < min n , P ni =1 1 d i o and p > max { d , · · · , d n } . Remark 3.4.
Note that τ ( fpt m ( f ) • f ) need not equal m if fpt m ( f ) < min n , P ni =1 1 d i o and p is less than or equal to some exponent [MY09, Proposition 4.2].3.3. On (higher) F -jumping numbers of Fermat hypersurfaces. Our final result com-putes higher jumping numbers of the degree d Fermat hypersurface. By Proposition 1.9, itsuffices to only consider those jumping numbers contained in (0 , f is the degree d Fermat hypersurface.
Corollary 3.5. If f is a L ∗ -linear combination of x d , · · · , x dd , then fpt m ( f ) = ( p l if p l ≤ d < p l +1 for some l ≥ − a − p if 0 < d < p and p ≡ a mod d with 1 ≤ a < d Theorem 3.6.
Suppose that p > d and write p = d · ω + a for some ω ≥ ≤ a < d . If f is a L ∗ -linear combination of x d , · · · , x d n n and a = 1, Corollary 3.5 implies that fpt m ( f ) = 1.We now assume a ≥ p < a ( d − fpt m ( f ) < ( d +1) · ω + ⌈ a/d ⌉ p ≤ F -jumping numbers in (0 , p > a ( d − fpt m ( f ) < F -jumping numbers in (0 , Remark 3.7. As a is strictly less than d , Theorem 3.6 implies that fpt m ( f ) and 1 are theonly jumping numbers of f in (0 ,
1] if p > ( d − . Example 3.8.
Suppose that d = 4, and p = 7. Then ω = 1, a = 3, and p < a ( d − d + 1) · ω + ⌈ a/d ⌉ = 5 + ⌈ / ⌉ = 7 = p . In this case, Theorem 3.6 provides no newinformation. Example 3.9.
Instead, let d = 6 and p = 11, so that ω = 1 , a = 5, and p < a ( d − d + 1) · ω + ⌈ a/d ⌉ = 7 + ⌈ / ⌉ = 9. Corollary 3.5 and Theorem 3.6 then imply fpt m ( f ) = 1 − a − p = , ( d +1) · ω + ⌈ a/d ⌉ p = , and 1 are F -jumping numbers of f contained in(0 , all of the F -jumping numbers of f in (0 , F -pure thresholds of diagonal hypersurfaces Notation 4.1.
Set δ i = d i , δ := ( δ , · · · , δ n ), and h δ i e := ( h δ i e , · · · , h δ n i e ). As in thestatement of Theorem 3.1, L = sup n N : δ ( e )1 + · · · + δ ( e ) n ≤ p − ≤ e ≤ N o . D will denote the diagonal matrix whose i th diagonal entry is d i , and we set ∆ := D − .Note that ∆ is also diagonal, with the i th diagonal entry being δ i . Throughout this chapter, e assume that f is a L ∗ -linear combination of x d , · · · , x d n n , and write f = u x d + · · · + u n x d n .Using multi-index notation,(3) f N = X | k | = N (cid:18) N k (cid:19) u k x D k . If λ ∈ R n , we use | λ | to denote the coordinate sum λ + · · · + λ n . When considering elementsof R n , (and ≺ ) will denote component-wise (strict) equality. Finally, { v , · · · , v n } denotesthe standard basis of R n , and n := (1 , · · · , Theorem 3.1: Part I. If L = ∞ , then fpt m ( f ) = | δ | . Proof.
Suppose that f ⌈ p e λ ⌉ / ∈ m [ p e ] . By (3), there exists k ∈ N with | k | = ⌈ p e λ ⌉ and D k ≺ ( p e − · n , so that k ≺ ( p e − · δ . Thus, p e λ ≤ ⌈ p e λ ⌉ = | k | < ( p e − | δ | , and so λ < | δ | . It follows from Lemma 1.11 that fpt m ( f ) ≤ | δ | .As L = ∞ , the entries of δ add without carrying (in base p ), and it follows that the entriesof p e h δ i e add without carrying for all e ≥
1. By Lemma 2.7, (cid:0) p e |h δ i e | p e h δ i e (cid:1) = 0 mod p , and as D h δ i e ≺ D δ = m , it follows that monomal x D h δ i e / ∈ m [ p e ] .Combining this with (3), shows that f p e |h δ i | / ∈ m [ p e ] , and Remark 1.2 then shows that f p d h δ i e / ∈ m [ p d ] for all d ≥ e . Lemma 1.11 shows that fpt m ( f ) ≥ | h δ i e | for all e , and theclaim follows by letting e → ∞ . (cid:3) Theorem 3.1: Part II. If L < ∞ , then fpt m ( f ) = | h δ i L | + p L . Proof.
The estimate for F -pure thresholds given in [Her11, Main Theorem] implies that(4) fpt m ( f ) ≥ h δ i L + 1 p L . If the inequality in (4) is strict, then Lemma 1.11 implies there exists e ≥ L such that(5) (cid:16) f p L |h δ i L | +1 (cid:17) p e − L = f p e |h δ i L | + p e − L / ∈ m [ p e ] . By Remark 1.2, it follows that f p L |h δ i L | +1 / ∈ m [ p L ]. Applying (3) shows there exists k ∈ N n such that | k | = p L | h δ i L | + 1 and x D k / ∈ m [ p L ]. This last condition implies that p L · k ≺ δ ,and applying Lemma 2.4 then shows p L · k h δ i L . Thus, | h δ i L | + p L = p L · k ≤ | h δ i L | , acontradiction. We conclude that equality holds in (4), and so we are done. (cid:3) Test ideals of diagonal hypersurfaces
We now prove Theorem 3.3 in three parts. As before, we assume f is a L ∗ -linear combi-nation of x d , · · · , x d n n : f = P ni =1 u i x d i i . We also continue to adopt Notation 4.1. Theorem 3.3: Part I. If fpt m ( f ) = 1, then τ ( fpt m ( f ) • f ) = ( f ). Proof.
Note that f p e = f p e ·
1, and that 1 ∈ B e . This, Γ e ( f p ) = f while Γ eµ ( f p ) = 0 for all1 = µ ∈ B e . It follows from this, and Lemma 1.7, that ( f ) = (cid:0) f p e (cid:1) [ pe ] = τ (1 • f ). (cid:3) o prove the remaining parts of Theorem 3.3, we will need Corollary 5.4 below. Lemma 5.1.
The natural number d i ( p e h δ i i e + 1 − p e δ i ) is less than d i . In particular, if d i ≤ p e , then D ( p e h δ i e + (1 − p e δ i ) · v i ) ≺ p e · n . Proof. As (cid:10) d (cid:11) e < d , it follows that d i ( p e h δ i i e + 1 − p e δ i ) < d i ( p e δ i + 1 − p e δ i ) = d i . (cid:3) In the proof of Lemma 5.2, we use • to denote the standard dot product on R n . Lemma 5.2.
Suppose that d i < p e and that d i is not a power of p . By Lemma 5.1, µ i := x D ( p e h δ i e +(1 − p e δ i ) · v i ) ∈ B e and Γ eµ i (cid:0) f p e |h δ i e | +1 (cid:1) = (cid:16)(cid:0) | p e h δ i e | +1 p e h δ i e + v i (cid:1) u p e h δ i e + v i (cid:17) /p e · x i . Proof.
To calculate Γ eµ i (cid:0) f p e |h δ i e | +1 (cid:1) , we must determine which (possibly) supporting mono-mials of f p e |h δ i e | +1 are R p e multiples of µ i . A monomial satisfying this condition is of form x D k for some k ∈ N with | k | = p e | h δ i e | +1 such that D k = p e a + D ( p e h δ i e + (1 − p e δ i ) · v i )for some vector a . Applying ∆ = D − then shows that(6) k = p e ∆ a + p e h δ i e + (1 − p e δ i ) · v i . If a i = 0, (6) shows that k i = p e h δ i i e + 1 − p e δ i , so that p e δ i ∈ N , which contradicts theassumption that d i is not a power of p . Thus, a i ≥
1. By summing the equation appearingin (6), we see that p e | h δ i e | + 1 = | k | = p e δ • a + p e | h δ i e | + 1 − p e δ i , and so(7) δ • ( a − v i ) = 0 . As a i ≥ a − v i < , and as the entries of δ are non-zero, follows from (6) that a = v i .Substituting this into (6) shows that the only (possibly) supporting monomial of f p e |h δ i e | +1 that is an R p e -multiple of µ i is x D ( p e h δ i e + v i ) = x D ( p e h δ i e +(1 − p e δ i ) v i ) · x D p e δ i = µ i · x p e i . (cid:3) Lemma 5.3. If n P i =1 δ ( e ) i ≤ p − (cid:0) | p e h δ i e | p e h δ i e (cid:1) = 0 mod p , then (cid:0) | p e h δ i e | +1 p e h δ i e + v i (cid:1) = 0 mod p . Proof.
Note that δ ( e ) i ≤ P ni =1 δ ( e ) i ≤ p −
2, which implies that both p e h δ i i e + 1 ≡ δ ( e ) i + 1and p e | h δ i e | + 1 ≡ P ni =1 δ ( e ) i are non-zero mod p . The claim by reducing the equality( p e h δ i i e + 1) · (cid:0) | p e h δ i e | +1 p e h δ i e +v i (cid:1) = (cid:0) | p e h δ i e | p e h δ i e (cid:1) · ( | p e h δ i e | + 1) mod p . (cid:3) Corollary 5.4.
Suppose that d i < p e and is not a power of p . If P ni =1 δ ( e ) i ≤ p − (cid:0) | p e h δ i e | p e h δ i e (cid:1) = 0 mod p , then x i ∈ (cid:0) f p e |h δ i e | +1 (cid:1) [ pe ]. Proof.
This follows immediately from Lemmas 5.2 and 5.3. (cid:3)
Theorem 3.3: Part II. If fpt m ( f ) = | δ | <
1, then τ ( fpt m ( f ) • f ) = m . Proof.
By Theorem 3.1, the entries of δ add without carrying (in base p ), so that no d i is a p th power (for else carrying would be necessary) and (cid:0) | p e h δ i e | p e h δ i e (cid:1) = 0 mod p for all e ≥ d i is a p th power and | δ | <
1, the denominator of | δ | is also not a p th power, and applying Remark 2.6 shows P ni =1 δ ( e ) i = | δ | ( e ) < p − e .Choose such an e so that additionally every d i is less than p e and τ ( | δ | • f ) = (cid:0) f ⌈ p e | δ |⌉ (cid:1) [ pe ] = (cid:0) f p e |h δ i e | +1 (cid:1) [ pe ] , here we have used Lemma 2.4 to obtain the equality ⌈ p e | δ |⌉ = p e h| δ |i e + 1 = p e | h δ i e | + 1.Applying Corollary 5.4 then shows ( x , · · · , x n ) ⊆ (cid:0) f p e |h δ i e | +1 (cid:1) [ pe ] = τ ( | δ | • f ). (cid:3) Theorem 3.3: Part III. If fpt m ( f ) < min { , | δ | } and p > max { d , · · · , d n } , then τ ( fpt m ( f ) • f ) = m . Proof.
Let L = max { N : P ni =1 δ ( e ) i ≤ p − ≤ e ≤ N } , and set λ := | h δ i L | + p L . Bydefinition, λ ( e ) = P ni =1 δ ( e ) i for 0 ≤ e ≤ L while λ ( e ) = p − e ≥ L + 1. As λ <
1, thereexists 1 ≤ l ≤ L such that λ ( l ) = P ni =1 δ ( l ) i ≤ p − λ ( e ) = p − e ≥ l . By Lemma2.7, our choice of l guarantees that (cid:0) p l |h δ i l | p l h δ i l (cid:1) = 0 mod p . As each d i is less than p , Corollary5.4 and Lemma 1.7 combine to show that ( x , · · · , x n ) ⊆ (cid:16) f p l |h δ i l | +1 (cid:17) h pl i = τ ( λ • f ). (cid:3) On (higher) F -jumping numbers of Fermat hypersurfaces Notation 6.1.
We now assume f is a L ∗ -linear combination of x d , · · · , x d n n , and write f = u x d + · · · + u d x dd . We continue to use δ to denote d . Remark 6.2.
Supposes p > d , and fix integers ω ≥ ≤ a < d such that p = d · ω + a .Isolating δ = d in this equation shows that δ = ωp + ad · p = ωp + ( aδ ) · p . From this, weconclude that δ (1) = ω and that δ ( e +1) = ( aδ ) ( e ) for all e ≥ Lemma 6.3.
Suppose that p > d > p ≡ a mod d . If a ≥
2, then ( d − · δ (2) ≥ p + 1. Proof.
Suppose, by means of contradiction, that ( d − · δ (2) ≤ p . As p is prime and both d − δ (2) are less than p , equality cannot hold. In particular,(8) ( d − · δ (2) ≤ p − . By Remark 6.2, we know ad = P e ≥ δ ( e +1) p e , and combining this observation with (8) shows( d − · ad = ( d − · δ (2) p + ( d − · ∞ X e =2 δ ( e +1) p e ≤ ( d − · δ (2) p + d − p (9) ≤ p − p + d − p = 1 + d − p . However, as a ≥
2, ( d − · ad ≥ ( d − · d = 1 + d − d , and comparing this with (9) shows d − d ≤ d − p , which implies that p ≤ d , a contradiction. (cid:3) Corollary 3.5.
We have the following formula for fpt m ( f ): fpt m ( f ) = ( p l if p l ≤ d < p l +1 for some l ≥ − a − p if 0 < d < p and p ≡ a mod d with 1 ≤ a < d Proof. If p l ≤ d < p l +1 , then p l +1 < δ ≤ p l . Consequently, δ ( e ) for 1 ≤ e ≤ l and δ ( l +1) = 0.Adding d copies of δ ( l +1) yields d · δ ( l +1) ≥ d ≥ p l ≥ p . In the notation of Theorem 3.1, wehave that L = l , and as h δ i l = 0, fpt m ( f ) = d · h δ i l + p l = p l . e now assume that p > d . If a = 1, the identities in Remark 6.2 imply δ ( e ) = ω for all e ≥
1. As d · δ ( e ) = d · ω = p −
1, it follows that d copies of δ add without carrying. ByTheorem 3.1, fpt m ( f ) = 1. Suppose now that a ≥ d > d · δ (1) = d · ω = p − a while d · δ (2) > p by Lemma 6.3. By Theorem 3.1, fpt m ( f ) = d · h δ i + p = d · ωp + p = p − a +1 p . (cid:3) In order to prove Theorem 3.6, we will need the following lemmas.
Lemma 6.4. If d is not a p th power, then x i ∈ (cid:0) f N (cid:1) [ pe ] if and only if 0 d · k − p e · v i ≺ p e · d and (cid:0) N k (cid:1) = 0 mod p for some k with | k | = N . Proof.
For every k ∈ N d , there is a unique element c k ∈ N d such that 0 d · k − p e · c k ≺ p e · d .If we set µ k := x d · k − p e · c k , it follows that µ k ∈ B e and that x d · k = x p e · c k µ k . Thus,(10) f N = X | k | = N (cid:18) Nk (cid:19) u k x d · k = X | k | = N (cid:18)(cid:18) N k (cid:19) u k (cid:19) /p e x c k ! p e µ k . Let I denote the ideal generated by the elements (cid:0)(cid:0) N k (cid:1) u k (cid:1) /p e x c k . Apparently, (10) showsthat f N is in I [ p e ] , and applying Lemma 1.4 then shows (cid:0) f N (cid:1) [ pe ] ⊆ I . If x i ∈ (cid:0) f N (cid:1) [ pe ],then x i ∈ I , and so x i must be a unit multiple of one of the monomial generators of I . Weconclude that x i = x c k for some k with (cid:0) N k (cid:1) = 0 mod p .Next, suppose that 0 d · k − p e · v i ≺ p e · d and (cid:0) N k (cid:1) = 0 mod p for some k with | k | = N ,so that x d · k = x p e i µ k is a supporting monomial of f N . To show that x i ∈ (cid:0) f N (cid:1) [ pe ], it sufficesto show that x d · k is the only supporting monomial of f N that is an R p e -multiple of µ k . Let x d · κ be another such monomial, so that x d · κ = x p e c µ k and x d · k = x p e i µ k . Solving for µ k inthese expressions shows µ k = x d · k − p e c = x d · κ − p e v i , and so(11) d · ( k − κ ) = p e · ( c − v i ) . As | k | = | κ | = N , it follows from (11) that | c | = | v i | = 1, so that c = v j for some j . If j = i , then (11) shows that d ( k j − κ j ) = p e , which contradicts the assumption that d is nota p th power. Thus, c = v i , and so k = κ by (11). (cid:3) Notation 6.5.
From now on, we assume p = d · ω + a for some ω ≥ ≤ a < d . Lemma 6.6.
We have the following inequalities:(1) p < d (2 ω + ⌈ aδ ⌉ − < p .(2) If p < a ( d − d + 1) · ω + ⌈ aδ ⌉ ≤ p .(3) If p > a ( d − p < d ( ω + a − < p . Proof.
The first point follows by applying the inequality 2 aδ ≤ ⌈ aδ ⌉ < aδ + 1 and theidentity p = d · ω + a . For the second point, note that p = d · ω + a < ad − a by hypothesis,and it follows that ω +2 aδ < a . Adding d · ω to both sides yields ( d +1) · ω +2 aδ < a + d · ω = p .The proof of the third point is similar, and is left to the reader. (cid:3) Theorem 3.6: Part I. If a ≥ p < a ( d − τ (cid:16)(cid:16) ( d +1) · ω + ⌈ aδ ⌉− p + p − p + · · · + p − p e (cid:17) • f (cid:17) = m for all e ≥
1, and τ (cid:16)(cid:16) ( d +1) · ω + ⌈ aδ ⌉ p (cid:17) • f (cid:17) = m .In particular, ( d +1) · ω + ⌈ aδ ⌉ p ∈ (0 ,
1] is an F -jumping number of f . Proof.
By Lemma 6.6, ( d +1) · ω + ⌈ aδ ⌉ p ∈ (0 ,
1] . By Lemma 6.3, ( d − · δ (2) ≥ p + 1. Thus, thereexists non-negative integers l , · · · , l d − such that P l i = p − l i ≤ δ (2) for 1 ≤ i ≤ d − i , which we are free to choose. In whatfollows, we assume that l d − < δ (2) . Fix e ≥
3, and set λ e := (cid:18) ωp + l p , · · · , ωp + l d − p , ωp + l d − p + p − p + · · · + p − p e , ω + ⌈ aδ ⌉ − p (cid:19) . By Remark 6.2, δ (1) = ω , and as l d − < δ (2) , the first d − λ e are less than orequal to h δ i . Set k := p e λ e . It follows that the first d − d · k are less than orequal to d · p e h δ i , and thus strictly less than p e while, by Lemma 6.6, the last entry of d · k is strictly between p e and 2 p e . Thus, d · k − p e · v d ≺ p e · d . By construction, the entriesof k add without carrying (in base p ), so (cid:0) | k | k (cid:1) = 0 mod p . Finally, | k | = p e · ( d + 1) · ω + ⌈ aδ ⌉ − p + d − X i =1 l i p + p − p + · · · + p − p e ! = p e · (cid:18) ( d + 1) · ω + ⌈ aδ ⌉ − p + p − p + p − p · · · + p − p e (cid:19) . We then apply Lemmas 6.4 and 1.7 to deduce that x d ∈ (cid:0) f | k | (cid:1) [ pe ] = τ (cid:18)(cid:18) ( d + 1) · ω + ⌈ aδ ⌉ − p + p − p + · · · + p − p e (cid:19) • f (cid:19) . As this argument is symmetric in the variables, the first claim follows.We now show that τ (cid:16) ( d +1) · ω + ⌈ aδ ⌉ p • f (cid:17) = m . By way of contradiction, suppose that x is in τ (cid:16) ( d +1) · ω + ⌈ aδ ⌉ p • f (cid:17) = (cid:0) f ( d +1) · ω + ⌈ aδ ⌉ (cid:1) [ p ]. By Lemma 6.4, there exists k ∈ N d with | k | = ( d + 1) · ω + ⌈ aδ ⌉ such that 0 d · k − p · v ≺ p · d . Restated, 0 p · k − δ · v ≺ δ · d ,and applying Lemma 2.4 shows p · k i ≤ h δ i = δ (1) p = ωp for all 2 ≤ i ≤ d . These sameinequalities also show p · k < δ , and summing these bounds shows( d + 1) · ω + ⌈ aδ ⌉ = | k | < ( d − · ω + 2 pδ. Gathering the multiples of ω and multiplying through by d implies 2 d · ω + d · ⌈ aδ ⌉ < p ,which this is impossible as 2 d · ω + d · ⌈ aδ ⌉ ≥ d · ω + d · aδ = 2( d · ω + a ) = 2 p . We concludethat x (and by symmetry, no variable) is in τ (cid:16) ( d +1) · ω + ⌈ aδ ⌉ p • f (cid:17) . (cid:3) Theorem 3.6: Part II. If a ≥ p > a ( d − τ (cid:16)(cid:16) p − p + · · · + p − p e (cid:17) • f (cid:17) = m for all e ≥
1. In particular, the only F -jumping numbers of f in (0 ,
1] are fpt m ( f ) and 1. Proof.
As in the proof of Theorem 3.6: Part I, Lemma 6.3 guarantees there exists non-negative integers l , · · · , l d − such that P d − i =1 l i = p − l i ≤ δ (2) for 1 ≤ i ≤ d −
1, with t least one inequality being strict. We again assume l d − < δ (2) . Fix e ≥
3, and let λ e := (cid:18) ωp + l p , · · · , ωp + l d − p , ωp + l d − p + p − p + · · · + p − p e , ω + a − p (cid:19) . Set k = p e · λ e . By construction, the entries of k add without carrying, so (cid:0) | k | k (cid:1) = 0 mod p .As in the proof of Part I of Theorem 3.6, one may verify that d · k − p e · v d ≺ p e · d , and | k | = p e · d · ω + a − p + d − X i =1 l i p + p − p + · · · + p − p e ! = p e · (cid:18) p − p + p − p + p − p · · · + p − p e (cid:19) . Once more, Lemmas 6.4 and 1.7 imply x d ∈ (cid:0) f | k | (cid:1) [ pe ] = τ (cid:16)(cid:16) p − p + p − p + · · · + p − p e (cid:17) • f (cid:17) .By the symmetry of this argument, we conclude that m ⊆ τ (cid:16)(cid:16) p − p + · · · + p − p e (cid:17) • f (cid:17) . (cid:3) References [BMS08] Manuel Blickle, Mircea Mustat¸ˇa, and Karen E. Smith. Discreteness and rationality of F -thresholds. Michigan Math. J. , 57:43–61, 2008. Special volume in honor of Melvin Hochster.2[BMS09] Manuel Blickle, Mircea Mustat¸˘a, and Karen E. Smith. F -thresholds of hypersurfaces. Trans.Amer. Math. Soc. , 361(12):6549–6565, 2009. 2[Dic02] L.E. Dickson. Theorems on the residues of multinomial coefficients with respect to a prime mod-ulus.
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