Pr obl em en

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Problemen NAW 5/18 nr. 2 juni 2017

149

Pr obl em en

Redactie:

Gabriele Dalla Torre Christophe Debry Jinbi Jin Marco Streng Wouter Zomervrucht Problemenrubriek NAW Mathematisch Instituut Universiteit Leiden Postbus 9512 2300 RA Leiden

problems@nieuwarchief.nl www.nieuwarchief.nl/problems

| Problem Section

This Problem Section is open to everyone; everybody is encouraged to send in solutions and propose problems. Group contributions are welcome. For each problem, the most ele- gant correct solution will be rewarded with a book token worth € 20. (To compete for the book token you should have a postal address in The Netherlands.)

Please send your submission by e-mail (LaTeX is preferred), including your name and address to problems@nieuwarchief.nl.

The deadline for solutions to the problems in this edition is 1 September 2017.

Problem A

Show that there are no infinite antichains for the partial order # on N^k defined by _^x x1 2, ,f,xk_h

, , , y y1 2 fyk

#_^ _h iff xi#yi for all i, 1 # # .i k

Problem B

A sector is a portion of a disk enclosed by two radii and an arc. For each pair of radii, there are two complementary sectors. If the two complementary sectors have unequal area, then we say that the larger sector is the major sector.

Let S be a subset of the plane. We say that x!S is virtually isolated if x is the only element of S in a major sector of a disk centered on x. Suppose that all elements of S are virtually isolated. Prove that S is countable.

Problem C (proposed by Hendrik Lenstra)

Let n be a natural number 12 . Suppose that for every prime p1 we have that pn ⁿ/ (p-1)ⁿ+1modn². Prove that n= .2

Edition 2016-4 We received solutions from Pieter de Groen, Alex Heinis and Toshihiro Shimizu.

Problem 2016-4/A (folklore)

For a finite sequence s=( ,s₁f, )s_n of positive integers, denote by ( )p s the number of ways to write s as a sum s=

/

_iⁿ₌₁a e_{i i}+

/

ⁿ_j₌^-₁¹b e_j( _j+e_j₊₁) with all a_i and b_j non-negative. Here e_i denotes the sequence of which the i-th term is 1 and of which all the other terms are 0.

Show that there exists an integer B> such that for any product F of (positive) Fibonacci 1 numbers, there exists a finite sequence s=( ,s₁f, )s_n with all s_i!{ , ,1 2f, }B such that

( ) p s = .F

Solution We received solutions from Pieter de Groen, Alex Heinis and Toshihiro Shimizu.

The book token goes to Pieter de Groen. The following solution is based on that of Alex Heinis.

Let s=( ,s₁f, )s_m be a finite sequence of positive integers. We say that s is left safe if s₁$s₂, right safe if s_m$s_{m 1}_- , and safe if they are both left safe and right safe. Let

( , , ),

s= s₁gs_m t=( ,t₁g, )t_n be finite sequences of integers with ,m n$1. We define the join s t0 of s and t as the sequence ( , ,s₁gs_m_-₁,s_m+t t_{1 2}, ,g, )t_n. Note that taking the join of sequences is associative.

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NAW 5/18 nr. 2 juni 2017 Problemen

Opl ossin gen

| Solutions Moreover, if ( , , )s s s_m₁f= is a finite sequence of positive integers, define

( ) ( , ) : ( ) ,

V s a b Zⁿ Zⁿ s a e_{i i} b e e

i n

j j j

j n

0 01

1 1

#

= ! $ $- = + +

= +

=

*

/ /

- 4

so that # ( )V s =p s( ), and define

( ) : ( )

W s b Zⁿ s b e_j _j e_j

j n 01

1 1 1

! $

= _$^- + ₊

=

*

/

- 4

(where we take the usual partial order on the set of finite sequences of some fixed

length). Then we have a bijection ( )V s "W s( ) given by ( , )a b 7b, with inverse

( ),

b s b ej j ej b

j n

1 1

7 - 1 + +

=

f

/

- p

, so also

# ( )W s =p s( ). We will then show the following.

Lemma. Let s be a right safe sequence, and let t be a left safe sequence. Then (p s t0 )= ( ) ( )

p s p t .

Proof. Write s=( ,s₁f, )s_m and t=( ,t₁f, )t_n. Since for any b!W s( )and c!W t( ), we have s_m+t₁$b_m_-₁+c₁, concatenation of sequences defines a map ( )W s #W t( )"W s t( 0 ). Sin- ce s is right safe, for any d!W s t( 0 ), we have s_m$s_m_-₁$d_m_-₂+d_m_-₁$d_m_-₁, so we have a map (W s t0 )"W s( ) sending ( )d_{i i}^{m n}₌^{+ -}₁ ² to ( )d_{i i}^m₌^-₁¹. By a similar argument, since t is left safe, we have a map (W s t0 )"W t( ) sending ( )d_{i i}^{m n}₌^{+ -}₁ ² to (d_m_{- +}₁ _{i i})ⁿ₌^-₁¹, and therefore also a map (W s t0 )"W s( )#W t( ), which by construction is the inverse of the concatena-

tion map. Hence (p s t0 )=p s p t( ) ( ).

□

Now we return to the problem. Let F_n be the n-th Fibonacci number, where we take as convention F₀=F₁= . Let F1 _iⁿ ₁F_m

=

%

₌ i be a product of Fibonacci numbers. We may as- sume that each F_m_i is at least 2, that F₂ occurs at most twice (since F₅= ), and therefore 8 that if m_i= , then i2 = or i1 = .n

Let s_i denote the sequence of m_i ones; it is well-known then that ( )p s_i =F_m_i for all i.

Then note that s₁ is right safe, that all s_i with 2# #i n- are of length at least 3 1 and safe, and that s_n is left safe. Moreover, as s_i has length 3 if 2# #i n- , we see 1 that s₁0g0s_i is right safe if 2# #i n- . Hence inductively applying the lemma gives 1

( ) ( )

p s₁0g0s_n =

%

_iⁿ₌₁p s_i =F, as required.

Problem 2016-4/B (folklore)

Let , be a prime number. For any group homomorphism :f A"B between abelian groups and for any integer n$0, denote by fn the induced homomorphism /A ,ⁿA"B/,ⁿB. Let ( )kn n 03= and ( )cn n 0³= be sequences of integers.

Show that there exist integers , ,N a b$0 and a group homomorphism :( /f Z ,^NZ)^a"( /Z ,^NZ)^b such that for all n$0 we have # ker fn=,k_n and # coker fn=,c_n if and only if k0=c0= 0 and the sequences (kn+1-kn n)³=0 and (cn+1-cn n)³=0 are non-negative, non-increasing, eventually zero, and there is a constant C such that for all n such that kn+1-kn and cn+1-cn are not both zero, their difference is C.

(Recall that the cokernel coker f of a group homomorphism :f A"B between abelian groups is the quotient of B by the image of f.)

Solution We received a solution from Alex Heinis, who is also rewarded the book token for this problem. The second part of the following solution is similar to that of Alex Heinis.

Write A=( /Z ,^NZ)^a and B=( /Z ,^NZ)^b, and write An=A/,nA and Bn=B/,nB. Moreover, for all n$0, let kn=log^(#logker_,fn⁾. We show they satisfy the required properties.

For all n$0, let in=log^(#logim_,fn⁾. Note that the quotient map B"Bn sends im f to im fn, and the induced map imf"imfn is surjective with kernel im f+ ,ⁿB. By the structure theo- rem for finitely generated abelian groups (and as im f is ,^N-torsion), there exist unique

, ,

t1ftN$0 such that im f,

+

_i^N₌₁( /Z ,ⁱZ)^tⁱ. Moreover, we have ,ⁿB=B[,^{N n}^- ] if n#N and B,ⁿ = otherwise, so 0 imf+ ,ⁿB=(imf)[,^{N n}^- ] if n#N and im f+ ,ⁿB= otherwise. 0

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Problemen NAW 5/18 nr. 2 juni 2017

151

Opl ossin gen

| Solutions We deduce that

( , )

( , ) ( ) ,

min

max

i jt j N n t

j n N0 t j n N t

n j

j N

j j

j j N

j N n j

1 1

= - -

= + - = + -

= =

= = - +

/ /

and that in j jtj N

=

/

₌1 otherwise.

Therefore, for all n$N we have in+1-in=0. Moreover, note that in in j N ntj N 1- =

+

/

= -

if n< , so that (N in+1-in) (- in-in-1)=tN n- for all 1#n<N.

First note that k0=c0= , and ,0 k cn n$0 for all n$0 (as , to that power is the order of a group). Now we use the isomorphisms An/ kerfn"imfn and Bn/ imfn"cokerfn to see that we have kn=min( , )n N a i- and n cn=min( , )n N b i- for all n. Therefore:n

– If 1#n<N, then

–

( ) ( ) ( ) ( )

( ) ( )

k k k k c c c c

i i i i t 0

n n n n n n n n

n n n n N n

1 1 1 1

1 1 #

- - - = - - -

= - - - = -

+ - + -

- + -

If n$N, then k_n₊₁-k_n=c_n₊₁-c_n=0;

– Moreover, k_N k_N a i_N i_N a ^N_j t

1 1 1 j

- _- = - + _- = -

/

₌ ^{, and c}N ^cN ^b j ^tj

N

1 1

- _- = -

/

₌

As j tj N

1

/

= is the minimum number of generators of im f and A maps surjectively to im f, it follows that kN-kN 1- $0. As j tj

N 1

/

= is the dimension over F_, of (im f , and )[ ] (im f)[ ], 3B[ ],, it follows that cN-cN 1- $0. Therefore we see that (kn+1-kn n)³=0 and (cn+1-cn n)³=0 are non-increasing sequences that are eventually zero, hence they are

non-negative as well.

Finally, note that for n< , we have (N kn+1-kn) (- cn+1-cn)= -a b, which shows that the sequences ( )kn n 03

= and ( )cn n 03

= satisfy the required conditions.

For the converse, suppose that ( )kn n 03

= and ( )cn n 03

= satisfy the conditions in the pro- blem. Let N be the smallest integer n$0 for which kn+1-kn=cn+1-cn=0. Let for

n N

0# < , rn=(kn-kn-1) (- kn+1-kn)=(cn-cn-1) (- cn+1-cn), which is non-negative as the sequences (kn+1-kn n)³=0 and (cn+1-cn n)³=0 are non-increasing. Moreover, let sk=kN-kN 1- and let sc=cN-cN 1- . Consider the map

( / ) ( / ) ( / ) ( / )

f Z ^NZ ^s Z ^NZ ^r Z Z Z Z

n

N N s N r

n N

1 1

k n" c n

| , 5 , , 5 ,

= -

=

+ +

-

of which the matrix in block form (with respect to the given splitting into summands) is the diagonal matrix with diagonal (0_s_c#_s_k,,$I_r₁,,²$I_r₂,f ,, ^N-¹$I_r_N ₁)

- .

A couple of observations: first note that if n$N, then f_n= . Moreover, we note that for f n< , the induced mapN

( / ) ( / ) ( / ) ( / )

f_n Z ⁿZ ^s Z ⁿZ ^r Z Z Z Z

i

N n s n r

i N

1 1

k i" c i

| , 5 , , 5 ,

= -

=

+ +

-

is given by the same matrix as the one defining f. And finally, we note that therefore

( )

kerfn ker0s s i ker I

N i

1 r 1

c k5 , $ i

= # =

+

- ^.

Let us show by induction on n that # ker f_n=,^kⁿ and # coker f_n=,^cⁿ for all n. First of all, for n= , note that f0 ₀ is the map 0"0, so k0=c0= . Now let 0 0 <n#N, and suppose that # ker f_n_-₁=,^k^{n 1}^- and # coker f_n_-₁=,^c^{n 1}^- . Note that # ker f_n=,^ns^k⁺/iN 1=-1^min^{( , )}^{i n r}i and that # ker f_n_-₁=,⁽ⁿ^-¹⁾^s^k⁺ i=1^min^{( ,}^{i n}^-¹⁾^ri

N 1-

/ . Therefore #kerfn/#kerfn s r

1 ^k ^{i n}

N i

, 1

- = +/₌^- ; by de- finition of s_k and the r_i, the exponent is by a telescoping sum equal to kn-kn 1_- . Hence

#kerfn k k #kerf

n k

n n 1 1 n

, ,

= ^- ^- _- = . Similarly, we show that # coker f_n=,^cⁿ.

Finally, we note that for n> , we have #N kerfn=#kerfN=,k_N=,k_n, as for all i> , we N have ki₊1-ki=0; and similarly, we have # coker f_n=,^cⁿ.

Problem 2016-4/C (folklore)

Let R be the polynomial ring over Z with variables x_i, y_i, z_i for all i!Z. Let S be the polynomial ring over Z with variables t_i for all i!Z. Let : Rx "R be the isomorphism of rings given by x_i7x_{i 1}₊ , y_i7y_{i 1}₊ and z_i7z_{i 1}₊ .

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NAW 5/18 nr. 2 juni 2017 Problemen

Opl ossin gen

| Solutions Consider the morphism :f R S" of rings given by x t t t_i _i _{i i}₁ ₁7 _- ₊ , y t_i _i³7 and z t_i _i²7 . Does

there exist a finite number of elements , ,r₁fr_n!R such that the kernel I of f is generated as an ideal in R by {xⁱr i_j: !Z,j=1,f, }n?

Solution We received no solutions for this problem. The answer is no.

We assume monomials to be monic. Moreover, we will abuse notation by setting x|S"S to be the automorphism of rings sending t_i to t_i+1 for all i!Z. Let M3R denote the set of monomials of R, and let N3S denote the set of monomials of S. Then f maps M into N. Moreover, identifying (as groups) R with

+

_{m M}_! Z$m and S with

+

_{n N}_! Z$n^{, we see} that f R| "S is given by

,

x m x n

( ) m

m M m

m f n

n N ¹

7

! ! f ! - p

/ / /

for xm zero for all but finitely many m!M. Finally, we note that for all x!R, we have

( ) ( )

f xx =xf x.

Let, for x=

/

m M_! x mm !R and n!N, the n-th homogeneous part be xn= m f ¹_{( )}n x mm

! ^-

/

.

Then note that x!I, if and only if for all n!N, we have xn!I, and x=

/

n N_! xn, where we note that xn is zero for all but finitely many n!N, as the same is true for the xm for m!M. So let us say that an element of the form _{m f} ₁_{( ) m}_n x m

! ^-

/

in R is homogeneous

with respect to n.

Write R_n=

+

_{m f}! ^-¹_{( )}_nZ$m. Then note that |f R_n is injective if and only if #f^-¹( )n #1. So consider the set 'N 3N of n!N such that #f^-¹( )n $2, where we take the partial order on N given by divisibility by an element of f(M), i.e. n#n’ if there exists an m!M such that ( )nf m =n'.

Define, for integers k$1 and i, the monomial nk i, t ti i3 ti ti k ti k 1 2g 3 1 3 3

= + + + - + . We deter-

mine f-1(nk i,). Let m!f^-1(nk i,). Since n only has two exponents of at least 2, we see that the total y,z-degree of m is at most 2. Moreover, the degree of n is k3 + , so com-5 paring degrees modulo 3, we see that m must be divisible by either y zi i+3k or yi+3k iz. In the first case, write m=m y z' i i+3k, and note that ( )f m' =ti+1gti+3k. By induction on k, one can show that m’ must be equal to xi₊2xi₊5gxi₊3k_-1, therefore m must be mk i, ,1=xi+2xi+5gxi+3k-1y zi i+3k. In the same way, one shows in the second case that m must be equal to mk i, ,2=xi+1xi+4gxi+3k-2yi+3k iz. Since both of them do map to nk,i, we find that f 1(nk i,) {mk i, , ,mk i, , }

1 2

- = . In particular, nk i, !N', and ker f|R_n_{k i}_, is generated by mk i, ,1-mk i, ,2.

Moreover, nk,i is minimal in N’; if n!N' divides nk,i in f(M), then for ,m m m1 2, '!M such that m1!m2, ( )f m1 =f m( 2)= and ( ) ( )n f m f m1 ' =f m f m( 2) ( )' =nk i,, we have {m m m m1 ', 2 '}={mk i, ,1,mk i, ,2}. Since gcd m( k i, ,1,mk i, ,2)= , it follows that 1 m’= and there-1 fore that n=nk i,.

Now we can prove our answer. Let G be any subset of R-{ }0 such that I is generated by {x^ag a: !Z,g!G}. We show that G is infinite. Let G’ denote the set of non-zero homo- geneous parts of elements of G, and note that G is finite if and only if G’ is. Then I is also generated by {x^ag a': !Z,g'!G'}.

Note that for all k$1, we have mk, ,0 1-mk, ,0 2!I, and write mk, , mk, , ,' a g, ' ag'

0 1- 0 2=

/

a ga x , for some aa b, !R that are zero for all but finitely many pairs (a,b); we may assume that the a_a,b are homogeneous, by taking the nk,o-th homogeneous part of this identity if necessary.

If (a,g’) is such that aa g, '!0, and if g’ is homogeneous with respect to n, then x^an!N’ and na n,

# k 0

x , so by minimality of nk,0 it follows that na n,

x = k 0. Moreover, ker f|R_n_{k 0}_, is generated by mk, ,0 1-mk, ,0 2, so x^ag' is a non-zero integer multiple of mk i, ,1-mk i, ,2. It fol- lows that for any k$1 there exists ’gk!G’ such that g’k is a non-zero integer multiple of

(m, , m, , )

a k0 1 k0 2

x - for some a!Z. Now x preserves total degrees, so the total degree of g’k

is that of mk,0,1 and mk,0,2, which is k 2+ . In particular, G’ contains elements of every total degree that is at least 3, so G’ (and therefore G) is infinite, as desired.

Rectification

In the list of solvers of 2016-3/B, Rob van der Waall was missing, but should have been there. We apologise for this omission.