Algebraic games - playing with groups and rings

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Algebraic games playing with

groups and rings

Martin Brandenburg

December 15, 2014

Abstract

Two players alternate moves in the following combinatorial game:

Given a finitely generated abelian group A, a move consists of pickingsome 0= a A. The game then continues with the quotient groupA/a. We prove that under the normal play rule the second playerhas a winning strategy if and only ifA is a square, i.e. A = B2 forsome abelian groupB . Under the misere play rule only minor modifi-cations concerning elementary abelian groups are necessary. We also

compute the nimber of 2-generated abelianp-groups. In principle, thegame can be played with arbitrary algebraic structures. We study

some examples of non-abelian groups and commutative rings.

1 Introduction

Consider the following two-person game: Given a finitely generated abeliangroup A, a move consists of picking some 0=a A and replacing A by thequotient group A/a. Under the normal/misere play rule, the player withthe last possible move wins/loses: WhenA = 0, the next player cannot moveand therefore wins under the misere play rule and loses under the normal

play rule. For which A does the first player has a winning strategy, i.e. Ais anN-position? And when is it a P-position, i.e. the second player has awinning strategy? This question can be asked both for the normal as well asfor the misere play rule.

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More generally, if A is some algebraic structure, i.e. a (non-empty) set

equipped with various functions with various arities, we can play the fol-lowing game: A move identifies two non-equal elements a = b of A, i.e.replacesA by the algebraic structure of the same type A/(ab), which isdefined to be A/, where is the congruence relation on A generated by(a, b). The game ends whenA has only one element.

In generality we cannot expect any non-trivial analysis of this game. How-ever, this comes into reach if we consider specific types of algebraic structures,for example sets, vector spaces, (abelian) groups and rings. We analyze thegame in these cases. This was inspired by the game on rings as proposed byWill Sawin onmathoverflow.net([M]). These games might be calledalgebraic

games in contrast to the well-studied topological games.For example, we may start with a group G. A move consists of replacingGbyG/a, where 1=a G and a denotes the normal subgroup generatedbya. The game can be also seen as a sequence of elements a1, a2, . . . such thatai+1 / a1, . . . , ai. This is similar to the game considered in [AH]with theweaker conditionai+1 / {a1, . . . , ai}. When we consider some abelian groupA, every subgroup is normal, so that a move replacesA byA/a. Our maintheorem states:

Theorem 1.1. LetA be a finitely generated abelian group.

Under the normal play rule, A is a P-position if and only if A is asquare, i.e. A=B2 for some abelian group B.

Under the misere play rule, A is aP-position if and only ifA is

either a square, but not elementary abelian of even dimension

or elementary abelian of odd dimension.

Actually our proof will include a winning strategy. For example, Z/4 Z/8is a normal N-position: Player I mods out 0 4 to obtain Z/4 Z/4. Nomatter what Player II does, he will produce something isomorphic to Z/4 orZ/2Z/4. In the first case Player I mods out the generator ofZ/4 and wins.In the second case Player I mods out 0 2, so that Player II gets Z/2 Z/2.He can only react with something isomorphic to Z/2, whose generator PlayerI mods out and therefore wins. It is also a misere N-position: From Z/4Player I mods out 2 and from Z/2 Z/4 he mods out 0 1. In each case

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Player II has to play with Z/2 and does the last move, so that he loses under

the misere play rule.We also compute the nimbers of some finitely generated abelian groups.

Proposition 1.2. Letp be a prime number. Letn m be natural numbersandk := m n. The nimber of the game with the abelian group Z/pn Z/pm

equals n+m n k

k+ (n k 1 modk+ 1) n >k

We gather some results on the game of rings:

Proposition 1.3. IfR is PID which is not a field, thenR isN. Besides,R[x]/(x2) isP and thereforeR[x] isN.

Acknowledgements. For various discussions and suggestions I would like tothank Will Sawin and Kevin Buzzard. Special thanks goes to Diego Monterowho corrected some errors in a preliminary version and simplified the proof ofProposition3.3. Finally I would like to thank Jyrki Lahtonen for suggestingthe formula in Proposition1.2.

2 The game in general

2.1 Basics in combinatorial game theory

In this subsection we recall briefly some basic notions in combinatorial gametheory. We only sketch them and refer the reader to textbooks such as [F]or [MNW] for details.

We consider impartial two-person combinatorial games. This means thatPlayer I (who starts) and Player II alternate making moves, each havingthe same set of options for a given position in the game. No chance or

probabilty is involved, the game is purely combinatorial. Every game has aset of terminal positions, in which it ends. We require the ending condition,which asserts that the game has to end after some finite number of moves.However, we allow infinitely many positions. Formally, a game is just a (well-

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founded) set, the positions being the elements of this set, which are games

again.The first player who cannot move loses under the normal play rule. He winsunder themisere play rule. Thus, under the normal play rule one wants to bethe last one to move, whereas under the misere play rule one actually wantsto prevent this. Often misere games are more complicated than normal ones.The reader who is only interested in the game on algebraic structures underthe normal play rule can safely ignore every remark about misere games(however, see Example2.6).

We call a position in the game anN-positionif the next player to move hasa winning strategy. If the previous player has a winning strategy, we call

it a P-position. This definition applies to both play rules. Every terminalposition is clearly a normal P-position and a misereN-position. One of thefirst basic observations in combinatorial game theory is the following: Underboth play rules, every position is either an N-position or a P-position. Infact, we can declare a position to beN orPby recursion as follows:

1. Every terminal position is a normalP-position / misere N-position.

2. A non-terminal position is normal/misereN, whensomemove from ityields a normal/misere P-position.

3. A position is normal/misere P, wheneverymove yields a normal/misere

N-position.

Intuitively, 1. declares the play rule, 2. asserts the existence of a winningmove forN-positions, and 3. denies it for P-positions. The ending conditioneasily implies:

Proposition 2.1. Under either play rule, the sets of P- and N-positionsare characterized by the above three properties.

Example 2.2. Consider the game Nim with just two piles: We have twopiles of counters. A move consists in reducing the number of counters in oneof the piles. Under the normal play rule, (x, y) is a P-position if and only ifx= y, i.e. it is a square. In fact, 1. the terminal position (0 , 0) is a square,2. every non-square can be moved to some square, and 3. squares cannotmove to squares. Under the misere play rule, theP-positions arealmost thesame: (0, 0), (1, 1) are misere Nand (1, 0), (0, 1) are misereP, but the rest is

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as before. We have mentioned this example since the game on abelian groups

will be quite similar.Remark 2.3(Sprague-Grundy Theorem). Every combinatorial gameG un-der the normal play rule is equivalent to a Nim-pile with just one pile of size(G), the nimber of G. This is an ordinal number which may be definedrecursively by

(G) = mex{(H) :H is a move in G}.

Here, mex(S) denotes the smallest ordinal number not contained in S. Ob-serve that (G) = 0 if and only if G a P-position. Otherwise, we have(G) > 0. The nimber ofG carries much more information than just the

knowledge about which player wins. It is important to know this nimberwhenG is played in a sum of games.

2.2 The game on algebraic structures

Now let us introduce the game on algebraic structures:

Definition 2.4. Given some algebraic structureA ([BS,II,1]), i.e. a (non-empty) set equipped with various functions with various arities, a playerchooses two non-equal elements a=b in A and replaces A by the algebraic

structure of the same type A/(a b), which is defined to be A/, whereis the congruence relation on A generated by (a, b) ([BS, II, 5]). The gameends when A has only one element. One might call A/(a b) a principalquotient ofA.

Example 2.5. Given a group G, a move consists of replacing G by G/afor some element 1 =a G, where a is the normal subgroup generatedby a. We only need one element since identifying a with b is the sameas identifying ab1 with 1. If we play with an abelian group A (writtenadditively), we replace A by A/a for some 0 = a A. More generally,given an R-module Mover a commutative ring R, a move replaces M by

M/(m) for some 0=m M.

Example 2.6. Given a commutative ring R, a move consists of replacingR by R/(a) for some element 0 = a R, where (a) is the principal idealgenerated by a. In this game every ring = 0 is a normal N-position since

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only ifB is.

We will often use this result. The following two examples illustrate that thegame is easy to understand when some dimension or size classifies the wholestructure:

Example 2.11. We play with a set without any additional structure.The ending condition holds if and only if the set is finite. Since only theisomorphism type matters, we only have to look at the cardinality. Everymove reduces the cardinality exactly by one. The terminal sets are those withexactly one element. By induction it follows that a finite set is a normalP-position if and only if its cardinality is odd. Otherwise, it is a normalN-position. The misere positions are vice versa.

Example 2.12. We play with a vector space V over a fixed field. Theending condition holds if and only if V is finite-dimensional. Everythingonly depends on the dimension ofV. Every move reduces it by one. Theterminal vector spaces are those of dimension zero. By induction it followsthatV is a normalP-position if and only if its dimension is even. Otherwiseit is a normalN-position. The misere positions are vice versa.

2.3 Selective compound games

We review the concept of selective compound games. The general theory isdue to Smith ([S], Sect. 7,8). Our application to the game on algebraic struc-tures (Corollary2.16) will become useful later when dealing with productsof structures.

IfG1, . . . , Gn are games (as above, we only consider impartial ones), we canplay a new game G1 . . . Gn, called theselective compound ofG1, . . . , Gn.A position in that game is a tuple of positions in the games Gi. A moveconsists of picking a non-empty subset ofG1, . . . , Gn and making a move ineach of the chosen games. IfGi is already over, i.e. happens to be a terminalposition, then of course we continue with G

1 . . . Gi . . . Gn (with Giremoved).

Proposition 2.13. The selective compound G1 . . . Gn is a normalP-position if and only if everyGi is a normalP-position.

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Proof. It is clear that the terminal P-position satisfies this. If every Gi is

a normal P-position, then G1 . . . Gn can only move to G

1 . . . G

n,where either Gi =Gi is normal P or Gi is a move in Gi, which is therefore

normalN. And the latter happens for at least onei. Thus, it doesnt consistentirely of normal P-positions. If, on the other hand,G1 . . . Gn is suchsome Gi are normal N, the winning move is to choose some move Gi therewhich makes it normal P. IfGi is already normal P, we just let G

i = Gi.

Thus, we can move G1 . . . Gn toG1 . . . Gn where each G

i is normal

P.

Proposition 2.14. The selective compound G1 . . . Gn is a misere P-position if and only if either

all games except one, say Gi, are over (i.e. terminal), and Gi is amisereP-position,

at least two of the games are not over yet, and each Gi is a normalP-position.

The reader is invited to read the example2.18first as a piece of motivation.

Proof. Let us call G1 . . . Gn a P-position if it satisfies the condition in

the proposition, i.e. every Gi is normal Pwhen at least two are not overyet, or only one Gi is still playing and misere P. We have to prove that

P

satisfies the defining properties ofP in Proposition2.1. First of all, theterminal position is not P. Next, every non-terminal position which is notP has some (winning) move to a position which is P: If all games exceptfor Gi are over, then we continue to play only with Gi, which is misere Nand therefore has some move to a miserePposition, which is therefore P.If at least two games are not over yet, then some of the games is normal N.Now move in every one of these normal Ngames to some normalP-positionand leave the normalPgames untouched. We obtain the game G1 . . . G

n

where each Gi is normal P. If at least two Gi are not over yet, this is P

and we are done. Otherwise, everyGi which is normal N can be ended inone move and the other ones are already over. Now we choose the following

winning move: Pick some Gj which is normal N. If it is misere P, end allthe other Gi and keep Gj. If it is misere N, choose some move Gj Gjsuch thatGj is misere P, and combine this move with ending all other Gi. Ineach case, we arrive at a single active game which is miserePand therefore

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P.

Finally, we have to prove that a P position cannot move to a P position.This is clear when only one game is active. When at least two games areactive, then every activeGiis normal Pand therefore cannot be ended in onemove, but rather can only be moved to some normal N position Gi. Thus,after every move in the selective compound we still have at least two activegames, but one of them must be normal Nand therefore cannot be P.

Proposition 2.15. Given some type of an algebraic structure, assume thatfor allA1, . . . , An and alla, b A := A1 . . . An the canonical homomor-phism

A/(a b) A1/(a1 b1) . . . An/(an bn)

is an isomorphism. Then, for al lA1, . . . , An, the game onA1 . . . An isthe selective compound of the games on theA1, . . . , An.

More generally, assume that there are classes of algebras T1, . . . , Tn, eachbeing stable under quotients and containing the terminal algebra. Assumethat for allAi Tianda, b A := A1. . .Anthe canonical homomorphism

A/(a b) A1/(a1 b1) . . . An/(an bn)

is an isomorphism. Then, the game onA is the selective compound game ofthe games on theA1, . . . , An.

Proof. This follows from the definitions. The requirement a = b in thedefinition of a move means that ai =bi for at least one i, i.e. that we movein at least one factor.

Corollary 2.16. In the situation of Proposition2.15, we have that the prod-uctA= A1 . . . An is

normalP if and only if everyAi is normalP

miserePif and only if all factors except one, sayAi, are terminal, andAi is misereP, or at least two factors are non-terminal, and everyAi

is normalP.Example 2.17. LetR1, . . . , Rnbe commutative rings and letRdenote theirproduct. Then for every a R the induced homomorphism

R/(a) R1/(a1) . . . Rn/(an)

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relation | to the simple relation of. Writeni =pki with ki 0. Then we

claim that there are natural numbers mi 0 such that m1 k1 m2 k2 m3 . . . such that A/x is isomorphic to Z/pm1 . . . Z/pms. Nowconsider xi Z/pki and lift it to some natural number, also denoted by xi.We may write xi = p

riui for some unique 0 ri ki and ui with p ui.Since multiplication withuiinduces an automorphism ofZ/p

ki, we may evenassume that xi = p

ri.

Next, we give a recursive description of the quotient

Ak,r := (Z/pk1 . . . Z/pks)/(pr1 , . . . , prs).

This can be also written as the abelian group defined by generatorse1, . . . , es,

relationspkiei = 0 for 1 i s, as well as the relation

pr1e1+. . .+prses= 0.

Choose 1 l s in such a way that rl becomes minimal. If we define

el :=i

prirlei= el+i=l

prirlei,

the above relations becomes prlel = 0. In terms ofel, the relation p

klel = 0becomes

pkl

e

l =i=l

pkl+rirl

ei.

The left hand side vanishes because ofprlel = 0 and rl kl. Thus, we cansplit offel

= Z/prl. Also, sincepkiei= 0, we could equally well replace thecoefficient ofei byp

ri, where

ri:= min(kl+ri rl, ki).

For i < l we have ri =ki, so that we may split offei= Z/pki. Thus, if we

defineki = ki for i > l, we obtain the recursive expression

Ak,r = Z/prl Z/pk1 . . . Z/pkl1 Ak,r.

Let us add to the induction hypothesis that rl is the smallest exponent inthe decomposition, i.e. rl = m1. Applying the induction hypothesis toAk,r we get numbers ml+1 kl+1 ml+2 . . . ks such that Ak,r =

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Z/pml+1 . . . Z/pms. Besides, ml+1 is the minimum of the ri, which is

kl. Now let us define m1 = rl and mi = ki1 for 1 < i l. ThenAk,r = Z/pm1 . . . Z/pms and we have

m1 k1= m2 k2= m3 . . . kl1= mlklml+1kl+1. . . ks,

as required.

Proposition 3.4. The game on finite abelian groups is equivalent to thefollowing number-theoretic game: Positions are divisor sequences n1| . . . |nsof natural numbers 1, where those ni = 1 may be removed. There is amove fromn1| . . . |ns tom1| . . . |ms if and only ifm1|n1|m2|n2| . . . |ms|ns and

for at least one1 i s we havemi < ni. The only terminal position isthe empty sequence.

Proof. This follows from the Proposition3.3.

Proposition 3.5. In the number-theoretic game described above, n1| . . . |nsis a normal P-position if and only if it is a square in the following sense:Either s is even and n1 = n2, n3 = n4, . . . , ns1 = ns, or s is odd andn1= 1, n2= n3, . . . , ns1 = ns.

Proof. Clearly the terminal position, which is P, is a square with s = 0. Wehave to prove that every non-square moves to some square, and that a squarecannot move to another square.

Assume that a square n1| . . . |ns moves to some square m1| . . . |ms. We mayassume that s is even; otherwise add 1 on the left. For even i 2 we haveni= ni1|mi|ni, thusmi = ni. Since both sequences are squares, this alreadyimpliesmi= ni for alli. This is a contradiction.

Assume that n1| . . . |ns is not a square. Ifs is even, define mi :=mi+1 :=nifor all oddi. Then we have m1 = n1 = m2|n2|m3= n3= m4| . . . . Thenm isa square. In particular,m=n. Hence, m is a winning move. The case thats is odd can be reduced to this case by adding 1 on the left. The winning

move is here m1 := 1 and mi := mi+1:= ni for all eveni >1.

Theorem 3.6. LetA be a finite abelian group.

1. A is a normalP-position if and only ifA is a square, i.e. A=B2 forsome finite abelian group B.

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2. IfA= Z/n1 . . . Z/ns withni|ni+1 is not a square, then a winning

move is

x=

0 n1 0 n1

n3n2

0 . . . 0 n1 n3n2

. . . ns1ns2 ifs is even

1 0 n2n1

0 . . . 0 n2n1

. . . ns1ns2

else

In fact, we then have

A/x=

(Z/n1 Z/n3 . . . Z/ns1)2 ifs is even(Z/n2 Z/n4 . . . Z/ns1)2 else

Proof. 1. follows from Propositions 3.4 and 3.5, and 2. follows from an

inspection of the proofs of Propositions3.5 and 3.3.Example 3.7. For example, Z/4 Z/8 Z/40 is a normal N-position.Player I mods out 1 0 2, since 8/4 = 2. The quotient is isomorphic to thesquare Z/8 Z/8. Player II has many choices, but he loses in any case. Letus demonstrate this for the element 4 0. Then Player I gets Z/4 Z/8 andof course he mods out 0 4. Now Player II has the smaller square Z/4Z/4.If he wants to postpone his inevitable defeat, he could try 2 2 with quotient= Z/2 Z/4. The next moves are Z/2 Z/2 by Player I, Z/2 by Player IIand finally 0 by Player I, who wins.

Theorem 3.8. Let A be a finite abelian group. Then A is a misere P-position if and only ifA is

either elementary abelian of odd dimension,

or a square, but not elementary abelian

Thus, the only difference to the normal P-positions are the elementary abeliangroups (Z/p)s, which become misere Pif and only ifs is odd.

Proof. According to Proposition3.2 and Corollary 2.16, it suffices to treatthe case that A is a finite abelianp-group, sayA = Z/pk1 . . . Z/pks.

We say that A is P if it is either elementary abelian of odd dimension, orit is a square, without being elementary abelian. We have to show the threeproperties characterizing misereP-positions (Proposition2.1). The terminalposition is elementary abelian of dimension 0, thus it is not P. Next, we haveto show that ifA = 0 is notP, then there is some move which makes it P.

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Even if this formula was corrected and a similar formula was found for arbi-

trary finitely generated abelian p-groups, it is not so clear how to computethe nimber of any finitely generated abelian groups. The problem is that thenimber of a selective compound game does not only depend of the nimbersof the individual games.

4 The game on groups

4.1 Some classes of groups

Lets play the game on groups under the normal play rule. In every step, agroup G is replaced by the quotient group G/a for some 1 = a G. Ingeneral, the normal subgroupagenerated bya is quite large compared tothe cyclic subgroupa. This will be responsible for a variety ofN-positions.

For example, there are many groups which can be normally generated bya single element, which are therefore N ([B]). Every knot group has thisproperty. For example, the Wirtinger presentation of the trefoil knot is

G= a,b,c: a1ca= b, c1bc= a, b1ab= c

and we see G= a.

Now let us treat some classes of finite non-abelian groups. We may use ourresults on abelian groups.

Proposition 4.1. For everyn 3 the following groups areN:

1. the symmetric groupSn

2. the alternating groupAn

3. the dihedral groupDn

4. the group Dn Z/2

Proof. 1. Since Sn is generated by 2-cycles and every 2-cycle is conjugatedto (12), it follows that Sn= (12).

2. The abelian group A3 = Z/3 is N. The only nontrivial normal subgroupof A4 is the Klein Four Group V4, which does not contain (123), so that

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A4 =(123). For n5, we even have An = for every 1 = An

since An is simple.3. The dihedral group has the presentation Dn= r, s: rn =s2 = (rs)2 = 1.If n is even, then Dn/r2 = D2 = (Z/2)2, which is square abelian andthereforeP. Therefore,Dn isN. Ifn is odd, thenDn/s= r: rn =r2 =1= {1}.

4. This follows from (Dn Z/2)/(r, 0)=Dn/r Z/2=(Z/2)2.

Next, recall that the dicyclic group Dicn is defined by the presentation

Dicn= a, x: a2n = 1, an =x2, axa= x.

It has order 4n. For n= 2 this is the Quaternion group Q= {1, i, j, k}.

Proposition 4.2. For everyn 2, the dicyclic groupDicn isN. The sameis true forDicn Z/2.

Proof. From the presentation we see Dicn /x = a: an =a2 = 1, whichis trivial when n is odd. When n is even, mod out a2. The quotient is

a, x: a2 =x2 = (ax)2 = 1=(Z/2)2.

This shows that Dicn isN. As for the product, observe that

(Dicn Z/2)/(a, 0)=Dicn /a Z/2=(Z/2)2.

Proposition 4.3. Letp, q be two primes and letG be a non-abelian groupof orderpq. ThenG isN.

Proof. It is well-known that G=x, y :xq =yp = 1, yxy1 = xr for somer (Z/q). Hence, we haveG/y = x : xq = xr1 = 1, which is cyclicof order d = gcd(q, r 1). Since r (Z/q) has order p, it follows q r 1,

thus d = 1.

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4.2 Small orders

There are various online resources for the classification of groups of smallorder, for example [G]. For the general theory and development of thisclassification, see for example [HBE].

Proposition 4.4. Every non-abelian group of order15 isN.

Proof. We have already dealt with groups of orders pq for primes p, q, andgroups of prime order are cyclic. This only leaves the orders 8,12. Thereare 2 non-abelian groups of order 8, namely the dihedral group D4 and thequaternion group Q, which are N (Propositions4.1 and4.2). There are 3

non-abelian groups of order 12, namely A4, D6 and Dic3, which are also Naccording to the same Propositions.

Next, there are 14 groups of order 16 (see [W]). We denote them via theirID in GAPs SmallGroup library ([GAP]). Thus,Gn is SmallGroup(16,n).Since G1, G2, G5, G10, G14 are abelian, we only need to consider the other 9non-abelian ones.

G3 = a,b,c: a4 =b2 =c2 = 1, ab= ba, bc= cb, cac1 =ab=(Z/4 Z/2) Z/2 with (c) = (aab, bb).

G4 = a, b: a4 =b4 = 1, ab= ba3= Z/4 3 Z/4

G6 = s, t: a8 =b2 = 1, ab= ba5= Z/8 5 Z/2

G7 = D8

G8 = a, b: a8 =b2 = 1, ab= ba3= Z/8 3 Z/2

G9 = Dic4

G11 = D4 Z/2

G12 = Dic2 Z/2

G13 = a,x,y: a4 =x2 = 1, a2 =y2,xax= a1, ay= ya,xy= yx

We already know that G7, G9, G11, G12 are N (Propositions 4.1 and 4.2).Now, observe that G6/a2 =a, b : a2 = b2 = 1, ab= ba=(Z/2)2. Thesame argument shows G8/a2=(Z/2)2. We also have

G13/a= x, y: x2 =y2 = 1, xy= yx=(Z/2)2.

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2 groups of order 128 with IDs 175, 476,

9 groups of order 144 with IDs 92, . . . , 95, 100, 102, 103, 194, 196,

1 group of order 196 with ID 11.

4.3 The game on subgroups

The game on groups has disproportionally many N-positions because thenormal closure is rather large. A more interesting and balanced game couldbe the following one, which we will only sketch here:

Start with a group G. A position in the game is a subgroupU G. Theinitial position is the trivial subgroup, the terminal position is the wholegroup. A move picks some g G \ Uand replacesUby the subgroupU, g.

This game satisfies the ending condition if and only ifG is noetherian, i.e.the partial order of subgroups ofG satisfies the ascending chain condition.Equivalently, every subgroup of G is finitely generated. For example, thishappens when G is finite. Let us restrict to the normal play rule. When isG a P-position?

Remark that this resembles the game proposed in [AH], but still differs fromthat. Actually this game is (almost) a special case of the game on algebraic

structures, namelyG-sets, starting with the G-setG. In fact, for a subgroupUG, a move from the G-setG/Upicks someg G \ Uand replaces G/UbyG/U, g. The only difference is the following: Two G-setsG/U,G/V areisomorphic if and only ifU, Vare conjugate not necessarily equal.

Observe that whenG is abelian, we have just the game on the abelian groupG and we may use Theorem 3.6. More generally, when G is Hamiltonian(i.e. every subgroup is normal), we have the game on the groupG. But forarbitraryG, these games differ dramatically. Many moreP-positions arise.Some examples include D3 = S3, D5 and A4. However,D4 and D6 are N.Let us verify this for S3: If Player I starts with some 2-cycle (resp. 3-cycle),

then Player II responds with any 3-cycle (resp. 2-cycle). Since a 2-cycle anda 3-cycle already generateS3, Player II wins. The quaternion group Q isNas before because it is Hamiltonian.

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some element a R \ {0} and replacing R byR/(a), where (a) is the ideal

generated by R. The algebro-geometric picture will be quite useful: Givensome noetherian affine scheme, we replace it by some closed subscheme cutout by some single non-trivial equation in the coordinate ring.

Remark 5.1. Observe that the zero ring isNand therefore that fields are P.Hence, if a commutative ring R has some principal maximal ideal = 0, thenR is N. This applies in particular to principal ideal rings (not necessarilydomains) which are no fields, such as k[x], Z, or quotients thereof. It alsoshows, for example, that k[x, y]/(xy) and k[x, y]/(xy 1) are N (mod outx 1 in each case). Geometrically, we intersect the union of the coordinateaxes resp. the hyperbola with the line x = 1, which results in a single simple

point, which is thereforeP.

We had already seen in Example2.17:

Lemma 5.2. If a commutative ringR isP, thenSpec(R) is connected (i.e.R is not a direct product of two non-trivial rings).

The 1-dimensional smooth case is rather easy to understand:

Proposition 5.3. Let R be a Dedekind domain. If R has some principalmaximal ideal, thenR isN. Otherwise, R isP.

Proof. The first part is clear. Now assume that R has no principal maximalideal. Recall the fact that for every idealIR and every 0 = a I thereis some b R such that I= (a, b). If 0=a R is not a unit, then there issome maximal idealIcontaining a. Now R/(a) isNbecause if we choose bas above, we have b /(a), so that we may move to R/(a, b) =R/I, which isa field.

Corollary 5.4. Letk be an algebraically closed field. Thenk[x, y] isN.

Proof. Choose an elliptic curve E over k and some closed point E.Then E\ {} is affine and its coordinate ring R is Pby Proposition5.3.

Since R is the quotient ofk[x, y] by some Weierstrass equation defining E,the claim follows.

A more general result below (Proposition 5.12) will show this even when kis an arbitrary field.

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5.2 Zero-dimensional rings

After having considered smooth curves, the next step is to consider an exam-ple of a non-smooth curve such as the cusp with coordinate ringk[x, y]/(y2 x3). We will show that it is P (giving another reason why k[x, y] is N).However, we will need some results on zero-dimensional rings first, whichappear as intersections of the cusp with curves through the origin.

Lemma 5.5. LetVbe a vector space over some fieldk of finite dimension.Then the ringk[V] :=k V with multiplicationV2 = 0 isN if and only ifdim(V) is odd; otherwise it isP.

Proof. For V = 0 this is true. Now induct on dim(V). If dim(V) is odd,choose some 0=v V. The ideal generated by 0 v equals 0 kv and thequotient is k V/kv, which is Pby the induction hypothesis. Now assumethat dim(V) is even and 0 = a+v k[V] is some element. Ifa = 0, thena + v is invertible, so that the quotient is zero, which is N. Otherwise,a = 0and the quotient is k V/kv, which is N by induction hypothesis.

Corollary 5.6. Ifk is a field, thenk[x, y]/(x2,xy,y2) isP.

Lemma 5.7. Letk be a field andu 0. Thenk [x, y]/(y2 x3, xu+1, xuy) isP.

Proof. Let us call this ring Bu. ThenB0 = k and B1 = k[x, y]/(x2, y2, xy)

are P. Now let u 2 and assume that the Lemma is true for all < u.Observe that 1, . . . , xu, y , x y , . . . , xu1y is ak-basis ofBu. Choose some non-zero element b B , we want to show that Q:= Bu/b is N. Ifb has some1-coefficient, it is a unit and we are done. Otherwise write

b= r1x+. . .+ruxu +s1y+. . .+sux

u1y

withri, sj k, not all zero. Choose some minimal 1 d uwithri= si= 0for all 1 i < d. Thus, we have

b= rdxd +. . .+ruxu +sdxd1y+. . .+suxu1y,

and at leastrdor sdare non-zero. Now let us consider the case d= u, so thatb= rux

u + suxu1y. Whenru = 0, we have Q = k[x, y]/(y

2 x3, xu+1, xu1y)and thereforeQ/xu= Bu1, which is Pby the hypothesis. This proves that

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Q isN. Now assume ru = 0. Then 0=xu1y in Q and Q/xu1y= Bu1,

which isPby the hypothesis.So let us assume d u 1. In Bu we compute:

xud1b = rdxu1 +rd+1x

u +sdxu2y+sd+1x

u1y

xudb = rdxu +sdx

u1y

xud1yb = rdxu1y

Now compute in the quotient Q, where b = 0. When rd = 0, the thirdequation shows xu1y = 0 in Q, which in turn gives xu = 0 by the secondequation. But then we can consider b as an element b Bu1 and obtain

Q=Bu1/b

, which isNby the hypothesis. When rd = 0, we have sd= 0,so that the second equation gives again xu1y = 0, and the first equationreads asrd+1x

u + sdxu2y= 0. We see xu1 = 0 inQ and Q/xu1=Bu2,

which isPby the hypothesis, so that Q isN.

Corollary 5.8. Letk be a field andu 0. Thenk[x, y]/(y2 x3, xuy) andk[x, y]/(y2 x3, xu+1) areN. For example, k[x, y]/(x3, y2) isN.

Proposition 5.9. Letk be an algebraically closed field. Thenk[x, y]/(y2 x3) isP.

Proof. Let R := k[x, y]/(y2 x3) and consider some 0=fR, representedby f k[x, y]\ (y2 x3). Assume first that f / (x, y) and write f =a0 + a1x+ a2x

2 +. . .+ b0y + b1xy + b2x2y + . . . with a0 = 0. We claim

that x is invertible in R/f. This is clear ifb0 = 0. Otherwise, let g be thesame polynomial as f, but with a0 replaced bya0. InR/fwe have 0 =f gand in that product the y has disappeared, but the constant term is stillinvertible. Thus, we may repeat the argument. Sincex is invertible, we getan isomorphism R/f = (Rx)/f. But the normalization map : R k[t]defined by x t2 and y t3 induces an isomorphism on localizations, sothat R/f = k[t]t/(f) = k[t]/(f) and (f) is some polynomial of degree2. Now apply Remark5.1 to conclude that R/f is N.

Now let us assumef(x, y). The intersectionV(f) V(y2 x3) A2(k) iszero-dimensional. Thus,R/(f) is a direct product of local artinian rings. Inorder to show that it is N, we may even assume that it is local by Lemma5.2This means that there is a unique k such that (f)() =f(2, 3) = 0.

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Algebraic games - playing with groups and rings

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