Comments On " Orbits of automorphism groups of fields"
aa r X i v : . [ m a t h . A C ] A p r COMMENTS ON ”ORBITS OF AUTOMOPHISMS GROUPSOF FIELDS”Pramod K. Sharmaemail: [email protected] of Mathematics, Sikkim Central University,Gangtok-73710, INDIAABSTRACT
Let R be a commutative k − algebra over a field k . Assume R is a noetherian,infinite, integral domain. The group of k − automorphisms of R ,i.e. Aut k ( R )acts in a natural way on ( R − k ).In the first part of this article, we studythe structure of R when the orbit space ( R − k ) /Aut k ( R ) is finite.We notethat most of the results, not particularly relevent to fields, in [1, §
2] hold inthis case as well. Moreover, we prove that R is a field. In the second part,we study a special case of the Conjecture 2.1 in [1] : If K/k is a non trivialfield extension where k is algebraically closed and | ( K − k ) /Aut k ( K ) | = 1then K is algebraically closed. In the end, we give an elementary proof of[1,Theorem 1.1] in case K is finitely generated over its prime subfield. Let
K/k be a non trivial field extension. Authors in [1 , §
2] conjecture thatthe orbit space ( K − k ) /Aut k ( K ) is finite if and only if either both K and k are finite or both are algebraically closed. From the results of the authors, itis clear that K is finite if and only if k is finite. Moreover,if K is algebraicallyclosed then so is k . The converse is open, and several results are proved in [1]for this case. In this note, we prove that if R is an infinite, noetherian integraldomain which is an algebra over a field k such that | ( R − k ) /Aut k ( R ) | < ∞ then most of the results, not particularly relevent to fields, in [1, §
2] hold inthis case as well. We, infact, note that R is a field in case charateristic of R is p > R is integrally closed. Further,we note that if K is algebraicallyclosed then | ( K − k ) /Aut k ( K ) | = 1. Note that if the conjecture is true, | ( K − k ) /Aut k ( K ) | = 1 should imlpy that K is algebraically closed. Weare not able to prove this but observe that f ( K ) = K for every polynomial f ( X ) ∈ k [ X ]. It is also proved that if | ( K − k ) /Aut k ( K ) | < ∞ , then for any1on constant f ( X ) ∈ k [ X ] with degree ≤ , f ( K ) = K . Throughout,we assume that R is an infinite commutative k − algebra over afield k which is a noetherian integral domain such that | ( R − k ) /Aut k ( R ) | < ∞ . Theorem 2.1. The field k is infinite and integrally closed in R .Proof. If | k | < ∞ , then as | ( R − k ) /Aut k ( R ) | < ∞ , | ( R/Aut k ( R ) | < ∞ . Therefore | R/Aut ( R ) | < ∞ . Hence by [2 , Corollary 16], R is a finitefield.This contradicts our assumption that R is infinite. Hence | k | = ∞ .Now, let α ∈ ( R − k ) be integral over k . Then for each a ∈ k, aα is integralover k , moreover { aα | a ∈ k } is an infinite subset of ( R − k ). Note thatif β ∈ ( R − k ) is integral over k , then for any σ ∈ Aut k ( R ) , σ ( β ) is integralover k . Hence orbit of β , i.e. O ( β ) = { σ ( β ) | σ ∈ Aut k ( R } is a finite set.This implies | ( R − k ) /Aut k ( R ) | = ∞ , a contradiction to the assumption that | ( R − k ) /Aut k ( R ) | < ∞ . Hence k is infinite and integrally closed in R .Theorem 2.2. If characteristic of k is p >
0, then k p = k , and R p = R .Moreover, R is a field.Proof. By Theorem 2.1 , k is intergrally closed in R . Hence ( R − k ) p ⊂ ( R − k ). Consequently( R − k ) ⊃ ( R − k ) p ⊃ . . . ⊃ ( R − k ) p m ⊃ . . . is a chain of orbit closed subsets of ( R − k ) under the action of Aut k ( R ).Since | ( R − k ) /Aut k ( R ) | < ∞ , there exists n ≥ R − k ) p n = ( R − k ) p ( n +1) . Thus for any λ ∈ ( R − k ), there exists µ ∈ ( R − k ) such that2 p n = µ p ( n +1) ⇒ ( λ − µ p ) p n = 0 ⇒ λ = µ p ⇒ ( R − k ) = ( R − k ) p ⊂ R p Now, if λ ∈ ( R − k ) , a ∈ k , then λ, ( λ − a ) ∈ ( R − k ) = ( R − k ) p ⊂ R p . Assume λ = α p , λ − a = β p for α, β ∈ R . Then( α − β ) p = α p − β p = a ∈ R p ⇒ k ⊂ R p ⇒ R = R P since ( R − k ) = ( R − k ) p . Now as R = R P and k is integrally closed, k = k p . We shall now show that R is a field. Let I be a non-zero radical ideal in R . As R = R p , for any a ∈ I , there exists b ∈ R such that a = b p . As I is a radical ideal b ∈ I . Con-sequently I = I p , i.e. I p − I = I . As R is noetherian, there exists α ∈ I p − such that (1 − α ) I = (0). This is not possible as I = (0) as well as α = 1.Hence R has no non-zero radical ideal. Thus R is a field,.Theorem 2.3. For any x ∈ ( R − k ) and a ∈ k , there exists σ ∈ Aut k ( R )such that σ ( x ) = x + c .Proof. The proof is similar to the proof of [1, lemma 2.6].Theorem 2.4. If R is integrally closed, then ( R − k ) l = ( R − k ) for all l ≥
1. Moreover, R is a field, and R l = R as well as k l = k .Proof. It suffices to prove the statement assuming l is prime. In view ofTheorem 2.2, we can assume that l is other than the characteristic of k ifthat is prime. As in theorem 2.2,( R − k ) ⊃ ( R − k ) l ⊃ . . . ⊃ ( R − k ) l m ⊃ . . .
3s a chain of orbit closed subsets of ( R − k ). As | ( R − k ) /Aut k ( R ) | < ∞ ,there exists n ≥ R − k ) l m = ( R − k ) l ( m +1) for all m ≥ n . Thus for any λ ∈ ( R − k ), there exists µ ∈ ( R − k ) such that λ l m = µ l ( m +1) ⇒ ( λµ − l ) l m = 1 ⇒ λµ − l ∈ R since R is integrally closed. Further, as k is integrally closed in R, λµ − l ∈ ( k ∗ ) l m , the subgroup of ( l m ) th roots of unity in k . Therefore λ ∈ ( R − k ) l ( k ∗ ) l m for all m ≥ n . Consequemtly ( R − k ) ⊂ ( R − k ) l ( k ∗ ) l m . Next, forany m ≥ n ( R − k ) l m = ( R − k ) l m . Hence as above, we conclude that for any λ ∈ ( R − k ) , λ ∈ ( R − k ) l m ( k ∗ ) l m .Thus ( R − k ) ⊂ ( R − k ) l m ( k ∗ ) l m . We, now, consider two cases:Case 1. ( k ∗ ) l m $ ( k ∗ ) l ( m +1) . In this case, for any c ∈ ( k ∗ ) l m , there exists b ∈ ( k ∗ ) l ( m +1) such that c = b l .Hence,as ( R − k ) ⊂ ( R − k ) l ( k ∗ ) l m , we have( R − k ) ⊂ ( R − k ) l ( k ∗ ) l m = ( R − k ) l ⊂ ( R − k ) . Consequently ( R − k ) = ( R − k ) l .Case 2. ( k ∗ ) l m = ( k ∗ ) l ( m +1) for all m ≥ n. We have ( R − k ) ⊂ ( R − k ) l ( k ∗ ) l m ⊂ ( R − k ) k ∗ ⊂ ( R − k ) . Consequently ( R − k ) = ( R − k ) l ( k ∗ ) l m for all m ≥ n. Further, since( R − k ) ⊂ ( R − k ) l m ( k ∗ ) l m ⊂ ( R − k ) k ∗ ⊂ ( R − k ) ⇒ ( R − k ) = ( R − k ) l m ( k ∗ ) l m ⇒ ( R − k ) = ( R − k ) l m ( k ∗ ) l ( m +1) since ( k ∗ ) l m = ( k ∗ ) l ( m +1) ⇒ ( R − k ) l = ( R − k ) l ( m +1) ( k ∗ ) l ( m +1) m > n , then m − ≥ n . Hence( R − k ) l = ( R − k ) l m ( k ∗ ) l m = ( R − k ) . We shall now show that R is a field. Let I be a non-zero radical ideal in R ,then as I ∩ k = (0) , I − (0) ⊂ ( R − k ) . Let a ( = 0) ∈ I, then a = λ l since R − k ) l = ( R − k ). Thus as I is radical ideal, λ ∈ I. Consequently I l = I ⇒ I = I ⇒ (1 − λ ) I = (0)for some λ ∈ I . As λ = 0 ,
1, it is a non trivial idempotent . This is not pos-sible, hence I = (0). Therefore R is a field. The last part of the statementfollows by [ 1, Proposition 2.10]Remark 2.5. If U is the group of units in R , and U ∩ ( R − k ) = φ , then R l = R and k l = k . This gives an alternative proof of the last part of thestatement.We shall first prove that U l = U . Let v ∈ U ∩ ( R − K ) = U ∩ ( R − k ) l .Then v = λ l for some λ ∈ ( R − k ). Clearly λ ∈ U . Hence v ∈ U l . Therefore U ∩ ( R − k ) ⊂ U l . Next, note that U = ( U ∩ ( R − k ) ∪ ( U ∩ k ∗ ) = ( U ∩ ( R − k )) ∪ k ∗ ⊂ U l ∪ k ∗ Thus to prove U = U l , it suffices to show that k ∗ ⊂ U l . Let a ∈ k ∗ and λ ∈ U ∩ ( R − K ). Then λa ∈ U ∩ ( R − K ) . Thus λ − ( λa ) = a ∈ U l since( U ∩ ( R − k )) ⊂ U l . Hence k ∗ ⊂ U l . Thus U = U l . Now, let a ∈ k ∗ . Thenthere exists b ∈ U such that b l = a . As k is integrally closed in R, b ∈ k ∗ . Hence k = k l . This implies R l = ( R − k ) l ∪ k l = ( R − k ) ∪ k = R Hence the assertion holds. 5heorem 2.6. If R is integrally closed, then k = R Aut k ( R ) = { λ | σ ( λ ) = λ for all σ ∈ Aut k ( R ) } .Proof. Let λ ∈ R Aut k ( R ) , λ / ∈ k. By [2, Lemma 5], λ is a unit. Therefore L = R Aut k ( R ) is a field containing k . By Theorem 2.4, R − k ) l = ( R − k ) forany l ≥
1. Thus since λ / ∈ k, X I − λ has no root in k . Moreover, by Theorem2.3, for any a ∈ k and a root µ of X l − λ, µ + a is also a root of X l − λ sincefor any σ ∈ R Aut k ( R ) ,σ ( µ ) is also a root of X l − λ . As | k | = ∞ , this is notpossible. Hence k = R Aut k ( R ) .Remark 2.7. (i) If charateristic of k is p >
0, then we can drop the con-dition that R is integrally closed.This can be seen by taking l = p .(ii) Under the conditions of the theorem, | O ( λ ) | < inf ty if and only if λ ∈ k .If O ( λ ) = { λ , . . . , λ t } , then p ( X ) = ( X − λ ) . . . ( X − λ t ) ∈ k [ X ]. Thus each λ i , i = 1 , . . . t is integral over k , and consequently λ ∈ k . The converse is clear.Theorem 2.8. If λ ∈ ( R − k ), then S λ = { a ∈ k ∗ | σ ( λ ) = aλ } for some σ ∈ Aut k ( R ) } is a subgroup of finite index in k ∗ .Proof. Let a, b ∈ S λ . Then there exist σ, τ ∈ Aut k ( R ) such that σ ( λ ) = aλ, τ ( λ ) = bλ . Therefore στ ( λ ) = abλ and σ − ( λ ) = a − λ . Hence ab, a − ∈ S λ . Thus S λ is a subgroup of k ∗ . Assume [ k ∗ : S λ ] = ∞ . Choose an infiniteset { b , . . . , b n , . . . } in k ∗ such that b i S λ = b j S λ for all i = j . We claim O ( b i λ ) = O ( b j λ ) whenever i = j . If not, then there exist i = j such that O ( b i λ ) = O ( b j λ ) ⇒ σ ( b i λ ) = b j λ for some σ ∈ Aut k ( R ) ⇒ σ ( λ ) = b − b j λ ⇒ ∈ S λ ⇒ b i λ = b j λ As b i S λ = b j S λ for all i = j , the claim follows. This cotradicts the assump-tion that | ( R − k ) /Aut k ( R ) | , ∞ . Hence [ k ∗ : S λ ] , ∞ .Remark 2.9. (i) Let k be algebraically closed, and let [ k ∗ : S λ ] = m < ∞ .Then since ( k ∗ ) m = k ∗ , S λ = k ∗ . Hence for any c ∈ k ∗ , there exists σ ∈ Aut k ( R ) such that σ ( λ ) = cλ .(ii) Assume R is integrally closed and ( R − k ) ∩ U = φ , where U is the6roup of units of R . Then also the assertion of the theorem holds, i.e. weneed not assume that k is algebraically closed in this case. This follows since( k ∗ ) m = k ∗ by remark 2.5. We shall first give an elementary proof of [1, Theorem 1.1] in case the field K is finitely generated over its prime subfield. Then we study a particularcase of the Cojecture 2.1 in [1]:”Let K/k be a non trivial extension of fields. Then the number of orbits of
Aut k ( K ) on ( K − k ) is finite if and only if either both K and k are finite orboth are algebrically closed.”The authors in [1] have proved that if the number of orbits of Aut k ( K ) on( K − k ) is finite, then K is finite if and only if k is finite.Further, it is notedthat if K is algebraically closed then so is k . Thus it remains to show thatunder the given condition if k is algebraically closed then so is K . Basedon the conjecture we ask : If K/k is a non trivial field extension where k is algebraically closed, then is it true that K is algebraically closed if and Aut k ( K ) on ( K − k ) has one orbit? Before we look into this, we prove:Theorem 3.1. Let K be a field with | K/Aut ( K ) | < ∞ . If K is finitelygenerated over its prime subfield, then K is finite.Proof. It is noted in [1]that characteristic of K is p > K p = K . Let F p be the prime subfield of K . As K is finitely generatedover F p , K has finite transcendence degree over F p . Let S be a transcdencebasis of K | F p . Then K | F p ( S ) is finite algebraic. If S = φ , then clearly K is finite. Further if S = φ , then as K is perfect K = F p ( S ). As K is perfect, Fr¨ o benius endomorphism of K (say) σ is an automorphism. Therefore as[ K : F p ( S ), σ ( F p ( S ) = F p ( S ). This however is not true. Consequently S = φ ,and K is finite.Here after, we assume that K/k is a non trivial field extension where k is algebraically closed.Lemma 3.2. If K is algebraically closed then action of Aut k ( K ) over7 K − k ) has one orbit.Proof. Let x, y ∈ ( K − k ). Since k is algebraically closed, x, y are tran-scendental over k . Choose a transcendental basis S of K/k containing x,and a transcendental basis T of K/k containing y. As S and T have samecardinality, there exists a bijection from S to T mapping x to y. This extendsto a k − isomorphism σ from the field k ( S ) to the field k ( T ) whuich mapsx to y. As K is algebraic closure of k ( S ) as well as k ( T ), σ extends to anautomorphism τ of K such that τ ( x ) = y . Hence the result follows.Lemma 3.3. A field K is algebraically closed if and only if for every non-constant polynomial f ( X ) ∈ K [ X ] , f ( K ) = { f ( λ ) | λ ∈ K } = K .Proof. Let K be algebraicallty closed and f ( X ) ∈ K [ X ] be a non-constantpolynomial. Note that for any λ ∈ K, g ( X ) = f ( X ) − λ is a non constantpolynomial in K [ X ]. Hence has a root in K . If a is a root of g ( X ) in K ,then f ( a ) = λ . Therefore f ( K ) = K . Conversely, let f ( X ) ∈ K [ X ] be anynon-constant polynomial. Then since f ( K ) = K , there exists b ∈ K suchthat f ( b ) = 0. Hence K is algebraically closed.Remark 3.4. For a field K to be algebraically closed, it is sufficient toassume that p ( K ) = K for every irreducible polynomial p ( X ) ∈ K [ X ].Theorem 3.5. If Aut k ( K ) has one orbit over ( K − k ), then for any non-constant f ( X ) ∈ k [ X ], f ( K ) = K .Proof. First of all, note that since k is algebraically closed and f ( X )is non-constant, for any a ∈ k , there exists b ∈ k such that f ( a ) = b. Now, let α ∈ ( K − k ) be any element, then f ( α ) = β ∈ ( K − k ). Thus,since Aut k ( K ) has one orbit over ( K − k ), for any z ∈ ( K − k ), there exists σ ∈ Aut k ( K ) such that σ ( β ) = z . Hence σ ( f ( α )) = f ( σ ( α ) = z . Conse-quently f ( K − k ) = ( K − k ). Therefore f ( K ) = K .Remark 3.6. If f ( X ) ∈ k [ X ], and λ ∈ K , then ( λf ( X ))( K ) = K as wellas ( λ + f ( X ))( K ) = K .Theorem 3.7. Let Aut k ( K ) has one orbit over ( K − k ). If x ∈ ( K − k ),and E = { α ∈ K | α : algebraic over k ( x ) } , the algebraic closure of k ( x ) in8 , then | ( E − k ) /Aut k ( E ) | = 1.Proof. Let y ∈ ( E − k ), then y is transcendental over k since k is al-gebraically closed. As | ( K − k ) /Aut k ( K ) | = 1, there exists σ ∈ Aut k ( K )such that σ ( x ) = y . Therefore σ ( k ( x )) = k ( y ). If σ ( E ) = F , then F is thealgebraic closure of k ( y ) in K . Note that the transcendental degree of E over k is 1. Hence F has transcendental degree 1 over k . Note that k ⊂ k ( y ) ⊂ E .As E has transcendental degree 1 over k , E | k ( y ) is algebraic. Therefore E ⊂ F . Now, as E ⊂ F, x ∈ ( F − k ), we can prove as above that F ⊂ E .Cosequently E = F . Hence σ ∈ Aut k ( E ) and | ( E − k ) /Aut k ( E ) | = 1. Thusthe assertion is proved.We now ask the following:Question. Let k be an algebraically closed field and X an indetermi-nate over k . Let E be an intermediate field such that k $ E ⊂ k ( X ) and | ( E − k ) /Aut k ( E ) | = 1. Then is E = k ( X )?Remark 3.8. In the above question, we can assume that k ( X ) ⊂ E , sinceif y ∈ ( E − k ), it is transcendental over k . Therefore k $ k ( y ) ⊂ E ⊂ k ( y ) = k ( X )since transcedence degree of k ( X ) over k is 1 ( Here k ( y )) is algebraic closureof k ( y ) in k ( X )). Hence we can assume k ⊂ k ( X ) ⊂ k ( X ) in the abovequestion.Lemma 3.9. If Aut k ( K ) acts transitively over ( K − k ), then for any sub-group H of finite index in G = Aut k ( K ), K H = k .Proof. If H is a subgroup of finite index in G , then there exists a normalsubgroup H of finite index in G contained in H . As K H ⊂ K H , to provethe result we can assume H is normal in G . Now, let g ∈ G, and h ∈ H .Then for any a ∈ K H , h ( g ( a )) = g ( g − h ( g ( a )) = g ( a )since g − hg ∈ H . Cosequently g ( K H ) ⊂ K H for all g ∈ G . Now, as G actstransitively over ( K − k ) , K G = k . Thus if K H = k , then K = K H . This9nplies H is identity subgroup. However G is infinite, hence H is infinite.Therefore K H = k .lemma 3.10. If Aut k ( K ) acts transitively over ( K − k ), then for an alge-braic closure K of K either [ K : K ] = ∞ or K = K .Proof. Assume 1 < [ K : K ] < ∞ . Then by Artin-Screir theorem, char-acteristic of K is 0, and K = K ( i ) where i = −