About some questions of differential algebra concerning to elementary functions

Pub . Ma . UAB Vol . 26 N° 1 Ma a 1982 ABOUT SOME QUESTIONS OF DIFFERENTIAL ALGEBRA CONCERN I NG TO ELEMFJNTARY FUIJCTIONS San iago Za zuelaA mengou . Dp o . de Algeb a y Fundamen os, Uni e sidad de Ba celona Rebu el 2 de juny del  l081 The z udy o6 elemen any 6unc .íon4, ha .íz, o6 hoee 6unc .íon4 buíl up by u4 .íng xa .íonal 6unc .íonz o en Q, Exponen- íalz, Logan .í hmz and Algebna .íc ope a .íon4, begán zomewha z .íz ema .cally w .í h he d .í enze wohk4 ha Jozeph L .íou .ílle d .íd in he 1830'4 . Al hough h .íz capi al a,ím waz o ob a .ín ome ezul . abou he .ín egna .íon by mean4 o6 elemen any 6une .íona, along - h .íz way he had o 4 udy azpec 4 mo se c .í cunzc ibed o he z nuc- une o6 hese 6une .íonz . Abou - he 6 .ínz po .ín we muz zay ha .í, ce a .ínly, L .íou .ílle ob a .íned a e4ul ha , e en .ímp o ed z .ínce ha ime, haz non changed e44en íally . I 4u6 .íce4, 6o~ ezemple, o compa e he wonk o6 Líou ílle g .í en in 121 and Rozenl .ích 'e - in [6) . lde can obaen e hen ha he ín oduc íon o6 new langua- ge and new hecníque haz only línealízed he p oblem, makíng clean whích pnopen íez o6 elemen a y 6une íon4 cha ac eníze hem . U4íng hí4 new language and new hecníque .í íz po4zí- ble deal mo e cleanly wí h 4ome quez íon4 abou elemen any 6unc- íonz . Fon exemple, heí í nedundancy, ha waz alneady ez abl .í 4hed by Líou ílle hímzel6 in [21 and Handy in [11 . Now, he ínne- dundancy íz a conzequence o6 an S cuc une heonem (1 .2) ha we can quíkly gí e by u4íng Ro4enlích 'z hecníquez . Alzo, he non zol abílí y by'meanz o6 elemen any 6unc íonz o6 cen aín clazeí- cal nanzcenden al equa íon4 can be eazíly ez ablízhed . (oe z u- dy híz in Sec íon 2 . Ano hen ea men o6 he S nuc une heo e , and ínnedundance quez íonz (bu non u4íng Rozenlích 'z hecníque4) can be 6ound in he Rízch'4 papen [5) . 1 .- AN STRUCTURE THEOREM . Fi s some de ini ions . All ields will be conmu a i e and o cha ac e is ic ze o . 1 .A .  Remembe ha i E is a ield, a map D :E -» Eis a de i a ion o E i :  (1) V x,  y E E D(x+y)  = D(x) + D(y),  (2) M x,y EE D(xy)  = xD(y) + yD(x) . I ollows om (2)  ha D(1) = 0, hence by  (1) D(z) -0  y z E z .  The se CD = {x E E/ D(x)  = 0} is he se o cons an s o D . Gi en ha  ~ x E E D(x n ) = nD(x)x n-1 , D(x -1 ) _ -D(x)x _ 1 we ha e ha CD is a sub ield o E . A ield E wi h a amily o de i a ions A is a di e en ial ield ; hen, C = l C is he ield o cons an s o he dí e en ial ield E . DEA D Le E C F wo di e en ial ields . The ex ension E C F is di e en ial i Y D E AF , DI E E AE . Al hough wo di e en de i a ions o AF can coincide o e E, we won' dis inguish be wen AE and AF . Le CE , CF be he espec- i e cons an ields . We ha e CE CC F . When he equali yholds we say ha he ex ension is wi h he same ield o cons an s . . EXEMPLES : ¢(X 1 ,  . . ., X n ) wi h A= (ó/d x . )i=1 .. .n is a di e en ial ield . 1 C(X) C O(X, eX ) is a di e en ialex ension wi h he same ield o cons an s . 1 .B .  The elemen a y na u e ís hen o mula ed in he nex way : le E be a di e en ial ield ; x, y E E . Then - y = Log(x)  -  Dy = Dx/ x D E A  (y is Loga i hm o x) - y = Exp(x)  b  Dy/ y = Dx  D E A  (y is Exponen ial o x) I E C Fisa di e en ial ex ension, y E F is Elemen a y o e E i and on- ly i - ei he y is algeb aic o e E - o y = Log(x) being x E E - o y = Exp(x) being x E E . The di e en ial ex ension E C F is Elemen a y i F - E(O 1 ,  . . ., O n ) wi h 0 1 elemen a y o e E, and Oi elemen a y o e E(O 1 ,  . . ., O i-1 )  i >2 . Then, Ca dá E = Ca dAF . 1 .C .  The ool wich allow us o liea ize he a gumen s is he Module o Di e en ials . A as cons uc ion o i (su icien o us) is he ollowing : le E C F be íelds and conside he F- ec o space gene a ed by he symbols {dx}x E F . Le us impose hem he ollowing ela ions : (1) d x,  y E F  d(x+y) = dx + dy (2)  x,  y E F  d(xy)  = xdy + ydx (3) ~xEE dx = 0 . Then we ge a F- ec o space called he Module o he Di e en ials o E CF . I s symbol is S2F E' Remembe oo ha i {xi}i=1 .. . a e elemen s o F, hen hey a e algeb aical ly independen o e Ei and only i he amily {dxi}i=l . . . is F-linea y independen on 2F/E " So T .deg . E F = dimF(nF/E) " (see [6], P op .3 ) his wo k : The nex esul , due o Rosenlich , is a undamen al one o 1 .1 .- THEOREM .  Le E C F be a di e en ialex ension wi h he same ield o cons an s . Le C be his ield and ake y l , . . ., Yn E F,  z i s  . .  ,  z C,F-{0} and  {ciJ " } i=l . . .nC C such ha Y i = i,  . . ., .n,  Y D E p j=l . . . cijDzj/z . + Dyi E J E . Then ei he T .deg . E E(y l , . . ., Y n , z l , . . ., z ) >n o he n elemen s o A F/E . JLl c ij 1/ z dz j +dyi , i - 1, . . ., n a e C-linea y  dependen . P oo : see Theo em 1 . o [6j . 1 .D .  Le F be a di e en ial ield . We say ha he equalí y Y - Log X has a solu ion in F i he e a e elemen s x, y É F e í ying ¡ . I is na- u al, hen, o ask haw many solu ions o his equalí y he e a e in an elemen a y ex ension E C F . The ollowing heo em, om wich Risch gi es ano he e sion in [S], answe s his ques ion . P e iously some no a ion : le E C F be an elemen a y di e en ial ield ex ension wi h he same ield o cons an s :  E C F = E(O 1 ,  . . ., e n ) .  Le y l = Log x l , . . ., y - Log x he no algeb aic casesamong he O i 's ; ha is, = T .deg . EF and o each O i no algeb aic ( o e he p eceeding subex ension ) he e exis s x j o y j such ha Oi- x j o y j dependeng on whe he 0 i is Exponen ial o Loga i hm . Suposse hey a e a anged acco ding o he o de o appe- ance and ha E is an algeb aic closu e o E . 1 .2 .- Theo em .  On he abo emen ioned hypo hesis i he equali y Y - = Log X holds in F, o any solu ion x, y he eexis c 1 ,  . . ., c E C  g E E nF,  and n 1 ,  .... n ,  n E Z such ha P oo : i he equali y holds in F we can conside he sys em n n y + c l y l + ... + c y = , x nx l l . . . x = Dy i- Dxi ./ x  = 0EE . 1 Dy -1/x Dx = 0 E E By Theo em 1 .1 we ge DEA - ei he T .deg . E E(y 1 , . . ., Y , y, x l , . . ., - o he elemen s o n F/E : (dy .i -1/x . dx i ), i - 1, . . ., , i (dy - 1/xdx) a e C-linea ydependen . So he eexis c l ,  . . ., c , c E C no all ze o such ha We can also ake c ~ 0 since o he wise Bu i y = 0 J . o some j, because o he elemen a i y o E C F, each dy i , dx . excep dy is a linea combina ion o he p eceeding -1 dO s wi h coe- J icien s in F . Bu hey a e F-linea y independen beacuse o 1 .C . So c =0 . The same happens i x =0i . o some i . Appliyng epea dly his a gumen we ge c 1 - . . . -c = 0, no possible . He e i is clea ha only he secondcondi ion is possible . (1) c(dy - 1/ x dx) +  ci(dyi .- 1/x . dx i ) =0 . ei (dyi - 1/ xidx i ) = 0 . i=1 Hence, di iding by c, we can assume g , (2) dy + c 1 dy 1 + ... + c dy - 1/ x dx + c 1 1/ xldx 1 + . . . + c 1/x dx Conside now a maximal Q-linea y independen sys em among he {1,  cl ,  . . .,  c }  :  {e l ,  . . .,  e k }  such ha e l -  1 .  Then k d i  :  c i -  L  qijej  , qij E Q Ji, j .  The e o e j=1 k 1/ x dx + c 1 1/ x dx 1 + . . . + c l/x dx - e 1 1/ x dx +  1 g lj e j dx 1 + . . . + 1   j=1 k   + 1 g j e j l/ x dx - e l (1/ x dx +  gill/xidxi) + . . . + e k ( z gikl/x dxi) j=1  i=1  i=1 i - e l l/ d l + ... + e k l/ d k , 1  k Then (2') d(y + c1dy1 + . . . + c dy ) - e 1 1/ d 1 + ... + e k l/ d k . 1  k By P op 4 . o 16] we ha e - y + cly1 + ... + c y - g EE 1F - i CE 1F ~i . So xx g11 .. . ., xg l EE¡1F . Bu i ~i qij - mil/1 ' mil' mEZ we ge m l-l  m  _ xmxl .. . x l = EE 1F, q .e .d . Some ímes i is possible o gi e a comple e desc ip ion o he solu ion o Y = Log X . This happens when E is a classical di e en ial ield 1 .3 .- Theo em .  . On he hypo hesis o Theo em 1 .2, supose mo eo e ha E = C(z), C he ield o cons an s o E and z iÉC such ha ~ D =,5  Dz E C . Then, any solu ion o he equali y can be w i cn in he o m y=clyl+ . . . +c y +c being 1 q 11 q l xx 1 . ., x " qlk q k k - x 1 . . . x x = x l l . . .  x c'  ,  being c l ,  . . .,  c E Q,  c,  c' EC . P oo : applying he same a gumen used in 1 .2 and aking he sys em we ge he e exis q l ,  . . ., q E Q such ha xx l 1 . . .  x E  l F . Bu any de i á ion has only one ex ension o an algeb aic ex ension o E ([8] Cap . 2, 17, Co . 2) . so C 1F is a ield o cons an s and gi en ha E C F is an ex ension wi h he same ield o cons an s we ha e C = C 1F . The e o e (1) x = x 1 1 . . . x c'  ,  c' E C . De i ing (1) yields q l q D E p  ,  Dy = Dx/ x ° D(x 1  . . ., x  )1(x11 '..  xq ) 1 g 1 Dx 1 / x + . . . + g Dx /x . So 1  y - g l y l + " '  + g y + c  ,  c E C,  q .e .d . Rema k : i can happen ha x 1 1 . . . x 5E F . Howe e , i is an algeb aic poin ha doesn' dis u b he elemen a i y o he p ocesa . 2 .- SOME CowsEQUFNCEs . 2 .A .  The i s conclusion we d aw om 1 . is ha we'll name The I e- dundance o Elemen a y Func ions . This means ha building up elemen a y ex ensions by means o algeb aic elemen s, loga i hm elemen s o exponen- ial elemen s a e comple ly independen p ocesses : no one o hem can be ob ained om he o he s . In o de o se he p oblem we'll use an adecua e language ; we say ha he di e en ial ex ension E C F is Algeb aic i he ield ex en- sion E C F so is ;  i la Loga i hmic i F = E(0 1 ,  . . .,  0 n )  such ha 0 1 = LogT 1 , T 1 EE, 0 1 = Log i , T i EE(0 11  . . ., 0 i-1 )  'Vi >2 . Changing Log by Exp we ha e an Exponen ial ex ension . 2 .1 .- Lemma .  Le E C F = E(0) be a di e en iaiex ension wi h he same ield o cons an s C and 0 Y- E . (1) I VD =á  DO( - =E, hen 0 is anscenden al o e E . (2) I VD E p  DO/ 0 C=E, hen 0 is .algeb aic o e E i and only i he e exis s n C =N such ha O n E E, and he i educible polynomial o 0 o e E is X n - On , n being he leas o he e na u als . P oo : assume 0 o be algeb aic o e E and le P(X) - )In +a 1 X n-1 + . . . + + a n-1 X + an be he i educible polynomial o 0 . Then . (*) On + a 1 0 n-1 + ... + a n-1 0 + a n = 0 . (1) De i ing (*) we ge 1 D E p,  (Da 1 + nD0)O n-1 + ... - 0 . Gi en ha P(X) is he i educiblepolynomial o 0 o e E we ha e ha V D E p Da 1 + nDO =0 .  So  V D C=¿ DO = D(-a 1 /n) and o+al/n is a cons an . Due o E C F ís wi h he same íeld o cons an s we ge O E E, no possible . (2) Now i su ices o p o e ha 0 n C=E . De i ing (*) we ge Y D E A nDO/ OOn + (Da 1 + (n-1)D0/0)O n-1 + ... + Da n = 0 .'Bu an¢ 0,  so Dan = nDO/ 0 aJ D E A . Hence Da n/an - nDO/ 0 .* Dan /a n = DO n / On - D(a n / On) - 0 YDEA . So a n / O nECCE, and On EE, q .e.d . 2 .2 .- Theo em .  Le E be a di e en ial ield wi h ield o cons an s C . Le E C F = (0 1 ,  . . ., 0 ) be an elemen aldi e en ialex ension wi h he same ield o cons an s . Then : (a) W hen F is Loga í hmic, E C F is a pu ely anscenden al ex- ension . I E C F is Exponen ial, E C F is pu ely anscenden al unless he- e exis n i ,  . . ., n E =- Z such ha  0n 1  . .  On E E . 1 '  Le x E E . (b) The equali y Y = Log(x) ne e holds in F-E i E C F is Alge- b aic o Exponen ial . (c) The equali y Y e Exp(x) ne e holds in F-E i E C F is Loga- i hmic, and i he e is a solu ion when E C F is Algeb aic hen he e e- xis s n E N such ha y nE E . Mo eo e , i E = C(z), z 9E C, Dz E C E A being C he ield o cons an s o E, C algeb aica ly,closed, hen he e a e no excep ions o he case (c) . P oo :  (a) The s amen is an easy consequence o Lemma 2 .1 o he Lo- ga i hmic case . Assume ha E C F is Exponen ial and no pu ely anscen- den al ex ension . By Lemma 2 .1 he e exis s Os , p E N such ha OS E E(O1, . . ., 0 s-1 ) .  Le Ok he i s o hem, ha is,  0 1 ,  . . ., O k-1 a e algeb aic independen o e E and Ok E E(O 1 , . . ., 0k-l) . Then by Theo em 1 .2 we ge he s amen . (b) Le y be a solu ion . Then,  Y D E p Dy - Dx/x . Since x E E we can Cake Che di e en ial ex ension E C E(y) . By 2 .1 y is no algeb aic o- e E . Suposse now E C F is Exponen ial . Then, by 1 .2 we ge he e exis . . ., c :  C, n, n , . . ., n E .  Z such ha cl 1 l k  i E  1  lk  n .  n . _  i 1 y + c . . + . . . + c . . EEnF, xn0 . 1 . . . 0 . k EEnF, i l i1  l k lk  il  lk being 0 . , . . ., 0 . a maximal algeb aically independen sys em o e E among il lk 0 1 9  . . ., .O líke in  1 .D . Bu xEE :  so n  , . . . ., n .  a e 0,  and looking in 1 .2  o Che i1  l k cons uc íon o he ena u als we ha e c . _ ... = c . = 11  lk 0 . Hence y E E nF, no possible as we ha e p o ed abo e . (c) The Lemma 2 .1 assu e us ha i y is algeb aic o e E hen he e exis n E N such ha y n CE . Suposse E C F is Loga i hmic . By 1 .2 we ha e he eexis n 1 ,  . . .,  n ,  n E Z,  c l ,  . . .,  c E =- C such ha _  nn _ x + c 1 0i + . . . + c 0 EEnF , y n 1 1 .. . EEnF . Now, by Lemma 2 .1, 0 1 , . . ., O is an algeb aically independen sys em o e E, so c 1 - ... = c - 0, and n 1 - . . . = n - 0 ( look o Che cons uc ion o n 1 ,  . . ., n in 1 .2 ) . Hence y E Én F= E, no possible . On Che assum íon ha E= C(z) ..... ce o aplying Theo em 1 .3 o yn E=- E = CM . Rema k : an example ha gi e us an excep ion o (a) is : E = Q(z)(Exp(2z+2z 2 )), F - E(Expz, Expz 2 ) . Then, Exp(z+z 2 ) E F-E and is algeb aic o e E . 2 .B .  The ques ion o whe he some anscenden al equa i o ns can be sol ed by means o elemen a y u n c i ons some imes can be answe edusing Che S uc u e heo em 1 .2 . Le us see wo classical examples : assume E C F is a di en ial ex ension wi h Che same ield o cons an s . Le C be his ield and E - C(z) such ha 1 D E á Dz E C, z §E C . Suposse C is algeb aically closed and E C F Elemen a y . Conside Che equa ion  aY = Log(BY)  a, B C - E . Che s amen is consequen- Suposse he e ís a solu ion in F, y . Using he same no a ion o 1 .3 e ge c, c E C, c 1 ,  . . ., c n E Q . Passing o he Module o di e en ials, 2F,/E, we ha e (-e í )1/ x . dx í . 1 1 y = cx 1 . . . x nn ,  y - c + ely1 + ... + cnyn , '((x 1 1 . . x n cn)e 1 1/x1 dx 1  +  . . .  +  (x 1 c1  . .  x n . cn a  )c n 1/ xn dx n )° a = e1dy1 + ... + cndyn , a' - ac/B . Bu aking he Module o di e en ials espec on he penul ima- e subex ensionno algeb aic and aking in o accoun 1 .C e ha e ha a' (x 1 1. . . x n n)cn1/x dx n= c n dy n , whé e dx n n se o he elemen a i y o E C F, being one o hem no ze o . The e o e cn= = 0 ; epea ing his a gumen we ha e cí = 0  i . Consequen ly, any solu- íon is i ial . As a pa icula case and aking E _ ¢(z) e ha e ha he equa- ion  Log(Y) = Y/ z  has no solu ion by means o elemen a y unc ions . Wi h he same hypo esis conside now he equa ion (*)  Y + a = BExp(yY),+ V Exp(-YY),  a,  0,0', Y E E . Le y be a solu ion, y E F . We can suposse also ha  Exp(yy) EF . Then by 1 .3 and wi h he same no a ion we ha e (**)  Yy = c + cly1 +  . . .  + cn yn ,  c E C,  c 1 ,  . . .,  en E Q . Subs uíng o yy in (*) we ge ha y + a = Bx 1 1. . . x n n.+ .B'x 1 c1 ... x n cn (whe e e ha e ope a ed adequa ly B, B') . Passing now o he Module o di e en ials RF/E e ge 0 o dy n = 0 becau- dy =  B(x c1 ... xnn)cí1/x i dx i +  B'(x 1 c1 ... xn cn )(-c i )1/ x . dx i . i  i  i Bu aking ín o accoun (**) we ha e ha 1/ Y (c1dy1 + . . . + c n dy n ) _  B(x e1 . . . x en )c i 1/ xi dx í +  B~(x 1 c1 . . . xn cn ) 1 3