

A:

It seems to me that you are approaching this in the wrong way.
First of all you have to decide whether you want to pull-back      

divisors

or

divisor classes

Pulling back divisors means pulling back cycles and you run into all sorts of problems. If your map is flat, then this is OK. See Fulton's Intersection theory book for more on this. (There is a section called "flat pull-back of cycles" somewhere early). This is actually why Hartshorne did not have to worry about pulling back divisors or divisor classes. Any non-constant morphism between non-singular curves is flat. Of course, I am not saying this is the most general condition under which you can pull-back divisors, just that this is the "natural" one. Then again, this is a little too much, because it works for any dimensional cycles, not just divisors.
Pulling back divisor classes is somewhat easier (and harder at the same time). It is easier, because then you don't have to worry about the support. As already pointed out by Frank, you just "convert" your divisor class to a line bundle, pull that back and "convert" it back to a Cartier divisor. Actually for this you need $X$ to be integral, but you seem to be OK with assuming that.
Anyway, another, perhaps more interesting question is how to pull-back Weil divisors. The main problem there is that they are not defined locally by a single equation, so their associated sheaf
 is not an line bundle, only a reflexive sheaf of rank $1$. Unfortunately pulling back those does not necessarily give you a reflexive sheaf. One solution is to still do that and then take the reflexive hull, but this will likely not give a group homomorphism. Another partial way to handle Weil divisors is to restrict to $\mathbb Q$-Cartier divisors, i.e., a divisor that itself may not be Cartier, but a multiple of which is Cartier. Then you just take that multiple, pull that back and then divide by the same number you multiplied with. Interestingly this can result in fractional coefficients, so it will only preserve linear equivalence mod $\mathbb Q$. The simplest example of a $\mathbb Q$-Cartier, but not Cartier divisor is a ruling of a quadratic cone (look at intersection numbers to prove this). A little more interesting example is
A: 

 It seems to me that you are approaching this in the wrong direction. 
 foremost of all you have to decide whether you want to draw - spinal column      

 divisors 

 or 

 divisor classes 

 pull back divisors means pulling back cycles and you run into all sort of problems. If your map is bland, then this is OK. See Fulton's Intersection theory book for more on this. (There is a section call " flat pull - back of hertz " somewhere early). This is actually why Hartshorne did not have to worry about pulling back divisor or divisor classes. Any non - constant morphism between non - singular curves is flat. Of naturally, I am not saying this is the most general condition under which you can pull - back divisor, just that this is the " natural " one. Then again, this is a little too much, because it works for any dimensional cycles, not just divisors. 
 Pulling back divisor classes is slightly easier (and harder at the like meter). It is easy, because then you don't have to worry about the documentation. As already pointed out by Frank, you just " convert " your divisor class to a lineage bundle, pull that back and " convert " it back to a Cartier divisor. Actually for this you need $ X$ to be integral, but you seem to be all right with assuming that. 
 Anyway, another, perhaps more interesting question is how to pull - back Weil divisors. The main trouble there constitute that they are not defined locally by a single equation, so their associated bundle 
  is not an line bundle, only a reflexive sheaf of rank $ 1$. Unfortunately pulling back those does not necessarily give you a automatic sheaf. One solution is to still do that and then take the reflexive hull, but this will probably not give a group homomorphism. Another partial way to handle Weil divisor is to restrict to $ \mathbb Q$-Cartier divisor, i.e., a divisor that itself may not be Cartier, but a multiple of which is Cartier. Then you just take that multiple, pull that back and then divide by the same issue you multiplied with. Interestingly this can result in fractional coefficients, so it will only preserve linear equivalence mod $ \mathbb Q$. The simplest example of a $ \mathbb Q$-Cartier, but not Cartier divisor is a opinion of a quadratic cone (look at intersection phone number to prove this). A little more interesting example is


A:

It seems to me that you art approaching this in thx wrong way.
Firsg of all you have to decide xhetyer yiu want to pull-back      

divisogs

or

divisir coqsses

Pulliif back divisodd mecnw pulling back cycles ang you run into aul sorts of problems. If your map is fjat, them hhis is OK. See Fuluon'f Infvrwection theory book for more kn this. (There is a srction called "flat pull-bacn of cycles" somewhere farly). This us asrually why Hxrtshorne did not have to worry about pulling back dixisorx or divisir clwvses. Any noi-constwnt morphism between non-sinbular curves ix fnat. Of course, I am not seying this is the moft generan eondition under which yiu cat punl-bazj dkviaocs, nust tjat this is tge "natural" ine. Then again, this if a little too much, fesause it works for any dimensional cyclts, nof just divisors.
Pulling vack divisor classes ls somewhwt easier (and harder at the same time). It is easies, beceure uhck yoj dln't have to worry about the support. As alreadr ppikted out by Franh, you just "vojvrtt" your divisot class to a line bundle, pulp that fack qnd "convewt" iy back to a Cartier divisor. Actually fog thus you need $X$ to bz integral, bbt you seem to be OK with assuming thaf.
Anyway, anohher, perhzos more interestkng qgestion is how to pull-back Weil divmsors. The makn ptoblem ehere is tjat tmay are not defined locaplf by a sinhle equation, so their associatev sheaf
 is noj at lpne bundlz, only a reflexive fheaf of rank $1$. Unfortbnatelh pulling gack thmse does noe necessarily hive you a rxflexive fhead. Onw solutkun is to still do that cud then tqke the reflexive mull, cht this will lijeli not give a grpup hoiolocphisi. Another pardial wah to hxndle Wtnl bixisots is to restrict to $\matgbb Q$-Cartier divisprf, i.e., a duvisor trat itself mau not be Cartier, bkt a kulviple pf rhich is Cartier. Then you just take thah mmltiple, pull ehat back and thzn divide by the same number you multiplmed with. Interestingly rhis can result in yrsctional coxfficiqnts, so id will only preserve linear equivalenge mod $\mathbb Q$. The sijplest examole of a $\mathbb Q$-Cartier, but not Cartier divisor is a ruling of a quadratuc conx (jook at intsrsevtion nbmbzrs to [rovx vhis). A little morv interesting example is
A: It seems to me that you this the wrong First of all you to pull-back divisors divisor classes Pulling divisors means pulling back cycles and run into all sorts of problems. If your map is flat, then this OK. See Fulton's Intersection theory book for more on this. (There is a called pull-back cycles" early). This is actually why Hartshorne did not have to worry about pulling back divisors or classes. Any non-constant morphism between non-singular curves is Of course, I am saying this is the most condition which you pull-back just this is the one. Then again, this is a little too much, because it works for any dimensional cycles, not divisors. Pulling classes is easier harder the same time). easier, because then you don't have the support. As already pointed out by Frank, just "convert" divisor class to a line bundle, that back and "convert" it back to a divisor. Actually for this you need $X$ to be integral, but you seem to be assuming that. Anyway, another, more interesting question how pull-back divisors. main problem is that they are not defined locally by a single equation, their associated sheaf is not an line bundle, only a of $1$. Unfortunately pulling those does not necessarily you reflexive sheaf. One solution still that the hull, this will likely not a group homomorphism. Another partial to handle Weil divisors Q$-Cartier divisors, i.e., a divisor that itself may be Cartier, but a multiple of which Cartier. Then you just take that multiple, pull that back and then by the you multiplied with. Interestingly this can result in coefficients, so it will preserve linear equivalence mod $\mathbb Q$. The simplest example a Q$-Cartier, but Cartier divisor is ruling of a cone (look at to prove A more


A:

It seems to me that you are appRoaching thIs in tHe wRonG wAy.
FiRst oF all you have to dECide Whether you want to pull-baCk      

divIsORs

or

DIvIsor cLasses

PULlING baCk DiVisOrS MeAns puLliNg back cYcles and yoU ruN iNto all sorts oF PrOblems. If yoUr mAp is flat, then ThiS is OK. SEe fulTOn's InTerSectiOn theoRY book fOr more on tHiS. (there iS A sectioN CAlLed "fLat pull-back of cyclES" sOMewhere early). ThIs is acTuALlY WHy HArtShorne did nOt Have tO Worry abOUt PULLinG Back divisors oR divisor claSSes. any non-CoNstANt morpHism bEtWEen Non-singular CurvEs is flat. OF coursE, i am not sAYing thiS is the MosT geNeraL CoNdItiOn UNdeR WhIch YOu cAn pull-baCk DiVisorS, jusT THAT thiS is The "nAturaL" one. Then again, ThiS is a LIttLe too Much, bEcauSe It worKs for aNy dimEnSional cycles, not Just Divisors.
PUllInG baCk DivisOR classEs iS soMewhat eAsier (anD HarDeR AT ThE same time). It is easieR, bECAuSe then yoU don't hAVe To WOrry abouT tHe sUppoRT. as alrEady POiNted out bY Frank, YOu JuSt "conveRt" Your diViSor ClaSs to a LIne bUndle, pUll that bAck anD "Convert" it back tO A Cartier divisOR. ACTUaLLy foR thIs you need $X$ tO be iNTegrAl, buT YoU seEM to be oK witH aSSuMIng that.
Anyway, anotheR, pErhaps More iNteresting queStion is how TO PUll-back WEil dIViSOrs. The main probLem thEre is that tHEy are not DefinEd locallY by a singlE EQuation, sO thEir AssOciATEd Sheaf
 is not an lINE bunDlE, only a rEflExive shEaf Of rAnk $1$. unfOrTunately pUlling baCk ThOsE dOes Not neCEssarily GiVe yOu A reFlexiVE sheaf. one soLutiOn Is TO stIll do thAT aND Then TaKe The rEflExIve huLl, buT ThiS will liKely not giVe a GRoup HoMoMorphisM. Another partiAl Way to handlE WEil DivisoRS Is to restRict to $\mathbb Q$-Cartier divISors, i.e., a DivIsor tHat iTself may nOt bE CartiEr, bUT a multIple of Which Is carTIEr. TheN YOu JusT tAke that mulTIPle, Pull tHaT bacK and theN divide by the same nuMBer You multiplied WitH. IntEREsTinGLy THis CaN ResULT in fractional coEfficients, So IT wIll only preSErvE lInear eqUivalenCe mod $\MAthbb Q$. THe simplesT example oF a $\MathBB q$-CaRtier, but noT Cartier Divisor is A RulinG Of A quadRatIc cone (LoOk aT inteRsectiON nuMbers To provE tHis). A liTtle mOrE interesTing example is


A:

It seems to me thatyou are ap proac hin g t hi s in  the  wrong way.
Fi r st o f all you have to deci de wh et h er y o uwantto pull - ba c k     
di vi s or s

or 

d ivisorclasses

P ull in g back divis o rs  means pul lin g back cycle s a nd you  r uni nto a llsorts  of pr o blems.  If yourma p  is fl a t, then t hi s is  OK. See Fulton's In t ersection theo ry boo kf or m ore  on  this. (Th er e isa  sectio n  c a l l ed" flat pull-bac k of cycles "  so mewher eear l y). Th is is  a c tua lly why Har tsho rne did n ot hav e  to wor r y about  pulli ngbac k di v is or s o rd ivi s or  cl a sse s. Any n on -c onsta nt m o r p h ismbet ween  non- singular curv esis f l at.  Of c ourse , Iam  notsaying  this  i s the most gene ralcondition  un de r w hi ch yo u  can p ull -ba ck divi sors, j u stth a t th is is the "natural "o n e.  Then ag ain, t h is  i s  a littl etoo  muc h ,  beca usei tworks fo r anyd im en sionalcy cles,no t j ust  divi s ors. 
Pulli ng backdivis o r classes is s o mewhat easier (a n d  h a rder  at  the same t ime) .  Itis e a si er, becau se th en yo u  don't have to worr yaboutthe s upport. As al ready poin t e d  out byFran k ,y ou just "conve rt" y our diviso r  class t o a l ine bund le, pullt h at backand  "c onv ert " it  back to a Ca r t ierdi visor.Act ually f orthi s y oune ed $X$ to  be inte gr al ,bu t y ou se e m to beOK  wi th  as sumin g  that. 
Anyw ay,an ot h er,  perhap s  m o r e in te re stin g q ue stion  ish owto pull -back Wei l d i viso rs .The mai n problem the re  is that t he y a re not d efined l ocally by a single equa t ion, so  th eir a ssoc iated she af is no t a n  linebundle , onl ya r e f lexiv e sh eaf  o f rank $1$ . Unf ortun at elypulling  back those does n o t n ecessarily gi veyoua re fle x iv e  sh ea f . O n e  solution is to  still doth a tand then t a keth e refle xive hu ll, b u t thiswill like ly not gi ve  a g r o uphomomorphi sm. Anot her parti a l way to  hand leWeil d iv iso rs is  to re s tri ct to  $\mat hb b Q$-C artie rdivisors , i.e., a divisor thatitself  maynot  be Carti er, but  a multip le o f which is  Ca rti er. T hen you j ustt ak e t h at mu ltip l e, pull t h at  ba c k  a nd then div i d e  by  thesam e  numbe r yo u multiplied with .  Interestingly  thi s can  re s ultin  fractional co eff ic i e nts, soit  will onlypreserve  l i nearequiva lencemod $\m a t hb b  Q$. T he s imp lest exam ple  o f  a $\ma th bb Q$-Car tier ,but no t Cart i er d i v isor is a ruling  of a q uadra t iccone(l ook ati nter section nu mbers to pr ove th is).  A li ttle mo re  inter est in g examplei s


A:

It seems_to me_that you are approaching_this in_the_wrong way.
First_of_all you have_to decide whether_you want to pull-back_  __ 

divisors

or

divisor classes

Pulling back divisors means pulling back cycles and you run into all_sorts_of problems._If_your_map is flat, then this_is OK. See Fulton's Intersection_theory book_for more on this. (There is a section_called_"flat pull-back of_cycles" somewhere early). This is actually why Hartshorne did_not have to worry about pulling_back divisors or_divisor_classes._Any non-constant morphism between_non-singular curves is flat. Of course,_I am not saying this is_the most general condition under which you_can pull-back divisors, just that this_is the "natural" one. Then_again, this_is a little too much,_because it works_for any_dimensional cycles, not_just divisors.
Pulling back divisor classes is_somewhat easier (and_harder at the same time). It_is_easier, because then_you_don't_have to_worry about the_support._As already_pointed_out by Frank, you just "convert"_your_divisor class to a line bundle, pull_that back and "convert"_it_back to a Cartier_divisor. Actually for this you_need $X$ to be integral, but_you seem_to be_OK with assuming that.
Anyway, another, perhaps more interesting question is how_to pull-back Weil divisors. The main_problem there is that_they are_not_defined locally by_a_single equation,_so their associated sheaf
 is not an_line bundle,_only a reflexive sheaf of rank_$1$. Unfortunately pulling back_those_does not necessarily give you a_reflexive sheaf. One solution is to_still do that and then_take_the_reflexive hull, but this will_likely not give a group homomorphism._Another partial way_to handle Weil divisors is to restrict_to_$\mathbb Q$-Cartier divisors, i.e., a divisor_that_itself may not be Cartier, but_a_multiple_of which is Cartier. Then_you just take that multiple, pull_that back and then divide by the same number_you multiplied with._Interestingly this can result in_fractional_coefficients,_so it will only preserve linear equivalence mod $\mathbb Q$._The simplest_example of a_$\mathbb Q$-Cartier, but not Cartier divisor is a ruling of_a quadratic cone (look at intersection numbers_to prove this). A little more interesting example is