What is a hereditary algebra?

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Paula Simpson
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1 What is a hrditary algbra? (On Ext 2 and th vanishing of Ext 2 ) Claus Michal Ringl At th Münstr workshop 2011, thr short lcturs wr arrangd in th styl of th rgular column in th Notics of th AMS: What is? Th following txt is a slightly xpandd vrsion of my rport I should strss that th titl may b considrd as bing slightly odd, it bttr should rad: What is a hrditary ring? sinc for bing hrditary, it is only th ring structur of an algbra which plays a rol But qustions about rings in gnral ar considrd now-a-days as bing obsolt W ar going to outlin a numbr of quivalnt conditions, th most comprhndibl sms to us condition (9); but unfortunatly, this condition hids th origin of th naming thus w bttr start with condition (1) W also will draw attntion to on class of xampls: th path algbras of finit quivrs In th ralm of commutativ algbra, only fw rings ar hrditary, th most prominnt ons ar th Ddkind domains 1 So, whn a ring R is said to b hrditary? Th proprty which is inhritd is th projctivity which is passd down from a modul to its submoduls: A ring R is hrditary if and only if any submodul of a projctiv modul is projctiv, and thr is also th dual proprty: injctivity is passd down to factor moduls: R is hrditary if and only if any factor modul of an injctiv modul is injctiv Submoduls of projctiv moduls ar somtims calld torsionlss, factor moduls of injctiv moduls ar calld divisibl, thus w can rformulat th two conditions by saying that torsionlss moduls ar projctiv, or that divisibl moduls ar injctiv A furthr quivalnt condition is th following: for any R-modul M, thr is an xact squnc 0 P 1 P 0 M 0 with P 0,P 1 projctiv Namly, on th on hand, givn M, thr is a fr (thus projctiv) modul P 0 which maps onto M, and if R is hrditary, thn th krnl P 1 of this map has to b projctiv, as wll On th othr hand, if U is a submodul of a projctiv modul P, lt M = P/U and us Schanul s lmma in ordr to conclud that U is projctiv Of cours, w also hav th dual condition: for any R-modul M, thr is an xact squnc 0 M I 0 I 1 0 with I 0,I 1 injctiv It is usual in homological algbra, instad of looking at an R-modul M, to look at a projctiv rsolution 0 P 1 P 0 0, this is a complx of projctiv moduls with a uniqu homology group, and this homology group is prcisly M Now such projctiv rsolutions ar usd in ordr to calculat drivd functors, for xampl th functors Ext i On obsrvs quit asily: R is a hrditary ring if and only if Ext 2 R = 0 1

2 Th quivalnc of th first two assrtions mntiond abov (that torisonlss moduls ar projctiv and that divisibl moduls ar injctiv) is usually shown by using as intrmdiat condition this vanishing of Ext 2 On th othr hand, th vanishing of Ext 2 is a convnint condition for dfining hrdity in mor gnral sttings, say for daling with ablian catgoris in gnral which may not hav sufficintly many projctiv (or injctiv) objcts To work with Ext 2 is somtims considrd as a burdn (latr w will try to show that on can handl th lmnts of Ext 2 quit asily), thus on oftn tris to avoid Ext 2 In trms of Ext 1, on gts th following rformulations: If g: Y Y is a surjctiv modul homomorphism, thn th inducd map Ext 1 (X,g) is surjctiv, for any modul X If f: X X is an injctiv modul homomorphism, thn th inducd map Ext 1 (f,z) is surjctiv, for any modul Z Bfor w continu, a warning is ncssary Up to now, w hav mntiond moduls without spcifying whthr w man lft moduls or right moduls, and indd in th formulations abov, w always hav to considr both lft moduls and right moduls If w only ar intrstd in say lft moduls, w arriv at th notion of a lft hrditary ring 2 Lft hrditary rings Lt us collct again th various proprtis which w hav listd abov, whr now all th moduls considrd ar lft moduls W also insrt som additional conditions A ring R is said to b lft hrditary providd th following quivalnt proprtis ar satisfid: (1) any submodul of a projctiv modul is projctiv, (1 ) any lft idal is projctiv, (2) any factor modul of an injctiv modul is injctiv (3) for any R-modul M, thr xists an xact squnc 0 P 1 P 0 M 0 with P 0,P 1 projctiv (4) for any R-modul M, thr xists an xact squnc 0 M I 0 I 1 0 with I 0,I 1 injctiv (5) Any lmnt of th drivd catgory D b (modr) (or vn D(ModR)) is isomorphic to its homology (6) Ext 2 R = 0 (7) If g: Z Z is a surjctiv modul homomorphism, thn th inducd map Ext 1 (X,g) is surjctiv, for any modul X (8) If f: X X is an injctiv modul homomorphism, thn th inducd map Ext 1 (f,z) is surjctiv, for any modul Z Condition (1 ) (a thorm of M Auslandr) assrts that in ordr to chck whthr R is hrditary, it is sufficint to considr in (1) just on singl projctiv modul, namly RR, th fr modul of rank 1 It was askd in Cartan-Eilnbrg (1956) whthr lft hrditary rings ar also right hrditary Th first countr xampl was givn by Kaplansky in 1958, asir xampls wr givn latr (1961) by Small, for xampl th matrix ring R = [ Q Q 0 Z 2 ],

3 with condition (1 ) on asily shows that R is lft hrditary Howvr, th opposit ring R op = [ ] Z Q, 0 Q is not lft hrditary, sinc th radical of R op is (flat, but) not projctiv But on should know that a lft and right nothrian ring is lft hrditary if and only if it is right hrditary 3 Path algbras of finit quivrs Lt k b a fild Claim: Th path algbra kq of a finit quivr Q is hrditary Thr ar various ways to prov this Lt us outlin a proof using th condition (7) Thus, ltq = (Q 0,Q 1,s,t)bafinitquivr, hrq 0,Q 1 arfinitstsands,t: Q 1 Q 0 arst-maps RcallthatarprsntationofQisofthformM = (M x,m α,withvctor spacsm x forallx Q 0 andlinarmapsm α : M s(α) M t(α) IfM,M arrprsntations of Q, a homomorphism f = (f x ): M M is givn by a family of linar maps f x with x Q 0 satisfying crtaincommutativityrlations To b prcis, th homomorphism spac Hom(M,M ) is th krnl of th following map x Q 0 Hom k (M x,m x ) α Q 1 Hom k (M s(α),m t(α) ), (f x ) x (M αf s(α) f t(α) M α ) α and it turns out (and is not difficult to s) that th cokrnl is prcisly Ext 1 (M,M ) Thus, w dal with th following xact squnc: 0 Hom(M,M ) x Q 0 Hom k (M x,m x) α Q 1 Hom k (M s(α),m t(α) ) Ext1 (M,M ) 0 Now assum that thr is givn a furthr rprsntation M of Q and a surjctiv homomorphism g = (g x ) x : M M (and th surjctivity mans that all th vctor spac maps g x ar surjctiv), thn w obtain th following commutativ diagram with xact rows: α Q 1 Hom k (M s(α),m t(α) ) Ext1 (M,M ) 0 Hom(M s(α),g t(α) ) Ext 1 (M,g) α Q 1 Hom k (M s(α),m t(α) ) Ext1 (M,M ) 0 Th vrtical map Hom(M s(α),g t(α) ) on th lft is surjctiv, thus also Ext 1 (M,g) is surjctiv 4 What is Ext 2? It is wll-known and oftn usd that for any ablian catgory th first drivd functor Ext 1 of th Hom-functor can b dfind using quivalnc classs of short xact squncs: th Bar-dfinition of Ext 1 (not that th Bar dfinition assums that w obtain rally a st, not a class but othrwis w go to anothr univrs ) 3

4 Similarly (at last for small catgoris), thr is a Bar dfinition of Ext n, for n 2 using long xact squncs For xampl, for Ext 2 (and this is what hr w ar intrstd in), w us long xact squncs of th form 0 Z M 2 X 0, again, w nd quivalnc classs givn by maps btwn such squncs which ar th idntity on X and on Z But not: such maps ar no longr invrtibl, thus w hav to tak maps in both possibl dirctions this is th origin of th construction of th drivd catgory (calculus of fractions), and to look at Ext 2 is a convnint playing ground for gtting familiar with quasi-isomorphisms Lt m strss that daling with such xact squncs is nothing ls than looking just at morphisms f: M 2 in th catgory (th clumsy notation of writing down a long xact squnc givs nams to th krnl and th cokrnl), or, to formulat it mor fancy, w considr complxs of th form with on map f and othrwis using zro moduls and ( 0 M 2 0 ) with fixd homology) Thus, lt us dscrib th vanishing of Ext 2 in trms of maps f: M 2 Hr it is: A ring R is lft hrditary if and only if (9) for any homomorphism f: M 2 with pi-mono-factorization Y u M2, thr is a pushout-pullback diagram of th form Y u u M M 2 Bfor w look at th proof, lt us indicat som way of visualizing of what is going on: Givn th map f: M 2, dnot by Z its krnl, by Y th imag and by X th cokrnl Thus, w dal with two short xact squncs 0 X Y 0, and 0 Y u M2 Z 0 and f = u W may dpict this as shown on th lft And w ar looking for a modul M and maps u, as shown on th right: X Y M 2 Y Z f = u X Y M Z M 2 Y Z u Now, w show th quivalnc of (8) and (9) First, lt us start with a homomorphism f: M 2 with pi-mono-factorization Y u M2 and dnot by Z th krnl 4

5 of f, thus of If th condition (8) holds, thn th lmnt [0 X Y 0] of Ext 1 (Y,Z) is in th imag of Ext 1 (Z,u), thus thr is a commutativ diagram of th form 0 Z Y 0 u u 0 Z M M 2 0, this provids th rquird pushout-pullback diagram Convrsly, assum condition (9) is satisfid, lt u: M M b an injctiv homomorphism W want to show that Ext 1 (u,z) is surjctiv Thus tak an lmnt [0 Z M 0] in Ext 1 (M,Z) and apply condition (9) to th map u: M with its givn pi-mono-factorization: w obtain a pushout-pullback-diagram u M M u M If w add on th lft Z as th krnl of both and, w obtain th following commutativ diagram with xact rows 0 Z M 0 u u 0 Z M M 0, which shows that th uppr squnc is th imag of th lowr squnc undr Ext 1 (u,z) Mor prcisly, on ss: Ext 2 (X,Z) = 0 if and only if (9) is satisfid for all maps f with krnl Z and cokrnl X It may b instructiv to point out in which way a pushout-pullback diagram as givn in condition (9) provids diffrnt rprsntativs of an lmnt of Ext 2 (X,Z) Thus, considr th following pushout-pullback diagram u M Y u M 2, and thrfor 0 Z Y 0 u u 0 Z M M 2 0, with xact rows In addition, thr is also th xact squnc (0 Y u M 2 Z 0) If w form th xact squnc inducd from it by u, w obtain an xact squnc which is split xact: 0 Y u 0 M 2 N u M 2 X 0 5 p X 0

6 (hr, p is a split pimorphism) Concatnation of th xact squncs yilds 0 Z M 2 X 0 0 Z M N p X 0, in this way w s that th two rows yild th sam lmnt in Ext 2 (X,Z), and sinc p is split pi, this is th zro lmnt of Ext 2 (X,Z) 5 Moduls of finit lngth Sinc th functor Ext 2 is half xact in both variabls, th vanishing Ext 2 (T,S) = 0 for all simpl moduls S,T implis that on has Ext 2 (X,Z) for all finit lngth moduls X,Z So what dos it man that Ext 2 (T,S) = 0 for simpl moduls S,T? Lt I(S) b th injctiv nvlop of S and lt us assum from now on that thr xists a projctiv covr P(T) (this holds tru in cas w dal with an artinian ring, or mor gnrally, a smi-prfct ring) Lmma Any lmnt in Ext 2 (T,S) has a rprsntativ 0 S M 2 T 0, whr is a (non-zro) submodul of I(S), and M 2 is a (non-zro) factor modul of P(T) To say that is a non-zro submodul of I(S) mans that is a modul with simpl ssntial socl S; dually, to say that M 2 is a non-zro factor modul of P(T) mans that M 2 is a modul with a small maximal submodul with factor modul T Proof: W start with th xact squnc 0 S m f q M2 T 0, and th inclusion v: S I(S) Sinc m is a monomorphism and I(S) is injctiv, thr is v : I(S) with v = v m Lt v = v h, whr h: is surjctiv and v : I(S) is injctiv Sinc v hm = v m = v is a monomorphism, also hm is a monomorphism Thus, thr is th following commutativ diagram with xact rows m 0 S h 0 S hm M 1 Y 0 which w can complt by insrting a cokrnl Y of hm to obtain: m 0 S h 0 S Y 0 h hm M 1 Y 0 6

7 Sinc by assumption h is surjctiv, also h is surjctiv Nxt, w start with th xact squnc 0 Y u M 2 T 0 and form th squnc inducd by h : m 0 Y M 2 h h 0 Y M 2 T 0 T 0 By concatnation, w obtain th following commutativ diagram with xact rows: m f 0 S M 2 T 0 h h 0 S hm M 1 M 2 T 0, thus, th two squncs yild th sam lmnt of Ext 2 (T,S) This shows that w can rplac th uppr squnc by th lowr squnc: hr, is a submodul of I(S) and in addition, M 2 is a factor modul of M 2 By duality, w similarly s that w can rplac th scond sris by a third on 0 S M 1 M 2 T 0, whr M 2 is a factor modul of P(T) and whr M 1 is a submodul of M 1, thus of I(S) This complts th proof It rmains to analyz th vanishing of Ext 2 (T,S) As w just hav sn, w hav to look at maps f: M 2 with krnl S and cokrnl T such that S is an ssntial submodul of and th imag of f is a small submodul of M 2 If th condition (9) holds, w obtain a modul M with as a submodul, such that M/S is isomorphic to M 2 Lt us assum in addition that f is non-zro: thn M is indcomposabl It is vn what w call a diamond: a modul M with submoduls M M such that both M and M/M ar simpl and such that any propr non-zro submodul of M contains M and is containd in M (Proof: Tak M = and M = soc, thn clarly both M and M/M ar simpl Lt N b a propr non-zro submodul of M Considr M N If M N = 0, thn N must b simpl and M = M N, and thrfor M 2 = M/M = M /M N But by assumption, M 2 is a modul with a small maximal submodul, thrfor M /M = 0, but thn f = 0, in contrast to our assumption This shows that M N 0, and thrfor N contains th socl M of M Th dual argumnt shows that N is containd in M ) W can rphras th prvious considrations as follows: Th vanishing of Ext 2 (T,S) mans th xistnc of all possibl diamonds with socl S and top T Convrsly, w strss that if Ext 2 (T,S) 0, this may kill not only on, but usually a lot of diamonds Finally, on should b awar that whn daling with quivrs with rlations, th rlations in qustion just corrspond to non-zro lmnts of Ext 2 (so that in this cas th 7

8 vanishing of Ext 2 may b dscribd by saying that thr ar no rlations) Rcall that a rlation ρ for th quivr Q is by dfinition a non-zro linar combination ρ of paths of lngth at last 2 starting in som vrtx x and nding in som vrtx y (such a rlation is an lmnt of th path algbra kq) For xampl, if ρ is a singl path, thn this is calld a monomial rlation or also a zro rlation; if ρ is th diffrnc of two paths, on says that ρ is a commutativity rlation 8

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