1. (a) If u (I : R J), there exists c 0 in R such that for all q 0, cu q (I : R J) [q] I [q] : R J [q]. Hence, if j J, for all q 0, j q (cu q ) =

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1 Math 615, Witer 2016 Problem Set #5 Solutio 1. (a) If u (I : R J), there exit c 0 i R uch that for all q 0, cu q (I : R J) [q] I [q] : R J [q]. Hece, if j J, for all q 0, j q (cu q ) = c(ju) q I [q], o that ju I = I. Sice thi hold for all j J, u I : R J. (b) If thi i fale, we ca fid a Noetheria dicrete valuatio rig V R uch that f/g / V. Sice, i V, f (gv ), it will uffice to how that gv i tightly cloed. If g = 0 thi i clear. Otherwie, chooe cd 0 uch that cf q g q V for all q 0. The ord (c) + q ord (f) q ord (g) ad ord (f) ord (f) ord (c)/q for all q 0. It follow that ord (f) ord (g), a required. (c) If f = 0, thi i clear. Aume f 0. (fi) (fr) = fr by (b). But if r R, fr (fi) iff for ome ozero c ad all q 0, c(fr) q (fi) [q] = f q I [q], i.e., cr q I [q]. Thu, fr (fi) iff r I, ad the reult follow. 2. (a) We omit the ubcript 1. Multiplcatio by y carrie R iomorphically oto yr (ice y i a ozerodivior) ad xr R iomorphically oto xyr. Hece, the map iduced by multiplicatio by y carrie R/xR iomorphically o yr/xyr R/xyR. I particular, the pecified map i a ijectio, ad mut take A x = A R/xR m ito A xy = A R/xyR m. Suppoe u i R i uch that mu xyr, ad u i it image i R/xyR. The xu xyr, ad ice x i a ozerodivior, u yr. Hece, u AyR/xyRm, ad the iomorphim R/xR = yr/xyr how that u mut be the image of a elemet of A x. (b) Uig the parethetical commet, we may reduce the problem at oce to the cae where the two ytem are the ame except for oe elemet. Sice the order of the parameter i ot relevat, we may aume that the ytem are x 1,..., x 1, x ad x 1,..., x 1 y. The iue i uaffected if we replace R by R = R/(x 1,..., x 1 )R ad x, y their image i R, each of which i a parameter for the oe-dimeioal Cohe-Macaulay rig R. The reult i the immediate from part (a). (c) It uffice to prove the reult whe all the t j except t i are equal to oe: oe may the apply that reult time, oe time for each x j, to obtai the deired fact. We may proceed by placig x j lat i the equece, ad workig modulo the other elemet of the equece, a i part (b), to reduce to the cae where = 1. The reult the follow from takig x to be the image of x j ad y to be x t j 1 j. 3. A i Problem 2., it uffice to coider the cae where the two ytem of parameter are the ame except for oe elemet, ad oe ca compare each of I = x 1,..., x 1, x ad x 1,..., x 1, y with J = x 1,..., x 1, xy. If a m-primary ideal I i ot tightly cloed, let u be a elemet of the tight cloure ot i I. The either mu I, are we ca chooe z 1 m uch that z 1 u / I, while it i till true that z 1 u I. Repeatig thi proce fiitely may time, we ca fid a elemet u I I uch that mu I: the proce caot cotiue more tha N tep, where m N I. Thu, I fail to be tightly cloed if ad oly if there i a elemet of I : R m that i i I I. By the reult of Problem 2., we have a bijectio betwee (I : m)/i ad (J : m)/i iduced by multiplicatio by y. Thu, it uffice to how for u I : m that u I if ad oly if yu J. But c(yu) q (x q 1,..., xq 1, xq y q ) iff cu q (x q 1,..., xq 1, xq y q ) : y q = (x q 1,..., xq 1, xq ) (oe may check thi workig mod

2 2 A = (x q 1,..., xq 1 ), where it follow from the fact that x ad y are both ozero divior mod A), ad the reult i ow immediate. 4. A i the bracketed uggetio, there i a hort exact equece 0 B/(X 1,..., X 1, Z) f B/(X 1,..., X 1, Y Z) g B/(X 1,..., X 1, Y ) 0 where g i the quotiet urjectio ad f i iduced by multiplicatio by Y o the umerator. Note that each ideal i a deomiator i geerated by a regular equece i B. If C = B/(X 1,..., X 1 ) = Z[X, Y ], thi i imply the hort exact equece 0 C/ZC C/Y ZC C/Y C 0 that oe get by otig that Y C/Y ZC = C/ZC. Now apply B M. The log exact equece for Tor give the required reult, oce oe ote that for each quotiet, the Kozul complex give a free reolutio, o that every Tor i the log exact equece ca be viewed itead a a Kozul homology module. 5. If we take miimal geerator for A ad miimal geerator for B, together they give miimal geerator for A B, ad we fid that a miimal module of yzygie for A B ca be obtaied a the direct um of miimal module of yzygie for A ad B. The firt miimal module of yzgie of K i m. To obtai the ecod miimal module of yzygie, ote that ax + by = 0 implie that a ad b are both i m (or ele x ad y are ot miimal geerator of m. But m kill x, o that thi i equivalet to a m ad by = 0, ad the b xr = K. Thu, the ecod module of yzygie i m K. Suppoe that we have how the i th module of yzygie i iomorphic to K a i m b i. The the i + 1 t module of yzygie i m a i (K m)b i = K b i m a i+b i. It follow by a traightforward iductio o i that the miimal i th module of yzgie of K i K f i 1 m f i, where f i i the i th Fiboacci umber, determied recurively by the rule f 0 = f 1 = 1 ad f i+1 = f i + f i 1 for i 1. The rak of the i th free module i a miimal reolutio i the ame a the leat umber of geerator of the i th module of yzgie, which f i 1 +2f i = f i 1 +f i +f i = f i+1 +f i = f i+2, ad thi i the ame a the dimeio of Tor R i (K, K) a K-vector pace. 6. If we map R R M o that (r, ) rx + y the kerel coit of {(r, ) R 2 : rx + y = 0}. Thi implie that i K[x, y], (r, ) i a multiple of v = (y, x), which implie that the kerel i (Rx + Ry)v = M It follow that all the miimal module of yzygie are = M, ad that the reolutio ha the form R 2 A R 2 A R 2 R 2 > A R 2 M 0, where the matrix of A i w(x y), i.e., ha colum xw ad yw, where w i the trapoe of v, i.e., v writte a a colum. Thu, ( ) for i 1, Tor R i (M, M) = Ker (h)/im (h), a where h i the map M 2 M 2 iduced by A. M b 2 i i the kerel iff xa+yb = 0, i.e., it i a elemet of M 2 that ha the form cw, where c R. The image of M 2 i all elemet of the form w(xm 1 + ym 2 ) where m 1, m 2 xr + yr, ad thi yield w(x 2, xy, y 2 )R. The quotiet i a copy of K paed by the image of w, o that Tor R i (M, M) = K, i (a) Thi follow from takig the alteratig um of all the legth i the log exact equece for Kozul homology (there are oly fiitely may ozero term). (b) By part (a), thi reduce to the cae where M = K. If ay of the x i i a uit, the all the Kozul homology vaihe, ice the x i kill the Kozul homology. Therefore, we

3 may aume that all the x i are i the maximal ideal, o that their actio o K i the ame a multiplicatio by 0, the 0 map. Hece, if 1 (ot jut 2), the Kozul homology for K i idetical with the Kozul complex, ad the alteratig um of the legth i j=0 ( 1)j( ( ) j), which i the ame a 1 + ( 1) = 0. (c) Let h i be the legth of H i (x 1, x 2 ; A). By part (a), h 1 = h 0 + h 2. Sice h 2 i the legth of aihilator of (x 1, x 2 ) i A, it i poitive. Thu, h 1 > h 0. If h 1 i cyclic, it would be a homomorphic image of A/(x 1, x 2 ), which ha legth h 0, ice it i a cyclic module killed by x 1 ad x 2. Thu, h 0 h 1, a cotradictio. 8. The iue i whether we ca chooe c R {0} uch that c(x a 1 ) q I [q] = (x a 1q 1,..., x a 1q 1 )R for all q = pe 0. Let a = a. If we write q = a r with 1 r < a, o that /q = 1/a+r/aq with r/a < 1. we ca rewrite thi a c(x a ) q x r I [q], ad ice 1, x,..., x a 1 i a free bai for R over the polyomial ubrig T = K[x 1,..., x 1 ] ad x a i the ame, up to ig, a g = x a xa 1 1 T, thi i equivalet to whether we ca chooe c R {0} uch cg q I q := (x q 1,..., xq 1 )T for all q > 0. Note that I [q] = I q R. Suppoe that we take c = 1 if α < 1 ad if α = 1 we take c = x d 1 1 xd 1 where the d i are choe o that r/a < 1 i=1 d i/a i. Firt uppoe α 1. For every partitio q = b b 1, if ay d i + a i b i q we have that the correpodig term i the multiomial expaio of cg i i I [q]. Otherwie, b i < q/a i, ad the q = 1 i=1 b i < 1 i=1 (q/a i d i /a i ), ad the 1 /q 1 i=1 d i/a i )/q + 1 i=1 1/a i. 1 (r/a d i /a i )(1/q) + α. If α < 1, thi will be fale for q 0. c = 1 thi will be fale for all q, ice the firt term o the right i egative. It remai to how that we caot chooe c for α > 1 ad p 0. i fact, we hall how thi wheever p > ( + 2)/(α 1) ad 4a, which we aume from ow o. We may replace K by it algebraic cloure: if there i o value of c that work i the larger domai, there i clearly o value that work i the origial rig. We may aume that p 0 doe ot divide ay a i, ad, i particular, we may aume that p > a. We ext how that if c R {0} ad cu q J [q], the we ca take replace c by x 2a 2 ad the ame tatemet hold. Let T = K[x 1,..., x 1 ]. Sice R i itegral over T, the elemet c ha a multiple i T 0 (the cotat coefficiet of a leat degree moic polyomial over T that c atifie), ad o we may aume without lot of geerality that c T. The moomial i x 1/p 1,..., x 1/p with all expoet at mot p 1 form a free bai for T 1/p over T. Sice x i a eparable elemet over the fractio field of T, the ame moomial pa a free R-module G 1 R 1/p (i fact, G 1 i a rig, ad i the ame a R[T 1/p ], although we wo t ue thi). We ext ote that x a 1 multiplie R 1/p ito G 1. It uffice to how that x a 1 x i/p G 1 for 0 i p 1, i.e., that ((x ) 1/p) ap p+i) G1, 0 i p 1. Sice (x 1/p ) a G 1 (it i the p th root of x a, which i equal to a elemet of T ), ad (x 1/p ) p = x G 1, it i eough to how that each ap p + i i i the emigroup geerated by a ad p. Sice p i ivertible mod a, oe the elemet 0, p, 2p..., (a 1)p ha the ame reidue a i mod a 1, ay jp, ad the ap p + i jp i a oegative multiple of a, ay k, which yield the reult ice ap p + i = jp + ak. Let G e deote the R-pa of the et M e of moomial i x 1/pe 1,..., x 1/pe 1 uch that every expoet occurrig i the moomial i at mot p e 1. (Agai, G e = R[B 1/pe ].) Exactly 3

4 4 a i the cae = 1, G i R-free o thee geerator. Let γ e = (a 1)(1 + 1/p + 1/p /p e 1 ). We ext how by iductio o that x γ e R 1/pe G e. We have already proved the cae where e = 1. Aumig the reult for a certai e, we may take p th root to obtai that x γ e/p R 1/pe i i the R 1/p -pa of the moomial {µ 1/p : µ M e }. If we multiply by x a 1, we obtai that x a 1+(γ e/p) R 1/pe+1 i the R-pa of the moomial {νµ 1/p : ν M 1, µ M e } = M e+1, ice we may replacig each x R 1/p by the R-pa of the ν M 1 uig the reult for e = 1. Sice a 1 + (γ e /p) = γ e+1, the reult follow. Sice γ e < 2 for all e, we ca coclude that for all q = p e, x 2(a 1) R 1/q G e, a free R-module. We write d = x 2(a 1) for brevity. Now uppoe that cu q J [q] for all q 0, where c B {0}. Let q = p e. The cu qq (J [q] ) [q ] for all q 0, ad c 1/q u q J [q] R 1/q. Thi yield c 1/q du q J [q] G e for all q 0, ice dr 1/q G e, ad o c 1/q J [q] G e : Ge du q for all q. Note that c 1/q G e becaue c T. Becaue G e i R-free, thi i the ame a (J [q] : R du q )G e which i cotaied i (J [q] : R du q )R 1/q. Takig q th power, we fid that for all q 0, c (J [q] : R du) [q ]. Sice c 0, thi i a cotradictio ulej [q] : R du q = R (i a Noetheria domai, the iterectio of the power of a proper ideal i 0, ice ay elemet will alo be i all the power of the maximal ideal of the local domai obtaied by localizig at a maximal ideal that cotai it). But thi mea that du q J [q]. ) q I [q] for all q 0, we may take c I [q] for q 0. We have (a 1)(q + 2) = a(q + 2) (q + 2). Write q + 2 = va r, where 0 r a 1. The the iue i whether (x a ) q v+2 x r I [q], ad, a earlier, thi i equivalet to akig Hece, if there i a choice of c 0 uch that c(x a 1 to x 2(a 1). That i we eed oly how that we caot have cx (a 1)(q+2) whether g q v+2 (x qa 1 1,..., x qa 1 )B =: I q. To prove thi doe ot hold, it uffice to how that there i a poitive iteger q v + 2 ad iteger 1,..., 1 N uch that ( i a i < ) q for 1 i 1, =, ad the multiomial coefficiet 1,..., 1 i ot diviible by p. If thee coditio hold, the g i ot i I [q], for the multiomial expaio of g will have a ( ) 1 1,..., 1 i=1 (xa i i ) i term that i ot i I q. Sice q v + 2, we alo have that g q v+2 / I q, a required. We hall chooe all of the i to have the form t i p e 1 where 0 t i p 1 ad, moreover, t = t t 1 p 1. Let t = (q v + 2)/p e 1 ), ad let w i = (p/a i ) < p/a i. Firt oberve that t i at mot oe more tha (q v + 2)/p e 1 = p + 2/p e+1 v/p e 1, where v = (q r)/a, o that v/p e 1 = (p/a) + (1 + r)/p e 1 a. Thu, t 1 + p p/a + 2/p e 1 + (1 + r)/p e 1 a (1 1/a)p /p e 1, which i maller tha p provided p/a > 4. Note that a i (w i p e 1 ) < a i (p/a i )p e 1 = q, ad that wp e 1 q v + 2. We do ot have that t = w w 1, but we have, i fact, that w w 1 t. To ee thi, ote that t ( q v + 2)/p e 1) + 1 ad that w i (p/a i ) 1. Hece, it will uffice if ( i=1 p/a i) ( 1) p+1 (v 2)/p e 1. But v = (q +2+r)/a, o, after we divide by p, thi come dow to the iequality ( 1)/p+ 1 i=1 1/a i 1+(1/p) (q+2+r)/aq+2/q. Thi become 1 i=1 1/a i ( 1)/p+1+(1/p) (2+r)/aq+2/q. The right had ide i at mot /p+2/q (+2)p Hece, if p (+2)/(α 1), we have that the w 1 + +w 1 w.

5 Therefore, by decreaig the w i uitably, we ca chooe oegative iteger t i uch that t i w i ad 1 i=1 t i = t. We ow take i = t i p e 1 ad = tp e 1. The argumet will be complete if we ca how that the multiomial coefficiet ( ) 1,..., 1 doe ot vaih. Thi i true becaue it i equal to the multiomial coefficiet ( ) t t 1,...,t 1. Oe ca ee thi by expadig (z1 + + z 1 ) tpe i two way: o the oe had, the coefficiet of z t 1p e 1 1 z t 1p e 1 1 i ( ) 1,..., 1. O the other had, we ca thik of the expaio a ((z z 1 ) t ) pe 1, which give the coefficiet of the required term a ( ) t p e 1 ( ) t 1,...,t 1, which i the ame a t t 1,...,t 1. But ow we ca ee that thi doe ot vaih, imply becaue t < p. 5