Endomorphisms of finetely generated projective modules over a commutative ring
Endomorphisms of finetely generated projective modules over a commutative ring
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交换环上精细生成的射影模的自同态
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发表时间：
1973
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通讯作者：
G. Almkvist
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作者：
G. Almkvist
这篇论文的起源是一个错误的(？)在Bourbak I中([4]，第156页，发挥13d)。这里有一个S公式，f是一个2·2-矩阵，在一个环中有多个条目，f2=0，其中(Trf)4=0，4是这个函数的最小整数。利用Cay ley-Hami L定理，我们得到f2_af~b L=0，其中a=T r f，b=d e t f。不是th a t f 2=0，而是t个比赛，我们得到了t r f=a 2=2b。用f乘以f得到bf=0，这意味着b9Trf=ba=0。因此，a a=2AB=0，所以3而不是4是上面的最小整数。用小m和n展开，很快就得出猜想：如果f是一个n，则(Trf)m“+L=0。这是在1.7中使用Exter Ior代数的更一般的集合中提出的。在第一节中，定义了自同态f：P~P中的特征为时间多项式的L 2t(F)，其中P是广义射影A-模(A是与1的交换环)中的f。如果P是自由的，则2t(F)=det(1~tf)。L指数t竞赛公式(A含Q)
The origin of this paper is a mispr int (?) in Bourbak i ([4], p. 156, Exerc ise 13 d). There it is s t a ted t ha t i f f is a 2 • 2-matr ix wi th entries in a c o m m u t a t i v e ring and f 2 = 0 t hen (Tr f )4 = 0 and 4 is the smallest integer wi th this p roper ty . Using the Cay ley -Hami l ton theorem we get f2 _ af ~b l = 0 where a = T r f and b = d e t f . Not ing t h a t f 2 = 0 and taking t races we g e t a T r f = a 2 = 2b. Mult iplying the f irs t equa t ion by f gives bf = 0 which implies b 9 Tr f = ba = O. Hence a a = 2ab = 0 so 3 and not 4 is the smallest integer above. Expe r imen t ing wi th small m and n one soon makes the conjecture: I f f is an n • wi th f~+i = 0 t hen ( T r f ) m"+l = 0. This is p roved in a somewhat more general set t ing in 1.7 using ex te r ior algebra. I n Sect ion 1 the character is t ic polynomia l 2t(f) is def ined for an endomorph ism f : P ~ P where P is a f in i te ly genera ted projec t ive A-module (A is a commu ta t i ve ring wi th 1). I f P is free then 2t(f) = det (1 ~tf). The exponent ia l t race formula (in case A contains Q)