Workshop on group schemes and p-divisible groups: Homework 1. 1
... (iii) Write the ring map corresponding to the Z-group map det : GLn → Gm , and use the irreducibility of det(tij ) over any field (proof?) to deduce that the only group scheme maps from GLn to Gm over a field are detr for r ∈ Z. (iv) What is the scheme-theoretic intersection of SLn and the diagonall ...
... (iii) Write the ring map corresponding to the Z-group map det : GLn → Gm , and use the irreducibility of det(tij ) over any field (proof?) to deduce that the only group scheme maps from GLn to Gm over a field are detr for r ∈ Z. (iv) What is the scheme-theoretic intersection of SLn and the diagonall ...
Exercises - Stanford University
... (1) Let R be a reduced commutative ring (no nonzero nilpotent elements). Classify all 1-dimensional commutative formal group laws over R which are polynomials. b a or G b m. (2) Give an example of a 1-dim polynomial group law over Z[]/2 which is not G (3) Show that formal group laws over R have fo ...
... (1) Let R be a reduced commutative ring (no nonzero nilpotent elements). Classify all 1-dimensional commutative formal group laws over R which are polynomials. b a or G b m. (2) Give an example of a 1-dim polynomial group law over Z[]/2 which is not G (3) Show that formal group laws over R have fo ...
NOETHERIAN MODULES 1. Introduction In a finite
... Proof. Let I be an ideal in R[X]. To prove I is finitely generated we assume it is not, so I 6= (0). We will construct a sequence in I recursively. Let f1 be any element of I with least degree. If we have f1 , . . . , fk in I then (f1 , . . . , fk ) 6= I since I is not finitely generated, so there i ...
... Proof. Let I be an ideal in R[X]. To prove I is finitely generated we assume it is not, so I 6= (0). We will construct a sequence in I recursively. Let f1 be any element of I with least degree. If we have f1 , . . . , fk in I then (f1 , . . . , fk ) 6= I since I is not finitely generated, so there i ...
