MATH 5b: Solutions to Homework Set 7
March 2014
1. Let A = (ai,j )i,j . We have that the characteristic polynomial is given by
n
pA (X ) = det(i,j x ai,j )i,j ) =
(i,(i) x ai,(i) )
sgn( )
Sn
i=1
wher
MATH 5b: Solutions to Homework Set 3
January 2014
1. We have n|m, so take m = p1 pk and n = p1 pk , with i i 0. Let : Z/mZ Z/nZ
1
1
k
k
be the natural surjective homomorphism. We have 1 : Z/mZ Z/pi Z
MATH 5b: Solutions to Homework Set 4
February 2014
1. (a) Lets show that 2 is irreducible in R = Z[ n]. Suppose we have 2 = . Then 4 = N ()N ( ).
Suppose N () = 2. This means that if = a + b n, a2 + n
MATH 5b: Solutions to Homework Set 2
January 2014
1. We are given : R S a homomorphism of commutative rings.
(a). Suppose P is a prime ideal of S . Lets rst show that 1 (P ) is an ideal. Let x, y 1 (P
MATH 5b: Solutions to Homework Set 5
February 2014
1. (a) Since Tor(M ) M and M is an R-module, we get that Tor(M ) is associative and abelian. It
is non-empty because 0 Tor(M ). If m, n Tor(M ), then
MATH 5b: Solutions to Homework Set 6
March 2014
1. (a) Let : N Rn given by (r1 x1 + r2 x2 + rn xn ) (r1 , , rn ). N is generated by (x1 , . . . , xn )
and these elements are linearly independent. So e
MATH 5b: Midterm Solutions
February 2014
1. (1) Let R be a local ring with maximal ideal M . Let x R and assume x M . There exists
R such that x1 x = 1. Since M is an ideal, x1 x M , so 1 M . But thi
Classification of modules over PIDs
First we recall the following useful statement about free modules.
Prop. 1 Let R be a ring. M, N R-modules, : M N a R-linear surjective
morphism. Assume N is free.
MATH 5b: Solutions to Homework Set 8
March 2014
1. We let A be a nite abelian group of order n. Since A is abelian, any subgroup is normal, so it has
a unique p-Sylow subgroup P , which by the classic
MATH 5b: Solutions to Homework Set 1
January 2014
1. Lets rst show that Z (R) is a subring. First note that 0 Z (R). We have that if x, y Z (R) and
r R arbitrary, then
(x y )r = xr yr = rx ry = r(x y