MATH 5b; Solutions to Homework Set 6
February 2009
1. Recall it suces to show that ker(TA ) = 0, so assume x = (x1 , . . . , xn ) = 0 is the
transpose of a vector in ker(TA ). Set A = (ai,j ). Then x is a solution to the system of
equations E = (Ei : 1 i
MATH 5b; Solutions to Homework Set 5
February 2012
1. (a) Let x, y T = T or(M ). Then there exist nonzero a A(x) and b A(y ). As
R is an ID, ab = 0 and (ab)(x + y ) = b(ax) + a(by ) = b 0 + a 0 = 0, so ab A(x + y )
and hence x + y T . Similarly a(x) = ax
MATH 5b; Solutions to Homework Set 4
February 2012
1. x is of degree 1 in Q(y )[x] and hence irreducible in that ring. By Theorem 9.7, Q[y ]
is a UFD with eld of fractions Q(y ), so by Prop. 9.6, x is irreducible in Q[y ][x] Q[x, y ].
=
Thus as Q[x, y ] i
MATH 5b; Solutions to Homework Set 3
January 2012
1. The group Z(N ) is a group under multiplication of congruence classes, and as
(d, (N ) = 1, d is in this group and has an inverse e; that is 1 = de, so 1 de
mod (N ). Thus de = 1 + k(N ) for some intege
MATH 5b; Solutions to Homework Set 2
January 2012
We rst prove (7). If f is a unit in R then f 1 exists in R and for x I , 1 =
f (x)f 1 (x), so f (x) = 0.
Conversely assume f R with f (x) = 0 for all x I . Dene g : I R by
g (x) = 1/f (x). Then f g = gf =
MATH 5b; Solutions to Homework Set 1
January 2012
1. As K is a eld, K # is a group under multiplication. Further by hypothesis (i),
: K # Z is a group homomorphism. Thus by Lemma 1F from rst term, (1) is the
identity 0 of Z and for a K # , (a1 ) is the i