Math 150b: Modern Algebra
Homework 9 Solutions
Exercise 12.1.3.(a)
Solution. Let R be a Bezout domain and let m, n R be relatively prime. There will exist
u, v R such that um + vn = 1. Suppose we are asked to find x R such that x = a in R/(u)
and x = b in
Math 150b: Modern Algebra
Homework 10 Solutions
Exercise 12.4.1(b)
Solution. The irreducible polynomials of degree less than five are listed in formula 12.4.4
of Artin. One immediately sees x and x + 1 are factors of x16 x since 0 and 1 are roots;
after d
Math 150b: Modern Algebra
Homework 4 Solutions
Exercise 3.7.1.
Solution. We have R = he1 , e2 , . . .i is the set of all sequences with only finitely many
nonzero terms. If we are allowed to add scalar multiples of w = (1, 1, . . .) to elements of
R then
Math 150b: Modern Algebra
Homework 7 Solutions
Exercise 11.3.5.(a).
Solution. We prove the Leibniz rule ( f g)0 = f 0 g + f g0 for polynomial multplication. Letting
m1
n1
n
k
0
k
0
k
f = m
k=0 ak x and g = k=0 bk x , we have f = k=0 (k + 1)ak+1 x and g =
Math 150b: Modern Algebra
Homework 2 Solutions
Exercise 3.2.1
Solution. This exercise involves checking that the subset Q( 2) = cfw_a+b 2 | a, b Q C
is a field under the operations of addition and multiplication it inherits from C. In particular,
one must
Math 150b: Modern Algebra
Homework 5 Solutions
Exercise 8.2.1.
Solution. Let A be a real, positive definite, symmetric bilinear form on Rn and ai j its associated matrix in the standard basis. We will argue that the maximal entries necessarily lie
on the
Math 150b: Modern Algebra
Homework 8 Solutions
Exercise 11.7.2.
Solution. Since R is a domain, by definition it has no zero divisors. Assume we have
p(x), q(x) R[x] such that p(x)q(x) = 0. If p(x) = an xn + a1 x + a0 and q(x) = bm xm +
+ b1 x + b0 then t
Math 150b: Modern Algebra
Homework 3 Solutions
Exercise 3.5.4.
Solution. For part (a), we need to create a bijection between bases for F2p and elements
of GL2 (F p ). Any basis cfw_v1 , v2 may be obtained by base change from the standard basis
cfw_e1 , e
Math 150b: Modern Algebra
Homework 1 Solutions
Exercise 3.2.2.
Solution. We first note that since 5 is not a multiple of any of the given primes, we expect an
inverse to exist. As the primes p = 7, 11, 13, 17 are small, one could have simply determined
in