Notes on Chapter 1
Page 35 There are some elementary facts about elds which will be frequently used.
Proposition 1. If (A, +, , 0, 1) is a eld, x A, and a A[a + x = a], then x = 0.
Proof. We have
x = x + 0 = 0 + x = 0.
Proposition 2. If (A, +, , 0, 1) is
Notes on Chapter 4
See the preliminaries in the book for the denition of commutative ring and associated
notions. We give some additions to that material.
(1) A useful source of examples is furnished by the rings Zm dened as follows. For each
Notes on Chapter 3
Page 90 We give a complete proof of Theorem 3.3. Let be the function which assigns
to each subspace T such that S T V the subspace T /S of V /S .
Clearly if S T T V then T /S T /S .
Suppose that S T V , S T V , and T /S T /S ; we claim
Notes on Chapter 2
Page 63 Bottom: Let V be the vector space of all polynomials over R. Dene (p(t) =
tp(t). Then is injective but not surjective. On the other hand, dene (p(t) = derivative
of p(t). Then is surjective but not injective
Page 64 Theorem 2.11
Notes on chapter 5
Page 135 middle. Note that S1 is a submodule of Rn . Let I = cfw_b R : (0, . . . , 0, b) S1 .
Then I is an ideal of R. For each b I let (b) = (0, . . . , 0, b). Then is an isomorphism
from I as an R-module onto S1 .
Page 136 For part 2)
Notes on Chapter 6
Pages 140-141 The equivalence indicated. First suppose that |r . Say r = s. Let
d = gcd(, ), and write d = u + v . Then rd = ru + rv = ru + sv = (ru + sv ),
and hence r = (ru + sv ), showing that |r .
If |r , write r = w. Then r = w
Page 125, no. 6 Take the module Z Z over itself with scalar multiplication as in example
4.5, page 116. Then cfw_(1, 2), (2, 1) spans but does not contain a basis. For, let (a, b) be
arbitrary. Then (a, 0)(1, 2) + (0, b)(2, 1) = (a, b). This shows that cf
Page 106, number 1 No. Let V be the vector space of all polynomials, and S the subspace
of all polynomials in which the coecients of each power x2m is 0. Let T be the complement
of S , consisting of all polynomials in which the coecients of each power x2m
Page 83, no. 1 Write A = (aij ). Let the columns of A be b1 , . . . , bn . Let bl1 , . . . , blk be a
basis for the column space of A. For j [1, n] write bj = f1j bl1 + + fkj blk . Let Y be
the k n matrix whose j -th column is
Note that for an
Page 56, number 4 Not in general, even under the additional assumption. For, let V
have the basis cfw_b0 , b1 , b2 , b3 . We consider the partition cfw_b0 , b1 , cfw_b2 , b3 . Let S =
cfw_b0 , b2 , b1 + b3 . Then S cfw_b0 , b1 = cfw_b0 and S cfw_b2 , b