Handout, #1 As suggested, we induct on m. When m = 1 we must prove the
statement. if p1, q1, . . . , qn ∈ D (n
∈Z+) are primes and p1 = q1q2 …qn then n
= 1. So,
suppose we have the stated hypotheses and as sume that n ≥2. Since p1 is prime and
q1…qn it divides q1(without loss of generality). So q1 = ap1 for some a ∈D. But
ir reducibility of q1 implies that a is a unit (since p1 is not). Therefore we
As we are working in a domain we can cancel p1 from both sides to obtain 1 =
implying that q2is a unit. As q2is prime this is a contradiction and we
that our assumption that n≥2 is false. Thus, n = 1 and p1
We now prove the induction step . Let m ∈Z+ be at least 2 and assume that the
of the problem is true for m - 1 and any n ∈Z+. Let p1,
. . . , pm,
q1, . . . , qn ∈D (n ∈Z+)
be primes with p1p2…pm = q1q2 …qn.
Since pm is prime and divides q1…qn it divides
(without loss of generality). Since pm and qn are both primes (and
hence irreducible) we
may argue as above and conclude that pm and qn are associate. Writing
qn = apm for some
unit a ∈D we have
Since D is a domain we can cancel pm to obtain p1…pm-
1 = (aq1)q2 …qn-1. Since
also prime, the induction hypothesis implies that m- 1 = n - 1 and (after
reordering) p1 is
associate to aq1 and pi is associate to qi for i = 2, . . . ,m
- 1. But this
means that m = n
and pi is associate to qi for every i. That is, the statement of the exercise is
true for m≥2
if it is true for m - 1.
Handout, #2 a. We use the ideal test. First,
Let a, b ∈I and
r ∈R. Then there are i, j ∈Z+ so that
Without loss of generality we can
assume that i≤j. Then
so that a ∈Ij . Since Ij is an ideal,
Since a, b ∈I and r ∈R were arbitrary, this proves that I is an ideal.
b. If R has an identity and each Ij is proper then
for every j ∈Z+. It follows that
and therefore that I ≠ R, i.e. I is a proper ideal.
p 335, #38 The ideals
p 340, #24 We start by noticing 13 = 32 + 22 = (3 + 2i)(3
- 2i). Since N(3 +
N(3- 2i) = 13 is prime, both 3+2i and 3- 2i are irreducible in Z[i], and so we
the desired factorization .
Now we note that
so that 5 + i = (1 + i)(3 - 2i). We have already seen that 3
- 2i is
irreducible and 1 + i is,
too, since N(1 + i) = 2. So, we're finished.
which implies a = b = c = 0. This proves that the three matrices in question
independent and completes the exercise.
p 348, #18 We have
which proves that P is a subspace of R3. To prove that the set
is a basis
for P it therefore suffices to prove that this set is linearly independent over R.
So let b, c ∈R
which implies b = c = 0 and proves that the vectors in
question are linearly independent.
p 349, #24 We first deal with
. This set is n onempty since 0 ∈U and 0 ∈W
since both U and W are subspaces
of V . Therefore . Furthermore, if a ∈F then
au ∈U and au ∈W, again
because both U and W are subspaces of V . It follows from the subspace test
class that is a subspace of V .
We now turn to U + W. As above, this set is nonempty since
0 ∈U and 0 ∈W implies
0 = 0 + 0 ∈U +W. Let x, y ∈U +W. Then there exist
since the fact that U and W are subspaces implies
if a ∈F then
since, again, . As
above, this proves that U +W is a subspace of V .