Math Homework #9 Solutions

Handout, #1 As suggested, we induct on m. When m = 1 we must prove the following
statement. if p₁, q₁, . . . , q_n ∈ D (n ∈Z⁺) are primes and p₁ = q₁q₂ …q_n then n = 1. So,
suppose we have the stated hypotheses and as sume that n ≥2. Since p₁ is prime and divides
q₁…qn it divides q ₁(without loss of generality). So q₁ = ap₁ for some a ∈D. But then the
ir reducibility of q ₁ implies that a is a unit (since p₁ is not). Therefore we have

As we are working in a domain we can cancel p₁ from both sides to obtain 1 = aq₂…q_n,
implying that q₂is a unit. As q₂is prime this is a contradiction and we conclude therefore
that our assumption that n≥2 is false. Thus, n = 1 and p₁ = q₁.

We now prove the induction step . Let m ∈Z⁺ be at least 2 and assume that the statement
of the problem is true for m  - 1 and any n ∈Z⁺. Let p₁, . . . , p_m, q₁, . . . , q_n ∈D (n ∈Z⁺)
be primes with p₁p₂…p_m = q₁q₂ …q_n. Since pm is prime and divides q₁…q_n it divides q_n
(without loss of generality). Since pm and q_n are both primes (and hence irreducible) we
may argue as above and conclude that p_m and q_n are associate. Writing q_n = ap_m for some
unit a ∈D we have

Since D is a domain we can cancel p_m to obtain p₁…p_{m- 1} = (aq₁)q₂ …q_n-1. Since aq₁ is
also prime, the induction hypothesis implies that m-  1 = n -  1 and (after reordering) p₁ is
associate to aq₁ and p_i is associate to q_i for i = 2, . . . ,m -  1. But this means that m = n
and p_i is associate to q_i for every i. That is, the statement of the exercise is true for m≥2
if it is true for m -  1.

Finally, mathematical induction al lows us to conclude that the statement of the exercise
holds for all

Handout, #2 a. We use the ideal test. First, , since . 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 ∈I_j . Since I_j is an ideal, and
Since a, b ∈I and r ∈R were arbitrary, this proves that I is an ideal.
b. If R has an identity and each I_j 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

work.

p 340, #24 We start by noticing 13 = 3² + 2² = (3 + 2i)(3 -  2i). Since N(3 + 2i) =
N(3- 2i) = 13 is prime, both 3+2i and 3- 2i are irreducible in Z[i], and so we have found
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.

p 347, #6 The given set of vectors is linearly dependent over any field since

p 348, #8 If is linearly dependent in a vector space V over F then there
exist , not all zero , so that By re ordering we
can assume that Then we have and multiplying both sides
by yields proving that is a linear combination
of

p 348, #16 We see that

which proves that V is a vector space. We claim that

is a basis for V . According to what we've already done, it suffices to show that this set is
linearly independent. Suppose that satsify

Then, after adding the matrices on the left, we have

which implies a = b = c = 0. This proves that the three matrices in question are linearly
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
and suppose

Then

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 implies
. Given 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 mentioned in
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 so that
Thus

since the fact that U and W are subspaces implies . Moreover,
if a ∈F then

since, again, . As above, this proves that U +W is a subspace of V .