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May 25th

May 25th

# Math Homework #9 Solutions

Handout, #1 As suggested, we induct on m. When m = 1 we must prove the following
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 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…qn,
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, . . . , pm, q1, . . . , qn ∈D (n ∈Z+)
be primes with p1p2…pm = q1q2 …qn. Since pm is prime and divides q1…qn it divides qn
(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 aq1 is
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.

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 ∈Ij . Since Ij 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 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

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 .

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