MATH 113 - HOMEWORK 8
DUE MONDAY 8/12
23.4.
5x2 −x +2
x4 +5x3
9x2
+5x
+10
−3x2
+0x
+0
− (45x4 −9x3 +18x2 )
3x3
− (25x3
+1x2
−5x2 +10x)
6x2
−
(50x
2
+1x
−10x +20)
2
We’ve found q(x) = 9x2 + 5x + 10 and r(x) = 2.
23.10. Let f (x) = x3 + 2x2 + 2x + 1. The most efficient way factor f into linear factors is
just to look for zeros in Z7 by trial and error.
It’s easy to see that 0 and 1 don’t work (f (0) = 1 and f (1) = 6). But 2 does: f (2) =
(2)3 + 2(2)2 + 2(2) + 1 = 21 = 0.
Factoring out (x − 2) gives f (x) = (x − 2)(x2 + 4x + 3). Then the second factor is easy to
factor again: f (x) = (x − 2)(x + 1)(x + 3).
23.34. To show that f (x) = xp + a is not irreducible for any prime p and any a ∈ Zp ,
I’ll show that f always has a zero in Zp , and hence has a linear factor. By Fermat’s Little
Theorem, (−a)p = −a in Zp , so f (−a) = (−a)p + a = −a + a = 0.
23.35. Let g(x) = an + an−1 x + · · · + a0 xn . Then g(a−1 ) = an + an−1 a−1 + · · · + a0 a−n .
Multiplying by an , we have an g(a−1 ) = an an + an−1 an−1 + · · · + a0 = f (a) = 0, so
an g(a−1 ) = 0. Since fields have no zero divisors, this implies that a = 0 or g(a−1 ) = 0. But
a 6= 0, so a−1 is a zero of g(x), as desired.
23.36. By polynomial long division, we obtain f (x) = (x − α)q(x) + r(x), with deg(r) <
deg(x − α) = 1. So r is a constant, i.e. r ∈ F .
Now evaluating both sides at α, we find f (α) = (α − α)q(α) + r = 0q(α) + r = r, so
r = f (α), as desired.
45.10. In Z[x]: Note that 2 is an irreducible in Z[x]. 4x2 − 4x + 8 = (2)(2)(x2 − x + 2).
The quadratic x2 − x + 2 is irreducible, since (by the quadratic formula, for instance) it has
no real roots, much less integer roots!
In Q[x]: In this ring, 2 is not an irreducible, since it is a unit. So 4x2 − 4x + 8 is already
maximally factored into irreducibles. Since Q is a field, if this quadratic were reducible,
it would factor into linear polynomials. But it has no roots in Q (again by the quadratic
formula, if you like).
In Z11 [x]: In this ring, on the other hand, the quadratic has roots. By inspection 5 is
a root, and the polynomial factors as 4x2 − 4x + 8 = (x − 5)(4x + 5). Both factors are
irreducible, since they are degre 1 polynomials over a field.
45.24. Consider (1, 0) ∈ Z × Z. Note that (1, 0) = (1, 0) · (1, 0) = (1, 0) · (1, 0) · (1, 0), etc.
Continuing to factor in this way, we never arrive at a product of irreducibles.
In fact, it is impossible to factor (1, 0) into a product of irreducibles. Given (1, 0) =
(a, b)(c, d), we must have a = c = ±1, and one of b or d must be 0, since bd = 0. So we have
(1, 0) = (±1, n)(±1, 0), where n can be anything. If we choose n prime in Z, then (±1, n)
will be irreducible, but the other factor, (±1, 0), behaves exactly like (1, 0).
Repeating, whenever we factor (1, 0) or (−1, 0), one of the factors must be one of these
two options. So they cannot be factored into a product of irreducibles.
45.25. Let D be an integral domain and p ∈ D be prime. Suppose p = ab. We’d like to
show that a is a unit or b is a unit.
We have p | ab, so since p is prime, p | a or p | b. Without loss of generality (swapping
a and b if necessary), we may assume that p | a. So pc = a for some c ∈ D. But then
p = (pc)b, so p(1) = p(cb), and since p 6= 0 and D is an integral domain, we can cancel the
p from both sides to obtain 1 = cb. Hence b is a unit.
45.26. Let D be a UFD and p ∈ D irreducible. Suppose p | ab. We’d like to show that
p | a or p | b.
Since p | ab, pc = ab for some c ∈ D. Now factor a, b, and c into irreducibles: a = p1 . . . pl ,
b = q1 . . . qn , c = r1 . . . rm . We have pr1 . . . rm = p1 . . . pl q1 . . . qn .
We have an equality between two products of irreducibles. Since D is a UFD, we must
have m + 1 = l + n, and p is associates with some pi or qj .
If p is associates with pi , then p | pi , and since pi | a, we have p | a. On the other hand, if
p is associates with qj , then p | qj , and since qj | b, we have p | b.
45.32. ACC =⇒ MC: Let S be a nonempty set of ideals in R, and suppose for contradiction that there is no ideal I ∈ S which is not properly contained in some other ideal J ∈ S.
Then for any I ∈ S, there exists J ∈ S with I ( J.
Since S is nonempty, we can pick some I0 ∈ S. By our assumption, there exists I1 ∈ S
with I0 ( I1 . By our assumption again there exists I2 ∈ S with I1 ( I2 . Continuing in this
way, we obtain an infinite ascending chain I0 ( I1 ( I2 ( . . . of ideals in R, contradicting
ACC.
MC =⇒ ACC: Suppose for contradiction that ACC is false. Then there is an ascending
chain of ideals in R, I0 ( I1 ( I2 ( . . . . Let S = {Ii | i ∈ N}, the set of ideals in the chain.
Then S contradicts the MC, since for any Ii ∈ S, Ii ( Ii+1 ∈ S.
FBC =⇒ ACC: This is very much like our proof in class that R is a PID implies R has
the ACC. Suppose for contradiction that ACC is false. Then
there is an ascending chain of
S∞
ideals in R, I0 ( I1 ( I2 ( . . . . By our lemma in class, I = n=0 In is an ideal in R. By the
FBC, I is generated by finitely many elements, b1 , . . . , bk . That is, in the language of the
problem, I is the intersection of all
S ideals in R containing b1 , . . . , bk .
Now for each bi , bi ∈ I = ∞
Letting m =
n=0 In , so bi ∈ Ini for some ni ∈ N.
max(n1 , . . . , nk ), we have Ini ⊆ Im for all i, so bi ∈ Im for all i.
But I is the intersection of all ideals containing all the bi , and Im is such an ideal, so
I ⊆ Im . Hence we have Im ( Im+1 ( I ⊆ Im , which is a contradiction.
T ACC =⇒ FBC: First I claim that given any finite set b1 , . . . , bn , the intersection I =
b1 ,...,bn ∈J J of all ideals J ⊆ R containing b1 , . . . , bn is an ideal in R.
We checked in class that the intersection a colleciton of subgroups of a group is a subgroup.
So it remains to check closure under multiplication. Let i ∈ I and r ∈ R. Then i ∈ J for all
ideals J containing b1 . . . , bn . Since each J is an ideal, ri ∈ J and ir ∈ J for each such J.
So ri ∈ I and ir ∈ I.
Suppose for contradiction that FBC is false. Then we can find an ideal I in R which has
no finite generating set: for any b1 , . . . , bn ∈ I, I is not equal to the intersection of all ideals
in R containing b1 , . . . , bn .
T
Let b1 ∈ I. Consider the intersection I1 = b1 ∈J J. By the claim above, I1 is an ideal.
Now I1 ⊂ I, since I is an ideal containing b1 , but I1 6= I, since b1 does not generate I. So
we can pick b2 ∈ I with
/ I1 .
T b2 ∈
Now we let I2 = b1 ,b2 ∈J J. Note I1 ⊂ I2 , since I2 is an ideal containing b2 , but I1 6= I2 ,
since b2 ∈ I2 \ I1 . Also, I2 ⊂ I, since b1 , b2 ∈ I, but I2 6= I, since b1 , b2 do not generate I. So
we can pick b3 ∈ I with b3 ∈
/ I2 and repeat the argument.
In this way, we obtain an ascending chain I1 ( I2 ( I3 ( . . . , contradicitng ACC.
45.33. The proofs here are just the proofs in the previous problem turned on their heads.
DCC =⇒ mC: Let S be a nonempty set of ideals in R, and suppose for contradiction that
there is no ideal I ∈ S which does not properly contain some other ideal J ∈ S. Then for
any I ∈ S, there exists J ∈ S with I ) J.
Since S is nonempty, we can pick some I0 ∈ S. By our assumption, there exists I1 ∈ S
with I0 ) I1 . By our assumption again there exists I2 ∈ S with I1 ) I2 . Continuing in this
way, we obtain an infinite descending chain I0 ) I1 ) I2 ) . . . of ideals in R, contradicting
DCC.
mC =⇒ DCC: Suppose for contradiction that DCC is false. Then there is a descending
chain of ideals in R, I0 ) I1 ) I2 ) . . . . Let S = {Ii | i ∈ N}, the set of ideals in the chain.
Then S contradicts the mC, since for any Ii ∈ S, Ii ) Ii+1 ∈ S.
45.34. Z is an example. Z is a PID, so it has the ACC. But it does not have the DCC.
Recall that hai ⊆ hbi if and only if b | a. So h1i ) h2i ) h4i ) h8i ) . . . is an example of an
infinite descending chain of ideals: h2n i ) h2n+1 i for all n, since 2n | 2n+1 , but 2n+1 - 2n .
46.12.
a. We know that Z is a UFD. So by Theorem 45.29, Z[x] is a UFD.
b. Let I = {a + xf (x) | a ∈ 2Z, f (x) ∈ Z[x]} ⊂ Z[x].
Let a + xf (x), b + xg(x) ∈ I. Then (a + xf (x)) + (b + xg(x)) = (a + b) + x(f (x) +
g(x)) ∈ I, since a + b ∈ 2Z and f (x) + g(x) ∈ Z[x].
0 = 0 + x0 ∈ I.
Let a + xf (x) ∈ I. Then −(a + xf (x)) = −a + x(−f (x)) ∈ I, since −a ∈ 2Z
and −f (x) ∈ Z[x].
Let a+xf (x) ∈ I and g(x) ∈ Z[x]. Write g(x) = a0 +a1 x+· · ·+an xn = a0 +xh(x),
with h(x) = a1 + a2 x + · · · + an xn−1 .
Then (a + xf (x))g(x) = (a + xf (x))(a0 + xh(x)) = aa0 + a0 xf (x) + axh(x) +
xf (x)h(x) = aa0 + x(a0 f (x) + ah(x) + f (x)h(x)) ∈ I, since aa0 ∈ 2Z and a0 f (x) +
ah(x) + f (x)h(x) ∈ Z[x].
Another way to see that I is an ideal is that I is the preimage of 2Z under the
evaluation at 0 homomorphism: I = ev−1
0 [2Z] = {f (x) ∈ Z[x] | f (0) ∈ 2Z}, and
the preimage of an ideal is an ideal.
c. I claim that Z[x] is not a PID, since the ideal I in part (b) is not principal.
Suppose for contradiction that I = hp(x)i. Note that 2 = 2 + x0 ∈ I, so p(x) | 2.
But then, since Z is an integral domain, deg(p) = 0, so p is a constant in Z. But
the only elements of Z dividing 2 are ±1 and ±2.
If p = 1 or p = −1, then I contains a unit, and hence I = Z[x]. But 1 ∈
/ I,
contradiction.
If p = 2 or p = −2, then every element of I is of the form 2g(x) for some
g(x) ∈ Z[x]. But 2 + x ∈ I, and 2 + x is not a multiple of 2 in Z[x], contradiction.
Hence I is not principal, and Z[x] is not a PID.
d. No. Every Euclidean domain is a PID, and Z[x] is not a PID, so Z[x] is not a
Euclidean domain.
46.20. We have a Euclidean domain D and two elements a, b ∈ D. Let I = {x ∈ D | ∃c ∈
D, x = ca and ∃d ∈ D, x = db} = hai ∩ hbi. The intersection of ideals in a ring is an ideal
(as shown in Exercise 45.32 above), so I is an ideal in D.
Now since D is a Euclidean domain, it is a PID, and I = hci for some c ∈ I. We need to
check that c is an lcm of a and b.
First of all, c ∈ I, so c ∈ hai ∩ hbi, and hence a | c and b | c. Now suppose d ∈ D is
divisible by both a and b. Then d ∈ hai ∩ hbi = hci, so c | d.
Hence c is an lcm of a and b (by definition).