Download MATH 103B Homework 6 - Solutions Due May 17, 2013

MATH 103B Homework 6 - Solutions
Due May 17, 2013
Version May 18, 2013
Assigned reading: Chapter 17 of Gallian.
Recommended practice questions: Chapter 17 of Gallian, exercises
27, 28, 35
Supplementary Exercises for Chapters 15-18
11, 13, 19, 25
Assigned questions to hand in:
(1) (Gallian Chapter 17 # 8) Suppose that f pxq P Zp rxs and f pxq is irreducible over Zp ,
where p is prime. If deg f pxq “ n, prove that Zp rxs{xf pxqy is a field with pn elements.
Solution: By Corollary 1 to Theorem 17.5, since Zp is a field and f pxq is irreducible over
Zp , then Zp rxs{xf pxqy is a field. It remains to argue that this field has pn many elements.
An element of this field is a coset
ppxq ` xf pxqy.
Since f pxq generates xf pxqy, it has minimal degree in this ideal. Therefore, we can find
minimal coset representatives of cosets of this ideal by computing the remainder upon
division by f pxq. The remainder, by the division theorem, is either zero or has degree
strictly less than n, the degree of f pxq. Thus, the set of possible remainders is obtained
by considering all possible combinations of coefficients
an´1 xn´1 ` ¨ ¨ ¨ ` a0
as ai P Zp . Since there are p choices for each coefficient, and n slots, there are pn choices.
Thus, there are pn distinct cosets of xf pxqy in Zp rxs and the field Zp rxs{xf pxqy has pn
elements.
(2) (Gallian Chapter 17 # 25) Find all the zeros and their multiplicities of
x5 ` 4x4 ` 4x3 ´ x2 ´ 4x ` 1
over Z5 .
Solution: Let f pxq “ x5 ` 4x4 ` 4x3 ´ x2 ´ 4x ` 1. Evaluating f pxq at a “ 1, we get
15 ` 4 ¨ 14 ` 4 ¨ 13 ´ 12 ´ 4 ¨ 1 ` 1 “ 1 ` 4 ` 4 ´ 1 ´ 4 ` 1 “ 0.
Therefore, this is a root. To compute its multiplicity, we divide f pxq by x ´ 1:
x5 ` 4x4 ` 4x3 ´ x2 ´ 4x ` 1 “ px ´ 1qpx4 ` 4x2 ` 3x ´ 1q
Then
14 ` 4 ¨ 12 ` 3 ¨ 1 ´ 1 “ 1 ` 4 ` 3 ´ 1 ‰ 0,
so px ´ 1q2 does not divide f pxq and a “ 1 has multiplicity 1.
However,
34 ` 4 ¨ 32 ` 3 ¨ 3 ´ 1 “ 81 ` 36 ` 9 ´ 1 “ 0,
so a “ 3 is another root of f pxq. To compute its multiplicity, we divide x4 ` 4x2 ` 3x ´ 1
by x ´ 3:
x4 ` 4x2 ` 3x ´ 1 “ px ´ 3qpx3 ` 3x2 ` 3x ` 2q.
Evaluating the quotient at a “ 3:
33 ` 3 ¨ 32 ` 3 ¨ 3 ` 2 “ 27 ` 27 ` 9 ` 2 “ 0,
so a “ 3 has multiplicity at least 2. Dividing again:
x3 ` 3x2 ` 3x ` 2 “ px ´ 3qpx2 ` x ` 1q.
However, 32 ` 3 ` 1 ‰ 0 so a “ 3 has multiplicity 2.
The polynomial gpxq “ x2 ` x ` 1 has no roots in Z5 :
‚ gp0q “ 1
‚ gp1q “ 3
‚ gp2q “ 2
‚ gp3q “ 3
‚ gp4q “ 1
Therefore, a complete factorization of f pxq is
f pxq “ px ´ 1qpx ´ 3q2 px2 ` x ` 1q.
Thus, f pxq has two zeros: a “ 1 with multiplicity 1 and a “ 3 with multiplicity 2.
(3) (Gallian Chapter 17 # 33) Let F be a field and let ppxq be irreducible over F . Show that
ta ` xppxqy : a P F u
is a subfield of F rxs{xppxqy isomorphic to F .
Solution: Consider the function ϕ : F Ñ F rxs{xppxqy defined by
ϕpaq “ a ` xppxqy.
We will show this is a ring isomorphism.
‚ Homomorphism: Let a, b P F .
– ϕpa ` bq “ pa ` bq ` xppxqy “ pa ` xppxqyq ` pb ` xppxqyq “ ϕpaq ` ϕpbq.
– ϕpabq “ pabq ` xppxqy “ pa ` xppxqyq pb ` xppxqyq “ ϕpaq ¨ ϕpbq.
‚ One-to-one: It’s equivalent to prove that Kerϕ “ t0F u. Let a P F and suppose
ϕpaq “ 0F rxs{xppxqy . By definition, this means that a ` xppxqy “ 0F rxs{xppxqy “ xppxqy.
By properties of cosets, this means a P xppxqy. Since ppxq is irreducible, it’s nonzero
and nonconstant. Therefore, the only constant in xppxqy is the zero polynomial.
That is, a “ 0, as required.
‚ Onto: Let a ` xppxqy be a coset in the set. Then a P F and ϕpaq “ a ` xppxqy, as
required.
(4) (Gallian Supplementary Exercises for Chapters 15-18 # 12) Is the homomorphic image
of a principal ideal domain a principal ideal domain?
Solution: No. Consider the rings Z, Zn for n an integer greater than 1 that is not prime.
We will prove that
I. Z is a PID.
II. there is a homomorphism from Z onto Zn
III. Zn is not a PID.
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Recall that a PID is an integral domain with the additional property that every ideal in
the ring is generated by some element in the ring. Proofs:
I. We have proved that Z is an integral domain (cf. Chapter 13, example 1). Let A
be an ideal of Z. If A “ t0u then A “ x0y so it is principal. Otherwise, if x P A is
negative then 0´x P A as well (by closure under subtraction and since 0 P A because
A is a subring and hence contains the additive identity). Thus, A must contain a
positive number as well. Let m be the smallest positive number in A. We will prove
that A “ xmy.
– Suppose n P xmy. Then n “ Cm for some C P Z. Since A is an ideal of Z, it
is closed under this type of multiplication so n P A.
– Suppose n P A. By the division algorithm, n “ qm ` r for some q P Z,
0 ď r ă m. Towards a contradiction, assume that r ‰ 0. Then r “ n ´ qm P A
is a positive number smaller than m, contradicting our choice of m. Therefore,
n “ qm and so n P xmy.
Thus, Z is a PID.
II. Consider the reduction mod n function. It maps Z onto Zn . Moreover, for x, y P Z,
px ` yq mod n “ x mod n ` y mod n and pxyq mod n “ px mod nqpy mod nq.
Thus, it is a ring homomorphism.
III. Zn is not a PID because it is not an integral domain. To prove that it’s not an
integral domain, let n “ mr be a nontrivial factorization of n. Then, m, r P Zn are
nonzero elements but mr “ 0 in Zn . Thus, Zn has zero divisors and is not a PID.
(5) (Gallian Supplementary Exercises for Chapters 15-18 # 20) For any integers m and n,
prove that the polynomial x3 ` p5m ` 1qx ` 5n ` 1 is irreducible over Z.
Solution: Let f pxq “ x3 ` p5m ` 1qx ` 5n ` 1. Applying the mod 5 irreducibility test,
we compute
f¯pxq “ x3 ` x ` 1.
This is a degree 3 polynomial, like f pxq. Moreover it is irreducible over Z5 if and only if
it has no roots in Z5 , by Theorem 17.1. To look for roots of f¯pxq in Z5 , we evaluate the
polynomial at the five elements of Z5 :
‚
‚
‚
f¯p0q “ 1
f¯p1q “ 3
f¯p2q “ 8 ` 2 ` 1 “ 1
‚
‚
f¯p3q “ 27 ` 3 ` 1 “ 1
f¯p4q “ 64 ` 4 ` 1 “ 4.
Thus, f¯pxq has no roots in Z5 and hence is irreducible over Z5 . By the mod 5 irreducibility test, we conclude that f pxq is irreducible over Q. Moreover, we showed that
f pxq has nontrivial factorizations in Qrxs if and only if it has no proper factorizations
in Zrxs. Therefore, f pxq has no proper factorizations in Zrxs. The only other nontrivial
factorization over Zrxs it may have are of the form cqpxq, where c P Z a nonunit and
qpxq P Zrxs with deg qpxq “ deg f pxq. However, no such factorizations are possible in
this case because the only divisors of 1 (the leading and constant coefficients of f pxq) are
˘1, the units of Zrxs. Thus, f pxq is irreducible over Z.
(6) (Gallian Supplementary Exercises for Chapters 15-18 # 23) Is x3y a maximal ideal in Zris?
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Solution: Yes, it is a maximal ideal. To prove this, we use Theorem 14.4, which states
that x3y is a maximal ideal in Zris if and only if Zris{x3y is a field. By exercise 26 in
Chapter 14, since x3y is a proper ideal and Zris is a commutative ring with unity, the
factor ring is a commutative ring with unity, 1 ` x3y. It remains to prove that every
nonzero element of this factor ring has a unit. We begin by listing the elements of the
factor ring:
‚ 0Zris{x3y “ x3y.
‚ 1Zris{x3y “ 1Zris ` x3y “ 1 ` x3y.
Note that the zero of a field does not have a multiplicative inverse and the unity of a ring
is its own inverse. The other cosets of x3y are determined by the remainder modulo 3 of
the real and imaginary parts of each of the coset representatives:
‚ 2 ` x3y
Mutliplicative inverse: 2 ` x3y.
‚ i ` x3y
Mutliplicative inverse: 2i ` x3y.
‚ 1 ` i ` x3y
Mutliplicative inverse: 2 ` i ` x3y.
‚ 2 ` i ` x3y
Mutliplicative inverse: 1 ` i ` x3y.
‚ 2i ` x3y
Mutliplicative inverse: i ` x3y.
‚ 1 ` 2i ` x3y
Mutliplicative inverse: 2 ` 2i ` x3y.
‚ 2 ` 2i ` x3y.
Mutliplicative inverse: 1 ` 2i ` x3y.
Since each nonzero element of the commutative ring with unity Zris{x3y has a multiplicative inverse, this is a field.
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