Download Homework 5 February 16, 2006 Math 522 Direction: This homework

Homework 5
February 16, 2006
Math 522
Direction: This homework is due on February 23, 2006. In order to receive full
credit, answer each problem completely and must show all work.
1. Consider the function φ : Z12 → Z6 defined by φ(x) = x mod 6. (a) Is φ a ring
homomorphism? (b) What is the kernel of φ? (c) What are the elements of the set φ(Z12 )?
(d) What are the elements of the factor ring Z12 /Kerφ? (e) Is Z12 /Kerφ isomorphic to
φ(Z12 )?
Answer: (a) Since φ(x + y) = (x + y) mod 6 = x mod 6 + y mod 6 = φ(x) + φ(y), and
φ(xy) = (xy) mod 6 = (x mod 6)(y mod 6) = φ(x)φ(y), therefore φ : Z12 → Z6 is a ring
homomorphism.
(b) The Kernel of φ is given by
Ker φ = {x ∈ Z12 | φ(x) = 0} = {x ∈ Z12 | x mod 6 = 0} = {0, 6} = h6i .
(c) The image of Z12 under the mapping φ is given by
φ(Z12 ) = {φ(x) | x ∈ Z12 } = {x mod 6 | x ∈ Z12 } = {0, 1, 2, 3, 4, 5} = Z6 .
(d) The factor ring Z12 /Ker φ is given by
Z12 /Ker φ = Z12 /< 6 > = {0+ < 6 >, 1+ < 6 >, 2+ < 6 >, · · · , 5+ < 6 >}.
(e) Define the mapping ψ : Z12 /< 6 > → Z6 by ψ(x+ < 6 >) = x for each x ∈ Z6 . Then
ψ is one-to-one, onto and preserves the ring operations. Therefore ψ is an isomorphism and
Z12 /Ker φ ' φ(Z12 ).
2. Let φ : Z → Z3 ⊕ Z4 be a function defined by φ(x) = (x mod 3, x mod 4). (a) Is φ a ring
homomorphism? (b) What is the kernel of φ? (c) What are the elements of the set φ(Z)?
(d) What are the elements of the factor ring Z/Kerφ? (e) Is Z/Kerφ isomorphic to φ(Z)?
Answer: (a) Since
φ(x + y) = (x + y mod 3, x + y mod 4)
= (x mod 3, x mod 4) + (y mod 3, y mod 4)
= φ(x) + φ(y)
and
φ(x + y) = (xy mod 3, xy mod 4)
= (x mod 3, x mod 4)(y mod 3, y mod 4)
= φ(x)φ(y),
therefore φ : Z → Z3 ⊕ Z4 is a ring homomorphism.
(b) The Kernel of φ is given by
Ker φ = {x ∈ Z | φ(x) = (0, 0)} = {x ∈ Z | x mod 3 = 0 = x mod 4} = h12i = 12Z.
(c) The image of Z under the mapping φ is given by
φ(Z) = {φ(x) | x ∈ Z} = {(x mod 3, x mod 4) | x ∈ Z} = Z3 ⊕ Z4 .
(d) The factor ring Z/Ker φ is given by
Z/Ker φ = Z/12Z = { 0 + 12Z, 1 + 12Z, 2 + 12Z, · · · , 11 + 12Z }.
(e) Define the mapping ψ : Z/12Z → Z3 ⊕ Z4 by ψ(x + 12Z) = x for each x ∈ Z12 . Then
ψ is one-to-one, onto and preserves the ring operations. Therefore ψ is an isomorphism and
Z12 /Ker φ ' φ(Z).
3. Determine all ring homomorphisms from the ring (Z, +, ·) to (Z, +, ·). Also, determine
all ring homomorphisms from the ring (R, +, ·) to (R, +, ·).
Answer: Since f : Z → Z is a ring homomorphism, therefore we have f (n) = nf (1) and
f (1) = f (1)2 . Renaming f (1) as a, we get f (n) = an for all n ∈ Z. Since a = a2 , therefore
either a = 0 or a = 1. Hence the ring homomorphisms from the ring Z to Z are f (n) = n
and f (n) = 0 for all n ∈ Z.
Finding all the homomorphisms from the rings R to R is a bit hard. Here are the
essential steps one needs to establish for finding these ring homomorphisms. First, show
that f (x + y) = f (x) + f (y) implies f (x) = xf (1) for all x ∈ Q, where Q is the set of rational
numbers. Second, use the fact f (xy) = f (x)f (y) to show f (t) ≥ 0 for all t ≥ 0. Using this
fact and f (x + y) = f (x) + f (y), show f (x) is an increasing function. Third, show every
increasing function that satisfies f (x + y) = f (x) + f (y) is linear, that is f (x) = xf (1) for
all x ∈ R. Fourth, use f (xy) = f (x)f (y) to show f (1) = 0 or f (1) = 1. Hence there are two
ring homomorphisms from the ring R to R, and they are f (x) = x and f (x) = 0. The detail
proof can be found in a manuscript due to Sahoo (1997).
4. Let φ : R[x] → C be a function defined by φ(f (x)) = f (i). Here R denotes the ring of
√
real numbers, C denotes the ring of complex numbers and i denotes −1. (a) Is φ a ring
homomorphism? (b) What is the kernel of φ? (c) What is the set φ(R[x])? (d) What is the
factor ring R[x]/Kerφ? (e) Is R[x]/Kerφ isomorphic to φ(R[x])? (Hint: see example 3 on
page 298.)
Answer: (a) Since
φ((f + g)(x)) = (f + g)(i) = f (i) + g(i) = φ(f (x)) + φ(f (y))
and
φ((f g)(x)) = (f g)(i) = f (i) g(i) = φ(f (x)) φ(f (y)),
therefore φ : R[x] → C is a ring homomorphism.
(b) The Kernel of φ is given by
Ker φ = {f (x) ∈ R[x] | f (i) = 0} = {(x2 + 1)g(x) | g(x) ∈ R[x]} = x2 + 1 .
(Note that f (i) = 0 implies f (x) = (x2 + 1) g(x) for some g(x) ∈ R[x].)
(c) The image of R[x] under the mapping φ is given by
φ(R[x]) = {φ(f (x)) | f (x) ∈ R[x]} = {f (i) | f (x) ∈ R[x]}
( n
)
X
k
=
ak (i) | ak ∈ R
k=0
= { A + Bi | A, B ∈ R }
= C.
(d) The factor ring R[x]/Ker φ is given by
R[x]/Ker φ = R[x]/ < x2 + 1 > = { ax + b + < x2 + 1 > | a, b ∈ R}.
(e) Define the mapping ψ : R[x]/ < x2 + 1 > → C by ψ(ax + b + < x2 + 1 >) = ai + b for
a, b ∈ R. Then ψ is one-to-one, onto and preserves the ring operations. Therefore ψ is an
isomorphism and R[x]/Ker φ ' φ(R[x]).
5. Let f (x) = 4x3 + 2x2 + x + 3 and g(x) = 3x4 + +3x3 + 3x2 + x + 4 be two polynomials
in the ring Z5 [x]. Compute f (x) + g(x) and f (x)g(x).
Answer: f (x)+g(x) = 3x4 +2x3 +2x+2 and f (x)g(x) = 2x7 +3x6 +x5 +2x4 +3x2 +2x+2.
6. Find all the zeros of the polynomial f (x) = x2 + 3x + 2 in Z6 . Find the zeros of the same
polynomial in Z11 .
Answer: Since f (x) = x2 + 3x + 2, and
x ∈ Z6
f (x) mod 6
0
2
1
0
2
0
3
2
4
0
5
0
the zeros of f (x) in Z6 are 1, 2, 4, 5. Similarly, the zeros of f (x) in Z11 are 9 and 10.
7. Let f (x) = 5x4 + 3x3 + 1 and g(x) = 3x2 + 2x + 1 be two polynomials in the ring Z7 [x].
Determine the quotient and the remainder upon dividing f (x) by g(x).
Answer: If f (x) = 5x4 + 3x3 + 1 and g(x) = 3x2 + 2x + 1, then
f (x) = q(x)g(x) + r(x)
in Z7 [x] whenever q(x) = 4x2 + 3x + 6 and r(x) = 6x + 2.
8. Let f (x) be a polynomial in R[x]. Prove that a is a root of a polynomial f (x) if and only
if x − a is a factor of f (x). (Hint: use Remainder Theorem.)
Answer: By Remainder Theorem, we have
f (x) = (x − a)g(x) + f (a),
(1)
where g(x) ∈ F [x]. Suppose a is a zero of f (x). Then f (a) = 0. Hence from (1), we get
f (x) = (x − a)g(x). This proves that (x − a) is factor of f (x).
Conversely, if (x − a) is factor of f (x), then the remainder f (a) must be zero. That is
f (a) = 0 and thus a is a zero of f (x).
9. A polynomial f (x) is said to be irreducible in the ring (Z5 [x], +, ·) if it has no zeros in
Z5 . Which of the following polynomials
f (x) = 2x3 + x2 + 4x + 1;
g(x) = x4 + 2;
h(x) = x4 + 4x3 + 2
are irreducible in Z5 [x].
Answer: Since f (x) = 2x3 + x2 + 4x + 1, and
x ∈ Z5
f (x) mod 5
0
1
1
3
2
4
3
1
4
1
0
2
1
3
2
3
3
3
4
3
0
2
1
2
2
0
3
1
4
4
therefore f (x) is irreducible in Z5 [x].
Since g(x) = x4 + 2, and
x ∈ Z5
g(x) mod 5
therefore g(x) is irreducible in Z5 [x]
Since h(x) = x4 + 4x3 + 2, and
x ∈ Z5
h(x) mod 5
therefore h(x) is not irreducible in Z5 [x]
10. Determine how many irreducible quadratics are there in Z5 [x]. Quadratics are polynomials of degree 2.
Answer: The number of quadratic polynomials of the form ax2 + bx + c where a, b, c ∈ Z5
and a 6= 0 is equal to (4)(5)(5) = 100.
The number of reducible polynomials in Z5 is equal to the number of distinct expressions
of the form a(x + α)(x + β) and a(x + α)2 for α, β ∈ Z5 .
In Z5 , the number of distinct expressions of the form (x + α)(x + β) is equal to
5
2
= 10.
Hence the number of distinct expressions of the form a(x + α)(x + β) is equal to 4(10) = 40.
Similarly, the number of expressions of the form a(x + α)2 is 4(5) = 20.
Thus, the total number of reducible polynomials in Z5 is equal to 40 + 20 = 60. Hence
the number of irreducible quadratics of the form ax2 + bx + c is equal to 100 − 60 = 40.