Name:
MA407 Test 3
Miss Bosko
Show all work. Good luck!
(1) Let φ : Z12 → Z30 be a group homomorphism defined by
φ(x) = 10x.
(a) Determine Ker(φ) (4pts)
Ker(φ) = {x ∈ Z12 | φ(x) = 0} = {0, 3, 6, 9}
(b) Determine φ−1 (10) (4pts) φ−1 (10) = {x ∈ Z12 | φ(x) =
10} = {1, 4, 7, 10} = 1 + Ker(φ)
(c) Determine the image of φ (i.e. φ(Z12 )). (4pts) φ(Z12 ) =
{φ(x) | x ∈ Z12 } = {10, 20, 0}
(2) Let G be an Abelian group of order 48. Find all isomorphism
classes of G and circle the one which is isomorphic to Z24 ⊕ Z2 .
(6pts)
Z2 ⊕ Z2 ⊕ Z2 ⊕ Z2 ⊕ Z3
Z2 ⊕ Z2 ⊕ Z22 ⊕ Z3
Z22 ⊕ Z22 ⊕ Z3
Z23 ⊕ Z2 ⊕ Z3
Z24 ⊕ Z3
And, Z24 ⊕ Z2 ∼
= Z3 ⊕ Z8 ⊕ Z2 since gcd(3, 8) = 1
(3) Let a belong to a ring R. Let S = {x ∈ R | ax = 0}. Show
that S is a subring of R. Let x, y ∈ S. Then ax = 0 = ay
so a(x − y) = ax − ay = 0 − 0 = 0 ⇒ x − y ∈ S. Also,
a(xy) = (ax)y = 0 · y = 0 ⇒ xy ∈ S. So, S is a subring of R.
(4) For each ring listed, complete the following chart: (8pts)
Integral
Units Zero Divisors Domain?(Y/N) Field?(Y/N)
Z5
1,2,3,4
none
Y
Y
Z8
1,3,5,7
2,4,6
N
N
Z2 ⊕ Z2 (1,1)
(0,1), (1,0)
N
N
(5) Let R = 3Z and A = 15Z.
(a) Verify that A is an ideal of the ring R. (4pts) Let 15a ∈ 15Z
and 15b ∈ 15Z. Then 15a − 15b = 15(a − b) ∈ 15Z since
a − b ∈ Z. Let 3x ∈ 3Z. Then (3x)(15a) = 15(3xa) ∈ 15Z
since 3xa ∈ Z. Also, R is commutative, so we have A is an
ideal in R.
(b) Determine the elements in R/A (4pts) R/A = {A, 3+A, 6+
A, 9 + A, 12 + A}
(c) Create 2 operation tables for the factor ring R/A, one for
addition and one for multiplication. (6pts)
1
2
+
A
3+A 6+A 9+A 12+A
A
A
3+A 6+A 9+A 12+A
3+A 3+A 6+A 9+A 12+A
A
6+A 6+A 9+A 12+A
A
3+A
9+A 9+A 12+A
A
3+A 6+A
12+A 12+A
A
3+A 6+A 9+A
A 3+A 6+A 9+A 12+A
·
A
A
A
A
A
A
3+A A 9+A 3+A 12+A 6+A
6+A A 3+A 6+A 9+A 12+A
9+A A 12+A 9+A 6+A 3+A
12+A A 6+A 12+A 3+A 9+A
(d) Using the multiplication table for R/A find the unity. (1pt)
6+A
(e) Using the multiplication table find the zero divisors. (1pt)
There are no zero divisors.
(f) Is R/A an integral domain, field, or neither? (2pts) Since
there are no zero divisors, it is an integral domain. Furthermore, every nonzero element is a unit (i.e. has a multiplicative inverse).
3
Name:
MA407 Test 3
Miss Bosko
Show all work. You may use your notes, textbook, homework assignments, or worksheets from this class. No other resources are permitted.
(1) Let G be the GROUP Z4 ⊕ Z12 .
(a) Determine the cyclic subgroup generated by the element
(2, 8) in G. Call this H. (4pts)
H = h(2, 8)i = {(2, 8), (0, 4), (2, 0), (0, 8), (2, 4), (0, 0)}
(b) Use Lagrange’s theorem to determine the order of G/H.
(2pts)
|G/H| =
|G|
4 · 12
=
=8
|H|
6
(c) Use the Fundamental Theorem of Finitely Generated Abelian
Groups to determine the isomorphism classes of G/H (3pts)
Since 8 = 23 , the partition of 3 are 1 + 1 + 1, 1 + 2, and 3.
Thus, the classes are
Z8
Z2 ⊕ Z4
Z2 ⊕ Z2 ⊕ Z2
(d) Which one of the groups above is isomorphic to G/H? Justify. (6pts)
Hint: You will have to find the orders of the elements in
G/H, which take the form (a, b) + H
(1, 0) + H = {(3, 8), (1, 4), (3, 0), (1, 8), (3, 4), (1, 0)}
(0, 1) + H = {(2, 9), (0, 5), (2, 1), (0, 9), (2, 5), (01)}
(1, 1) + H = {(3, 9), (1, 5), (3, 1), (1, 9), (3, 5), (1, 1)}
(0, 2) + H = {(2, 10), (0, 6), (2, 2), (0, 10), (2, 6), (0, 2)}
(3, 0) + H = {(1, 8), (3, 4), (1, 0), (3, 8), (1, 4), (3, 0)}
(0, 3) + H = {(2, 11), (0, 7), (2, 3), (0, 11), (2, 7), (0, 3)}
(1, 2) + H = {(3, 9), (1, 6), (3, 2), (1, 10), (3, 6), (1, 2)}
Now, o(H) = 1 and o((1, 0) + H) = 4 since (1, 0)4 + H = H
o((0, 1)+H) = 4 since (0, 1)4 +H = (0, 4)+H = H because
(0, 4) ∈ H
o((1, 1)+H) = 4 since (1, 1)4 +H = (0, 4)+H = H because
(0, 4) ∈ H
o((0, 2)+H) = 2 since (0, 2)2 +H = (0, 4)+H = H because
4
(0, 4) ∈ H
o((3, 0)+H) = 2 since (3, 0)2 +H = (2, 0)+H = H because
(2, 0) ∈ H
o((0, 3) + H) = 4 since (0, 3)4 + H = H
o((1, 2)+H) = 2 since (1, 2)2 +H = (2, 4)+H = H because
(2, 4) ∈ H
Since G/H contains elements of order 4, but not of order
∼
8, then G/H
√ = Z2 ⊕ Z4
(2) Prove that Q[ 2] is a FIELD by first showing it is a ring and
then showing√that each nonzero element is a unit. (9pts)
Hint: (a + b 2)−1 = a+b1√2 and you must show that this sim√
plifies to an element in Q[ 2].
√
Proof. Notice that Q[ 2] ⊂ R. So,
√ to show
√ it is a ring, it must
pass the
Then
√ subring test.
√ Let a + b 2, c + d 2√∈ Q[sqrt2].
√
(a + b√2) − (c +
√d 2) = (a − c) + (b − d) 2√∈ Q[ 2].
√ Also,
(a √
+ b 2)(c + d 2) = (ac + 2bd) + (ad + bc) 2 ∈ Q[ 2]. So
Q[ 2] is √
a subring and therefore, also a ring. Now suppose
0 6= a + b 2, then
√
√ −1
1
1
a−b 2
√ =
√ ·
√
(a + b 2) =
a+b 2
a+b 2 a−b 2
√
a−b 2
a
−b √
= 2
=
+
2
a − 2b2
a2 − 2b2 a2 − 2b2
√
2
Since a + b 2 6= 0. Either a or
√ b is not
√ zero. Therefore, a −
a
−b
2
2b 6= 0. Then, a2 −2b2 + a2 −2b2 2 ∈ Q[ 2]
(3) In the RING Z8 [x] find the following.
(a) An element that has degree greater than 0 that is a zero
divisor. Justify. (4pts) 2x · 4x = 8x2 = 0 so both 2x and
4x are zero divisors.
(b) An element that has degree greater than 0 that is a unit.
Justify. (4pts) (4x + 1)(4x + 1) = 16x2 + 8x + 1 = 1. So,
4x + 1 is a unit.
(4) Show that a2 − b2 = (a + b)(a − b) for all a and b in a RING R
if and only if R is commutative. (8pts)
Proof. Suppose a2 − b2 = (a + b)(a − b) for all a, b ∈ R. Then
a2 − b2 = (a + b)(a − b) = a2 − ab + ba − b2 . So, 0 = −ab + ba ⇒
ab = ba. Thus, R is commutative.
5
Suppose R is commutative. Then ab = ba and (a + b)(a − b) =
a2 − ab + ba − b2 = a2 − ab + ab − b2 = a2 − b2 .
(5) Let R be a RING with unity and let N be an ideal containing
a unit.
(a) Prove N = R. (6pts)
Proof. Let N be an ideal with a unit x. Then there exists
an x−1 ∈ R. So, x−1 x = 1 ∈ N . Now that 1 ∈ N, r · 1 ∈
N ∀r ∈ R. Then R ⊆ N . Since N ⊆ R by definition,
R = N.
(b) Use the statement above to prove the corollary:
An ideal of a field F is either the trivial ideal, {0}, or all
of F . (4pts)
Proof. Let N be an ideal of a field, F . If N 6= {0} then
N must contain a unit. By above, N = R. Therefore,
N = {0} or N = R.
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