Download Math 307 Abstract Algebra Homework 11 Sample solution 1. Find an

Math 307 Abstract Algebra
Homework 11
Sample solution
1. Find an multiplicative inverse of 2x + 1 in Z4 [x]. Is the inverse unique?
Solution. (2x + 1)(2x + 1) = 4x2 + 4x + 1 = 1. So, (2x + 1) is its own inverse.
Note that in a ring with unity, if b, c are the inverses of a, then b = b(ac) = (ab)c = c. So,
the inverse if always unique if it exists.
2. (a) Given an example to show that a factor ring of an integral domain may have zero-divisors.
(b) Give an example to show that a factor ring of a ring with zero-divisors may be an integral
domain.
Solution. (a) Let R = Z, A = 4Z. Then R/A ∼
= Z4 has a zero divisor.
(b) Let R = Z4 and A = h2i. Then R/A = {0 + A, 1 + A} ∼
= Z2 has no zero divisors.
3. Let R1 and R2 be rings, and φ : R1 → R2 be a ring homomorphism.
(a) Show that if A is an ideal of R1 , then φ(A) is an ideal of φ(R1 ).
(b) Give an example to show that φ(A) may not be an ideal of R2 .
Solution. (a) Note that φ(A) is a subgroup of R2 using the group theory result. So, it is a
subgroup of φ(R1 ). For any b ∈ φ(A) and y ∈ φ(R1 ), there are a ∈ A, x ∈ R1 such that
φ(a) = b and φ(x) = y. Because ax, xa ∈ A, we have by = φ(a)φ(x) = φ(ax) ∈ φ(A) and
yb = φ(x)φ(a) = φ(xa) ∈ φ(A).
(b) Suppose A = R1 = Z, R2 = Q, and φ(x) = x. Then φ is a ring homomorphism,
φ(A) = φ(R1 ) = Z so that φ(A) is an ideal in φ(R1 ). But φ(A) = Z is not an ideal in R2 = Q,
say, 1 ∈ Z, 1/2 ∈ Q and 1 · (1/2) ∈
/ Z.
4. (8 points) Let A = hx2 + x + 1i = {(x2 + x + 1)f (x) : f (x) ∈ Z2 [x]} ⊆ Z2 [x].
(a) Show that Z2 [x]/A = {a + bx + A : a, b ∈ Z2 } has 4 elements.
Proof: Note that Z2 [x]/A = {f (x) + A : f (x) ∈ Z2 [x]}. Now for every f (x) ∈ Z2 [x], f (x) =
a + bx + (x2 + x + 1)q(x) so that f (x) + A = a + bx + (x2 + x + 1)q(x) + A = a + bx + A. Thus,
Z2 [x]/A = {a + bx + A : a, b ∈ Z2 } = {A, 1 + A, x + A, 1 + x + A} has four distinct elements.
(b) Show that (a + bx + A)(c + dx + A) = (ac + bd) + (ad + bc + bd)x + A.
Proof: (a + bx + A)(c + dx + A) = ac + adx + bcx + bdx2 + A = ac + adx + bcx + bdx2 + bd +
bd + bdx + bdx + A(ac + bd) + (ad + bc + bd)x + bd(x2 + x + 1)A = (ac + bd) + (ad + bc + bd)xA.
(c) For each nonzero element a + bx + A ∈ Z2 [x]/A, show that there is c + dx + A ∈ Z2 [x]/A
such that (a + bx + A)(c + dx + A) = 1 + A, and deduce that F = Z2 [x]/A is a field.
Proof: Because Z2 [x]/A is a commutative ring with unity, we only need to show that every
nonzero element has an inverse. Then Z2 [x]/A is a field. Now, (1 + A)(1 + A) = 1 + A and
(x + A)(1 + x + A) = 1 + A. The result follows.
(d) Show that the nonzero elements in F form a cyclic group under multiplication.
Proof: Note that (x + A)1 = x + A, (x + A)2 = x2 + A = 1 + x + A, (x + A)3 = x3 + A =
x(x2 + x + 1) + (x2 + x + 1) + 1 + A = 1 + A. The result follows.
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5. (12 points) Let A = hx2 + 1i = {(x2 + 1)f (x) : f (x) ∈ Z3 [x]} ⊆ Z3 [x].
(a) Show that Z3 [x]/A = {a + bx + A : a, b ∈ Z3 } has 9 elements.
Proof: For similar reason in previous problem, Z3 [x]/A = {a + bx + A : a, b ∈ Z3 } has 9
elements.
(b) Show that (a + bx + A)(c + dx + A) = (ac + 2bd) + (ad + bc)x + A.
Proof: (a + bx + A)(c + dx + A) = ac + adx + bcx + bdx2 + A = ac + (ad + bc)x + bdx2 + bd −
bd + A = ac + (ad + bc)x + bd(x2 + 1) + −bd + A = ac + (ad + bc)x + bd(x2 + 1) + 2bd + A =
(ac + 2bd) + (ad + bc)x + A
(c) For each nonzero element a + bx + A ∈ Z3 [x]/A, show that there is c + dx + A ∈ Z3 [x]/A
such that (a + bx + A)(c + dx + A) = 1 + A, and deduce that F = Z3 [x]/A is a field.
Proof: It suffices to show that every nonzero (a+bx+A ∈ Z3 [x]/A has an inverse, i.e., we want
to find (c+dx) so that 1 = (a+bx)(c+dx) = (ac+2bd)+(ad+bc)x, i.e., ac+2bd = 1, bc+ad = 0.
Solving the linear system with c, d as unknowns, we get (c, d) = (a2 + b2 )−1 (a, 2b). Here note
that (a2 +b2 )−1 = (a2 +b2 ) ∈ {1, 2} if (a, b) 6= (0, 0). So, (a+bx+A)−1 = (a2 +b2 )(a+2bx)+A.
(d) Determine the multiplicative inverse of 1 + 2x + A ∈ F.
Proof: By (c), (1 + 2x + A)−2 = 2(1 + x + A) = 2 + 2x + A.
(e) Show that the nonzero elements in F = Z3 [x]/A form a cyclic group under multiplication.
Proof: Note that every element in F∗ has order 2, 4, or 8 under multiplication. Consider1 +
x+A. Then (1+x+A)4 = [(1+x+A)2 ]2 = [1+2x+x2 +A]2 = [2x+A]2 = 4x2 +A = −1+A.
So, 1 + x + A has order 8 and is a generator of F∗ under multiplication.
(f) Show that X = x + A is a zero of the polynomial X 2 + 1 ∈ F[X].
Proof: X 2 +1 = (x2 +A)+(1+A) = A, so X = x+A is a zero of the polynomial X 2 +1 ∈ F[X].
6. Let Z[i] = {a + ib : a, b ∈ Z} be the integral domain D of Gaussian integers. Let F = {[(a, b)] :
a ∈ D, b ∈ D∗ } be the field of quotients of D, and Q[i] = {x + iy : x, y ∈ Q}.
(a) Show the if [(a, b)] ∈ F then [(a, b)] = [(p + iq, m)] for some p, q ∈ Z, m ∈ N.
ac+bd+i(bc−ad)
a+ib
. Since a, b, c, d
c+id =
c2 +d2
2
d . Then p, q ∈ Z and m ∈ N.
Proof: Let [(a, b)] = [(a + ib), (c + id)].
let p = ac + bd, q = bc − ad, m =
c2
+
(b) Show that φ : F → Q[i] defined by φ([(a+ib, c+id)]) =
ac+bd
+ i(bc−ad)
c2 +d2
c2 +d2
∈ Z, c + id 6= 0,
is an isomorphishm.
Proof: Let A = [(a1 + ib1 , c1 + id1 )], B = [(a2 + ib2 , c2 + id2 )].
+ib1
+ib2
Well-defined: if A = B, then ac11+id
= ac22+id
, (a1 c2 + b1 d2 ) + i(a1 d2 − b1 c2 ) = (a2 c1 +
1
2
b2 d1 ) + i(a2 d1 − b2 c1 ). Then a1 c2 + b1 d2 = a2 c1 + b2 d1 and a1 d2 − b1 c2 = a2 d1 − b2 c1 .
φ(A) =
(a1 c1 +b1 d1 )+i(b1 c1 −a1 d1 )
,
c21 +d21
φ(B) =
(a2 c2 +b2 d2 )+i(b2 c2 −a2 d2 )
.
c22 +d22
One-one: for A, B ∈ F, if φ(A) = φ(B),
a1 c1 +b1 d1
c21 +d21
=
a2 c2 +b2 d2
c22 +d22
and
b1 c1 −a1 d1
c21 +d21
=
a2 c2 +b2 d2
,
c22 +d22
so
A = B.
Onto: for any y = p + iq ∈ Q[i], we may let p = a/m and q = b/m. Then for A = [(a + ib, m)]
we have φ(A) = am
+ ibm
= y.
m2
m2
2
Isomorphism: For any A, B ∈ F,
φ(A + B) = φ([(a1 + a2 ) + i(b1 + b2 ), (c1 + c2 ) + i(d1 + d2 )])
=
(a1 + a2 )(c1 + c2 ) + (b1 + b2 )(d1 + d2 ) + i(b1 + b2 )(c1 + c2 ) − (a1 + a2 )(d1 + d2 )
(c1 + c2 )2 + (d1 + d2 )2
=
(a1 c1 + b1 d1 ) + i(b1 c1 − a1 d1 ) (a2 c2 + b2 d2 ) + i(b2 c2 − a2 d2 )
+
c21 + d21
c22 + d22
= φ(A) + φ(B)
Alternatively
First, note that Q[i] = F1 = {(a + ib)/(c + id) : a + ib ∈ Z[i], c + id ∈ Z[i]∗ } under the map
φ1 ((a + ib)/(c + id)) = (ac + bd)/(c2 + d2 ) + i(bc − ad)/(c2 + d2 ), which is just the identity
map on the subset Q[i] = F1 in C. In particular, φ1 can be viewed as a field isomorphism.
Now, we show that F is isomorphic to F1 by the map φ2 ([a + ib, c + id)]) = (a + ib)/(c + id).
Well-defined: Clearly, [(x1 , y1 )] = [(x2 , y2 )] in F implies x1 y2 = x2 y1 . Applying φ to both
sides, we get x1 /y1 and x2 /y2 , which are equal in F1 .
One-one: If x1 /y1 = φ([(x1 , y1 )]) = φ([(x2 , y2 )]) = x2 /y2 , then x1 y2 = x2 y1 . So, [(x1 , y1 )] =
[(x2 , y2 )].
Onto: If x/y ∈ F1 , then f ([(x, y)]) = x/y.
Isomorphism: φ([(x1 , y1 )] + [(x2 , y2 )]) = φ([(x1 y2 + x2 y1 , y1 y2 )]) = (x1 y2 + x2 y1 )/(y1 y2 ) =
x1 /y1 + x2 /y2 = φ([(x1 , y1 ]) + φ([(x2 , y2 )]).
Also, φ([(x1 , y1 )][(x2 , y2 )]) = φ([(x1 x2 , y1 y2 )]) = (x1 x2 )/(y1 y2 ) = (x1 /y1 )(x2 /y2 )
= φ([(x1 , y1 )])φ([(x2 , y2 )]).
It follows that φ = φ2 ◦ φ1 is an isomorphism between F and Q[i].
Extra credit problems
1. Find an example of a commutative ring R such that a2 6= a for any nonzero elements.
Answer. R = 2Z.
2. Find an example of a non-commutative ring R such that a2 = 0 for all a ∈ R.
Answer: Unknown???
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