I II.1 Solutions Math 731 Fall 2013
3. Let R such that a2 = a for all a. Then
a + a = (a + a)2 = a2 + a2 + a2 + a2 = a + a + a + a,
hence a + a = 0 and a = a for all a. Now let a, b R. We have
a + b = (a + b)2 = a2 + ab + ba + b2 = a + ab + ba + b,
so ab
I II.5 Solutions Math 731 Fall 2013
i
i=0 ai x , g
1. (a) Let f =
i
i=0 bi x
=
R[x]. Then
(ai + bi )xi =
(f + g ) =
i=0
(ai )xi +
i=0
i=0
k
k
(f g ) =
(bi )xi = (f ) + (g ),
(ai bki )x = (
k=0 i=0
i
(bi )xi ) = (f )(g ).
(ai )x )(
i=0
i=0
If f R[x] then
I II.4 Solutions Math 731 Fall 2013
1. Since every nonzero element in Zn which is not a zero divisor is a unit, it follows that
S = U (Zn ) and S 1 Zn = Zn .
a
2a
2a
3. (a) Indeed, E is closed under multiplication. Since =
=
, it follows that E 1 Z = Q.
b
I II.3 Solutions Math 731 Fall 2013
1. We know that a maximal ideal in a commutative ring R with identity is prime. Let (p) be a
prime nonzero ideal, and assume (p) (q ) R. We get q | p, so p = qr. Since p | qr we get p | q
or p | r. In the rst case, p, q
I II.2(part one) Solutions Math 731 Fall 2013
1. Assume am = 0, bn = 0. Then (b)n = (1)n bn = 0 and
(a b)m+n = am+n +
m+n
1
am+n1 (b) + . +
m+n
m+n1
a(b)m+n1 + (b)m+n = 0
since in each term, either the exponent of a is m or the exponent of b is n. This pr
I II.2(part two) Solutions Math 731 Fall 2013
12. (a) Let (r, n), (s, m) A and let (t, p) S . Then we have (t, p)(r, n) = (tr + pr +
nt, pn), (r, n)(t, p) = (rt + pr + nt, np) and for x R
(r s)x + (n m)x = 0, (tr + pr + nt)x + pnx = t(rx + nx) + p(rx + nx
I II.6 part 1 Solutions Math 731 Fall 2013
1. (a) Note that (x, c) is a proper ideal in D[x], made of polynomials with constant term divisible
by c. Assume (x, c) = (f ) for some f D[x]. Then there are f1 , f2 D[x] with x = f f1 , c = f f2 .
Since c has d
I II.6 part 2 Solutions Math 731 Fall 2013
4. Let f = a0 + a1 x + + an xn , so C (f ) =gcd(a0 , a1 , ., an ) and C (af ) =gcd(aa0 , aa1 , ., aan ).
If d | C (f ), then ad | aaj for all j , hence aC (f ) | C (af ). Conversely, let q be a prime divisor of
C
IV.6 Solutions Math 731 Fall 2013
1. Let I be an ideal of R, which is also an R-module. Since for a, b I we have b a + (a) b = 0,
any two (nonzero) elements are linearly dependent. Since I is a free R-module, it must have a basis
with one element, i.e. I
VII.4 Solutions Math 731 Fall 2013
3. (a) Since the minimal polynomial of D is q = (x a)(x b)(x c) and det(xI D) =
(x a)3 (x b)2 (x c) is the product of the invariant factors which divide each other, it follows
that q3 = q, q2 = (x a)(x b), q1 = x a.
(b)
IV.3 Solutions Math 731 Fall 2013
1. Assume 1 R.
(a)(b). Since any R-module is projective, given a short exact sequence 0 A B C 0,
it follows that C is projective, so the exact sequence splits.
(b)(c). Let A be an R-module and let 0 A B C 0. Since this se
IV.2 Solutions Math 731 Fall 2013
1. (a) Assume there are ri R not all zero such that r1 x1 + + rn xn = 0. Let k be the
largest index such that rk = 0. Then xk = rk 1 (r1 x1 + + rk1 xk1 ). Conversely, if xk =
r1 x1 + + rk1 xk1 , then r1 x1 + + rk1 xk1 + (
IV.1 part 1 Solutions Math 731 Fall 2013
1. The left multiplication does not depend on representatives, since ka = ma is and only if
n | k m. We have k (a + b) = k (a + b) = ka + kb = ka + kb, (k + m)a = (k + m)a = ka + ma =
ka + ma, k (ma) = k (ma) = (km
IV.1 part 2 Solutions Math 731 Fall 2013
12. (a) Let fi : Ai Ai+1 and gi : Bi Bi+1 , i = 1, 2, 3, 4. Supoose a3 ker 3 . Since
4 f3 = g3 3 , we get 4 (f3 (a3 ) = g3 (3 (a3 ) = g3 (0) = 0. Since 4 is injective, we get
f3 (a3 ) = 0. But ker f3 = imf2 , so we