Math 202 - Assignment 6
Authors: Yusuf Goren, Miguel-Angel Manrique and Rory Laster

Exercise 14.8.1.
Proof. The discriminant of x4 + 1 is D = 256 = 28 . We have x4 + 1 ≡ (x + 1)4 (mod 2). Let p be an odd prime (so p D), and suppose the irreducible factors of x4 + 1 have degrees n1 , n2 , . . . , nk .
By Corollary 41, the Galois group of x4 + 1 contains an element with cycle structure (n1 , n2 , . . . , nk ).
Since the Galois group of x4 + 1 over Q is the Klein 4-group, in which every element has order dividing
2, it follows that each ni = 1 or 2. This gives the possibilities (1, 1, 1, 1), (1, 1, 2), (2, 2). However, D is a square and so the Galois group in contained in A4 ; in particular it contains no transpositions, so (1, 1, 2) is ruled out. This leaves the possibilities (1, 1, 1, 1), and (2, 2), which correspond to the factorization into 4 linear factors or 2 quadratic factors, respectively.
Exercise 14.8.3.
Proof. The polynomial f (x) = x5 + 20x + 16 is irreducible mod 3 and hence must be irreducible.
The Galois group is therefore a transitive subgroup of S5 . The discriminant of f (x) is 216 56 and hence a square; therefore the Galois group is a subgroup of A5 . Modulo 7, we have factorization into irreducibles as f (x) ≡ (x + 2)(x + 3)(x3 + 2x2 + 5x + 5) (mod 7). Therefore the Galois group contains a 3 cycle. From the table on page 643, we see that the Galois group must be isomorphic to A5 .
Exercise 14.8.6.
Proof. By Eisenstein at 3, we see that f (x) is irreducible, so the Galois group is a transitive subgroup of S5 . The discriminant is 210 34 55 , which is not a square, so the Galois group is not contained in
A5 . The only possibilities are therefore F20 and S5 . The associated polynomial g(x) (see Exercise 21 in 14.7) turns out to have constant term equal to 0, and hence g(x) has a rational root (namely 0).
Therefore, the exercise implies that the Galois group is isomorphic to F20 .

Proposition 1 ([DF], p342). Let R be a ring and let M be an R-module. A subset N of M is an
R-submodule of M if and only if
1. N is nonempty and
2. x + ry ∈ N for all r ∈ R and all x, y ∈ N .
Exercise 10.1.4. Let R be a ring with identity, let M be the R-module Rn with component-wise addition and multiplication, and let I1 , I2 , . . . , In be left ideals of R for some n ∈ N. The following are submodules of Rn :
a. N1 = {(i1 , i2 , . . . , in ) : ik ∈ Ik for all k ∈ {1, 2, . . . , n}} and
b. N2 = {(x1 , x2 , . . . , xn ) :

n k=1 xk = 0}.

Proof. To prove (a), it suffices to show, by Proposition 1, that N1 is nonempty and x + ry ∈ N1 for all r ∈ R and all x, y ∈ N1 . For the first condition, (0, 0, . . . , 0) ∈ N1 since Ik is a subgroup of R containing the additive identity 0 for all k ∈ {1, 2, . . . , n}. That is, N1 is nonempty.

For the second condition, let x = (ik )k∈Z+ , y = (yk )k∈Z+ ∈ N1 and let r ∈ R. Then, by definition of addition and scalar multiplication, x + ry

=

(ik )k∈Z+ + r(yk )k∈Z+

=

(ik )k∈Z+ + (ryk )k∈Z+

=

(ik + ryk )k∈Z+

∈

N1

since ik + ryk ∈ Ik for all k ∈ {1, 2, . . . , n} by the left ideal axioms. This gives us (a). n To establish (b), we apply a similar method as that used in (a). Since k=1 0 = 0, the element
(0, 0, . . . , 0) ∈ N2 . Thus N2 is nonempty. Moreover if x = (ik )k∈Z+ , y = (yk )k∈Z+ ∈ N1 and r ∈ R, then x + ry = (ik )k∈Z+ + r(yk )k∈Z+ = (ik + ryk )k∈Z+ .
Therefore, because n n

n

xk + ryk = k=1 xk + r k=1 yk

since I is a ring

k=1

since x, y ∈ N2

= 0 + r0
= 0, we have that x + ry ∈ N2 by definition.

Exercise 10.1.5. Let R be a ring with identity, let I be a left ideal of R and let M be a left R-module.
Define




IM := ai mi : ai ∈ I and mi ∈ M for all i .

 finite IM is an R-submodule of M .
Proof. It suffices to show, by Proposition 1, that IM is nonempty and x + ry ∈ IM for all r ∈ R and all x, y ∈ IM . For the former condition, observe that 0R ∈ I since I is an additive subgroup of R and
0M ∈ M because M is a group. Hence the finite sum 0R · 0M = 0M satisfies the membership condition of IM . Therefore IM is nonempty. n m
For the latter condition, let r ∈ R and let x = i=1 ai mi , y = i=1 ai mi ∈ IM such that n, m ∈ N, ai , ai ∈ I and mi , mi ∈ M . Then n m

ai mi + r ·

x + ry = i=1 n

=

m

ai mi + i=1 n

=

(rai )mi

since M is a left R-module

i=1 m ai mi + i=1 ai mi i=1 ai mi i=1 for ai = rai ∈ I. That is, x + ry is a finite sum of elements of the form am such that a ∈ I and m ∈ M . Thus x + ry ∈ IM .
Exercise 10.1.6. Let R be a ring with identity and let M be a left R-module. For any nonempty collection {Ni }i∈I of R-submodules of M , the intersection
N=

Ni i∈I is an R-submodule of M .

Proof. Observe that N is a subset of M since, for all n ∈ N , n is an element of some Ni with i ∈ I.
Hence n ∈ M . So it suffices to show, by Proposition 1, that N is nonempty and x + ry ∈ N for all r ∈ R and all x, y ∈ N . For the first property, 0M ∈ Ni for all i ∈ I because each Ni is an additive subgroup of M . Therefore 0M ∈ N = ∩i∈I Ni and, so, N is nonempty.
For the second property, let r ∈ R and let x, y ∈ N . Then x and y are elements of Ni for all i ∈ I by definition. Thus, by the submodule axioms, x + ry ∈ Ni for each i ∈ I. That is, x + ry ∈ N = ∩i∈I Ni .
Exercise 10.1.7. Let R be a ring with identity and let M be a left R-module. If N1 ⊆ N2 ⊆ . . . is an ascending chain of R-submodules of M , then
∞

N=

Ni i=1 is an R-submodule of M as well.
Proof. Suppose that N1 ⊆ N2 ⊆ . . . is an ascending chain of R-submodules of M . To prove that N is also an R-submodule of M , it suffices to show, by Proposition 1, that N is nonempty and that x + ry ∈ N for all r ∈ R and all x, y ∈ N .
Since 0 ∈ N1 , 0 is an element of the union N . Hence N is nonempty. For the remaining property, let r ∈ R and let x, y ∈ N . Because x and y are elements of N , each must be an element of a submodule. That is, x ∈ Nj and y ∈ Nk for some j, k ∈ N. By the ascending chain hypothesis,
Nmin(j,k) ⊆ Nmax(j,k) . Therefore both x and y are members of Nmax(j,k) . Moreover, by the submodule axioms, x + ry ∈ Nmax(j,k) . Hence, since Nmax(j,k) ⊆ N , we have that x + ry ∈ N .
Definition. Let R be a ring and let M be a left R-module. A torsion element is an element m ∈ M such that rm = 0 for some nonzero r ∈ R.
Definition. Let R be an integral domain and let M be a left R-module. The set
T or(M ) = {m ∈ M : m is a torsion element} is the torsion submodule of M .
Exercise 10.1.8. Let R be a ring with identity and let M be a left R-module.
a. If R is an integral domain, then T or(M ) is an R-submodule of M ,
b. there exists a ring R with identity and a left R-module M such that T or(M ) is not a submodule of
M and
c. if R has zero divisors, then every nonzero left R-module contains nonzero torsion elements.
Proof. To prove (a), we suppose that R is an integral domain. It suffices to show, by Proposition 1, that T or(M ) is nonempty and x + ry ∈ T or(M ) for all x, y ∈ T or(M ) and all r ∈ R.
For the former condition, 0 ∈ T or(M ) since 1 · 0 = 0. Hence T or(M ) is nonempty. For the final condition, let x, y ∈ T or(M ) and let r ∈ R. As torsion elements, there exist nonzero s, t ∈ R such that s · x = 0 and t · y = 0. Thus
(st) · (x + ry) = (st) · x + [(st)r] · y
= (ts) · x + [(sr)t] · y

by the R-module axioms by the commutativity of R

= t · (s · x) + (sr) · (t · y)

by the R-module axioms

= t · 0 + (sr) · 0

since s · x = 0 and t · y = 0

= 0.

Because R is an integral domain and s, t are nonzero, the product st is nonzero. Therefore, we have shown that (st) · (x + ry) = 0 for a nonzero st ∈ R. That is, x + ry ∈ T or(M ) and (a) is immediate.
To see that (b) holds, consider the ring R = M = Z/6Z and the elements 2, 3 ∈ R. R is a left
R-module with respect to addition and left ring multiplication. Moreover, since
2·3=6=0
and
3 · 2 = 6 = 0, we find that 2 and 3 are elements of T or(M ). However, since 2 + 3 = 5 ∈ T or(M ), M is not closed under addition. That is, T or(M ) is not a subgroup of M and, hence, it is not a submodule of M either. Thus there exists a ring R and R-module M with the desired properties.
For (c), suppose that R contains the zero divisors s and r such that sr = 0. Then, for any nonzero left R-module M with nonzero element m, either r · m = 0 or r · m = 0. In the first case of r · m = 0, m ∈ T or(M ) since r is nonzero by hypothesis. In the second case of r · m = 0, we find that r · m ∈ T or(M ) because s · (r · m) = (sr) · m
=0·m

by the R-module axioms by hypothesis

= 0.
In either case, there exists a nonzero element contained in T or(M ). This is the desired result.
Definition. Let R be a ring with identity and let M be a left R-module. The annihilator of a submodule
N of M is the set
Ann(N ) = {r ∈ R : r · n = 0 for all n ∈ N }.
Exercise 10.1.9. Let R be a ring with identity and let M be a left R-module. For any R-submodule
N of M , the annihilator of N in R is a two-sided ideal of R.
Proof. Suppose that N is an R-submodule of M , It suffices to show that Ann(N ) is nonempty and that x + rys ∈ Ann(N ) for all x, y ∈ Ann(N ) and all r, s ∈ R. For the former condition, consider the element 0R . Since
0R · n = 0N for any n ∈ N , we see that 0R ∈ Ann(N ).
For the remaining condition, we let x, y ∈ Ann(N ) and let r, s ∈ R. For any n ∈ N , we have that
(x + rys) · n = x · n + r · (y · (s · n))
=0+r·0

by the R-module axioms since x, y ∈ Ann(N ) and sn ∈ N

= 0.
Hence x + rys ∈ Ann(N ).
Definition. Let R be a ring with identity and let M be a left R-module. The annihilator of an ideal
I of R is the set
Ann(I) = {m ∈ M : i · m = 0 for all i ∈ I}.
Exercise 10.1.10. Let R be a ring with identity and let M be a left R-module. For any ideal I of R,
Ann(I) is an R-submodule of M .

Proof. Suppose that I is an ideal of R. To prove the desired result, it suffices to show, by Proposition
1, that Ann(I) is nonempty and x + ry ∈ Ann(I) for all x, y ∈ Ann(I) and all r ∈ R.
To see that Ann(I) is nonempty, observe that i · 0N = 0N for all i ∈ I. Thus 0N ∈ Ann(I).
For the remaining condition, let x, y ∈ Ann(I) and let r ∈ R. If i ∈ I, then ir ∈ I by the ideal axioms. Hence i · (x + ry) = i · x + (ir) · y
=0+0

by the R-submodule axioms since x, y ∈ Ann(I) and ir ∈ I

= 0.
Therefore x + ry ∈ Ann(I).

References
[DF]

Dummit, David and Foote, Richard. Abstract Algebra, 3rd edition.