math202hw6sols

7/24/2019 Math202hw6sols

1/5

Math 202 - Assignment 6

Authors: Yusuf Goren, Miguel-Angel Manrique and Rory Laster

Exercise 14.8.1.

Proof. The discriminant ofx4 + 1 is D = 256 = 28. We have x4 + 1 (x+ 1)4 (mod 2). Letp bean odd prime (so p D), and suppose the irreducible factors ofx4 + 1 have degrees n1, n2, . . . , nk.By Corollary 41, the Galois group ofx4 + 1 contains an element with cycle structure (n1, n2, . . . , nk).Since the Galois group ofx4 +1 overQ is the Klein 4-group, in which every element has order dividing2, it follows that each ni = 1 or 2. This gives the possibilities (1, 1, 1, 1), (1, 1, 2), (2, 2). However, Dis 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 thefactorization 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 2

1656 andhence a square; therefore the Galois group is a subgroup ofA5. Modulo 7, we have factorization intoirreducibles asf(x) (x+2)(x+3)(x3 + 2x2 + 5x+ 5) (mod 7). Therefore the Galois group containsa 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 thatf(x) is irreducible, so the Galois group is a transitive subgroupofS5. The discriminant is 2

103455, which is not a square, so the Galois group is not contained inA5. The only possibilities are thereforeF20 and S5. The associated polynomial g(x) (see Exercise 21in 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). LetR be a ring and letM be anR-module. A subsetN ofM is anR-submodule ofM if and only if

1. Nis nonempty and

2. x+ry N for allr R and allx, y N.

Exercise 10.1.4. LetR be a ring with identity, letM be theR-moduleRn with component-wiseaddition and multiplication, and letI1, I2, . . . , I n be left ideals ofR for somen N. The following aresubmodules ofRn:

a. N1 = {(i1, i2, . . . , in) : ikIk for allk {1, 2, . . . , n}} and

b. N2 = {(x1, x2, . . . , xn) :n

k=1xk= 0}.

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


2/5

For the second condition, let x = (ik)kZ+ , y = (yk)kZ+ N1 and let r R. Then, by definitionof addition and scalar multiplication,

x+ry = (ik)kZ+ + r(yk)kZ+

= (ik)kZ+ + (ryk)kZ+

= (ik+ryk)kZ+

N1

sinceik+ryk Ik for all k {1, 2, . . . , n}by the left ideal axioms. This gives us (a).To establish (b), we apply a similar method as that used in (a). Since

nk=10 = 0, the element

(0, 0, . . . ,0) N2. ThusN2 is nonempty. Moreover ifx = (ik)kZ+ , y = (yk)kZ+ N1 and r R,then

x+ry = (ik)kZ++ r(yk)kZ+ = (ik+ryk)kZ+ .

Therefore, because

nk=1

xk+ryk=n

k=1

xk+r

nk=1

yk

sinceIis a ring

= 0 +r0 sincex, y N2= 0,

we have that x +ry N2 by definition.

Exercise 10.1.5. LetR be a ring with identity, letIbe a left ideal ofR and letMbe a leftR-module.Define

IM :=

finite

aimi : ai I andmi M for alli

.

IM is anR-submodule ofM.

Proof. It suffices to show, by Proposition 1, that IM is nonempty and x +ry IM for all r R andall x, y IM. For the former condition, observe that 0R I sinceIis an additive subgroup ofR and

0MMbecauseMis a group. Hence the finite sum 0R 0M= 0Msatisfies the membership conditionofIM. ThereforeIM is nonempty.

For the latter condition, letr R and letx=n

i=1aimi, y =m

i=1a

im

i IMsuch thatn, m N,ai, ai I andmi,m

i M. Then

x+ry =ni=1

aimi+r

mi=1

aim

i

=

ni=1

aimi+

mi=1

(rai)m

i sinceMis a leftR-module

=ni=1

aimi+mi=1

aim

i

for ai = rai I. That is, x+ ry is a finite sum of elements of the form am such that a I andm M. Thus x +ry IM.

Exercise 10.1.6. LetR be a ring with identity and letM be a leftR-module. For any nonemptycollection{Ni}iI ofR-submodules ofM, the intersection

N=iI

Ni

is anR-submodule ofM.


3/5

Proof. Observe that Nis a subset ofM since, for all n N, n is an element of some Ni with i I.Hence n M. So it suffices to show, by Proposition1, that N is nonempty and x+ry N for allr R and all x, y N. For the first property, 0M Ni for all i I because each Ni is an additivesubgroup ofM. Therefore 0MN=iINi and, so, N is nonempty.

For the second property, let r R and let x, y N. Then x and y are elements ofNi for

all i I by definition. Thus, by the submodule axioms, x+ ry Ni for each i I. That is,x+ry N=iINi.

Exercise 10.1.7. LetR be a ring with identity and letMbe a leftR-module. IfN1 N2 . . . is anascending chain ofR-submodules ofM, then

N=i=1

Ni

is anR-submodule ofMas well.

Proof. Suppose that N1 N2 . . . is an ascending chain ofR-submodules ofM. To prove thatNis also an R-submodule ofM, it suffices to show, by Proposition 1, that N is nonempty and thatx+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 ofa 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 bothx and y are members ofNmax(j,k). Moreover, by the submoduleaxioms,x +ry Nmax(j,k). Hence, since Nmax(j,k) N, we have that x +ry N.

Definition. LetR be a ring and letMbe a leftR-module. A torsion element is an elementm Msuch thatrm= 0 for some nonzero r R.

Definition. LetR be an integral domain and letM be a leftR-module. The set

Tor(M) = {m M :m is a torsion element}

is the torsion submodule ofM.Exercise 10.1.8. LetR be a ring with identity and letM be a leftR-module.

a. IfR is an integral domain, thenT or(M) is anR-submodule ofM,

b. there exists a ringR with identity and a leftR-moduleM such thatT or(M) is not a submodule ofM and

c. ifR has zero divisors, then every nonzero leftR-module contains nonzero torsion elements.

Proof. To prove (a), we suppose that R is an integral domain. It suffices to show, by Proposition1,thatT 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 finalcondition, letx, y T or(M) and letr R. As torsion elements, there exist nonzeros, t R such that

s x= 0 and t y= 0. Thus

(st) (x+ry) = (st) x+ [(st)r] y by theR-module axioms

= (ts) x+ [(sr)t] y by the commutativity ofR

=t (s x) + (sr) (t y) by theR-module axioms

=t 0 + (sr) 0 since s x= 0 andt y= 0

= 0.


4/5

Because R is an integral domain and s, t are nonzero, the product st is nonzero. Therefore, we haveshown 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 leftR-module with respect to addition and left ring multiplication. Moreover, since

2 3 = 6 = 0

and3 2 = 6 = 0,

we find that 2 and 3 are elements ofT or(M). However, since 2 + 3 = 5 T or(M), M is not closedunder addition. That is, T or(M) is not a subgroup ofM and, hence, it is not a submodule ofMeither. 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 anynonzero left R-module M with nonzero element m, either r m= 0 or r m= 0. In the first case ofr m= 0, m T or(M) sincer is nonzero by hypothesis. In the second case ofr m= 0, we find thatr m T or(M) because

s (r m) = (sr) m by theR-module axioms

= 0 m by hypothesis

= 0.

In either case, there exists a nonzero element contained in T or(M). This is the desired result.

Definition. LetR be a ring with identity and letMbe a leftR-module. The annihilator of a submoduleN ofM is the set

Ann(N) = {r R : r n= 0 for alln N}.

Exercise 10.1.9. LetR be a ring with identity and letM be a leftR-module. For anyR-submoduleN ofM, the annihilator ofN inR is a two-sided ideal ofR.

Proof. Suppose that N is an R-submodule ofM, It suffices to show that Ann(N) is nonempty and

thatx +rys Ann(N) for all x, y Ann(N) and all r, s R. For the former condition, consider theelement 0R. Since0R n= 0N

for any n N, we see that 0R Ann(N).For the remaining condition, we letx, y Ann(N) and letr, s R. For any n N, we have that

(x+rys) n= x n+r (y (s n)) by theR-module axioms

= 0 +r 0 since x, y Ann(N) andsn N

= 0.

Hencex +rys Ann(N).

Definition. LetR be a ring with identity and letM be a leftR-module. The annihilator of an ideal

I ofR is the setAnn(I) = {m M :i m= 0 for alli I}.

Exercise 10.1.10. LetR be a ring with identity and letM be a leftR-module. For any idealI ofR,Ann(I) is anR-submodule ofM.


5/5

Proof. Suppose thatIis an ideal ofR. To prove the desired result, it suffices to show, by Proposition1,thatAnn(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 0NAnn(I).For the remaining condition, let x, y Ann(I) and let r R. If i I, then ir Iby the ideal

axioms. Hence

i (x+ry) = i x+ (ir) y by theR-submodule axioms

= 0 + 0 sincex, y Ann(I) and ir I

= 0.

Thereforex +ry Ann(I).

References

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

math202hw6sols

Documents