Section 10.1.
Homework #1
Masaya Sato
Let R be a ring with 1 and M a left (unital) R-module. 1. Prove that 0R m = 0M and (−1)m 1)m = −m for all m ∈ M .
Proof. Since 0R is the additive identity of the ring R, 0R = 0R + 0R . So for all m ∈ M
0R m = (0R + 0R )m ⇒ 0R m = 0R m + 0R m ⇒ 0R m + (−0R m) = 0R m + 0R m + (−0R m) ⇒ 0M = 0R m. Next for all m ∈ M 0M = 0R m = ((−1) + 1)m 1)m = (−1)m 1)m + 1m 1m = (−1)m 1)m + m since M is a unital R-module and thus 0M + (−m) = (−1)m 1)m + m + (−m) ⇒ −m = (−1)m 1)m as desired. 3. Prove that R× and M satisfy the two axioms in Section 1.7 for a group action of the
multiplicative group R× on the set M . Proof. Recall that R is a unital ring and thus M is a unital R-module. Therefore, according
to the definition of a module discussed on page 337 of the textbook, for all r, s ∈ R× and m ∈ M (rs) rs)m = r(sm) sm) by part (b) of (2) and 1m = m part (d) of (2) as desired. 4. Let M be the module Rn described in Example 3 and let I 1 , I 2 , · · · , I n be left ideals of
R. Prove that the following are submodules of M : (a) A = {(x1 , x2 , · · · , xn )|xi ∈ I i } (b) B = {(x1 , x2, · · · , xn )|xi ∈ R and x1 + x2 + · · · + xn = 0}. Proof. (a) We first claim that {(x1 , x2 , · · · , xn )|xi ∈ I i } is a subgroup of M = Rn. Then for
all (x (x1 , x2 , · · · , xn ) and (y (y1, y2 , · · · , yn ) in A (x1 , x2, · · · , xn) + (y ( y1 , y2 , · · · , yn) = (x1 + y1, x2 + y2 , · · · , xn + yn ) is included in A since each each I i is an ideal of R, i.e. xi + yi ∈ I i. In ad additi dition on 0 ∈ I i and (0, (0, 0, · · · , 0) ∈ A, and associativity of the operation is inherited from the one of M . Moreover for all (x ( x1 , x2 , · · · , xn ) ∈ A and r ∈ R r (x1 , x2, · · · , xn ) = (rx1 , rx2 , · · · , rxn ) ∈ A Abstract Abstract Algebra Algebra by Dummit and Foote 1
Section 10.1.
Homework #1
Masaya Sato
again because each I i is a left ideal, i.e. rxi ∈ I i . Therefore Therefore {(x1 , x2 , · · · , xn )|xi ∈ I i } is an R-submodule of M . (b) Similarly observe that B is a subgroup of M of M . For all (x (x1 , x2 , · · · , xn ) and (y (y1 , y2 , · · · , yn ) in B (x1 , x2 , · · · , xn ) + (y ( y1 , y2, · · · , yn) = (x1 + y1 , x2 + y2, · · · , xn + yn ) ∈ B since (x1 + y1 ) + (x2 + y2 ) + · · · + (xn + yn ) = (x1 + x2 + · · · + xn ) + (y1 + y2 + · · · + yn) = 0+ 0 = 0, 0, and 0 ∈ B since 0 ∈ R and and 0 + 0 + · · · + 0 = 0. Associa Associativ tivit ity y of the binary binary operatio operation n + : B × B → B is inherited from the one of M . Moreo Moreove verr for all (x ( x1 , x2 , · · · , xn ) ∈ B and r∈R r(x1 , x2 , · · · , xn ) = (rx1 , rx2 , · · · , rxn ) ∈ B because rxi ∈ R and rx1 + rx2 + · · · + rxn = r (x1 + x2 + · · · + xn ) = r0 = 0. Therefore {(x1 , x2 , · · · , xn )|xi ∈ R and x1 + x2 + · · · + xn = 0} is an R-submodule of M of M . 5. For any left ideal I of R define
I M =
a m |a ∈ I , m ∈ M i
i
i
i
finite
to be the collection of all finite sums of elements of the form am where a ∈ I and m ∈ M . Prove that I M is a submodule of M . Proof. First observe the I M is nonempty since 0M ∈ I M , i.e. 0M = 0R m ∈ I M for any
m ∈ M . Then for all m, n ∈ I M m = ai mi + · · · + ai mi and n = a j m j + · · · + a j m j . 1
1
k
1
k
1
l
l
and for all r ∈ I m + rn = (ai mi + · · · + ai mi ) + r(a j m j + · · · + a j m j ) = ai mi + · · · + ai mi + (ra (ra j )m j + · · · + (ra (ra j )m j . 1
1
1
1
k
k
1
k
k
1
1
1
l
l
l
l
Therefore m + rn can be expressed as the finite sum of elements of the form am, am, where m ∈ M and a ∈ I since I is a left ideal. Hence m + rn ∈ I M and thus I M is a submodule of M as desired. 6. Show that the intersection of any nonempty collection of submodules of an R-module is
a submodule.
Abstract Abstract Algebra Algebra by Dummit and Foote 2
Section 10.1.
Homework #1
Masaya Sato
Proof. For an arbitrary index set Λ, let {N λ |∅ N λ ≤ M ∀λ ∈ Λ} be the collection of
nonempty submodules of an R-module M . Th Then en we we claim claim that that λ∈Λ N λ is a submodule of M . Obse Observ rvee first first that λ∈Λ N λ is not empty since each N λ has the additive identity 0M . Th Then en for for all all m, n ∈ λ∈Λ N λ m and n are in N λ for every λ ∈ Λ. This This implies implies that that rm + n ∈ N λ for all r ∈ R since each N λ is a submodule. Therefore rm + n is an element of N λ for all λ ∈ Λ and included in λ∈Λ N λ . Hence λ∈Λ N λ is a submodule as desired.
if rm = 0 for some nonzero 8. An element m of the R-module M is called a torsion element if rm element r ∈ R. The set of torsion elements is denoted T or( or(M ) = {m ∈ M |rm = 0 for some nonzero r ∈ R}. (a) Prove that if R is an integral domain then T or( or(M ) is a submodule. (c) If R has zero divisors show that every nonzero R-module has nonzero nonzero torsion torsion elements elements.. Proof. (a) Recall that an integral integral domain R is a commutative ring with the unity 1 = 0
without any zero divisors. Since 0 = r0 for every r ∈ R, T or( or(M ) is not empty. Then for all m, n ∈ T or( or(M ) there are nonzero s, t ∈ R so that sm = 0, tn = 0, and st = 0 since R is an integral domain. Moreover for all r ∈ R (st)( (st))rn = (ts) ( rs)( )(tn st)(m m + rn) rn) = (st) st)m + (st ts)m + r(st) st)n = t(sm) sm) + (rs tn)) = 0 + 0 = 0. Therefore m + rn ∈ T or( or(M ) and thus T or( or(M ) is a submodule of M . (c) Suppose that R has zero divisors. divisors. So there exist exist some r = 0 and s = 0 in R such that rs = 0. Then for an arbitrary nonzero R-module M , choose a nonzero m ∈ M and consider 0, then sm ∈ M . If sm = 0, then m is an torsion elements. Otherwise if sm = r(sm) sm) = (rs) rs)m = 0R m = 0M and therefore m is a torsion element. Hence M has nonzero torsion elements. annihilator of N in R is defined to be {r ∈ R|rn = 9. If N is a submodule of M , the annihilator
0 for all n ∈ N }. Prove that the annihilator of N in R is a 2-sided ideal of R. Proof. Let A = {r ∈ R|rn = 0 for all n ∈ N }. Then for all r, s ∈ A r − s ∈ R and
(r − s)n = rn + (−s)n = rn − sn = 0 − 0 = 0. So A is a subgroup of R under addition. Then, for all r ∈ A and s ∈ R, sr ∈ R and (sr) sr)n = s(rn) rn) = s0 = 0 and thus A is a left ideal of R. Moreover sn ∈ N because N is a submodule and (rs) rs)n = r(sn) sn) = 0. This implies that A is a right ideal and therefore A is a 2-sided ideal of R. Abstract Abstract Algebra Algebra by Dummit and Foote 3
Section 10.1.
Homework #1
Masaya Sato
15. If M is a finite abelian group then M is naturally a Z-modu -module le.. Ca Can n this this actio action n be extended to make M into a Q-module? Solution: No. Suppse M = Z2 . Then M is a finite abelian group under ordinary addition + and moreover Z-module by the action ×, ordinary multiplication. However, the action of 1/2 ∈ Q Z cannot be extended to Q since for 1/
1/2 × 1 = 1/2 ∈ / M = {0, 1}. 18. Let F = R, let V = R2 and let T be the linear transformation from V to V which
is rotation clockwise about the origin by π/2 radians.. Sh Show ow that that V and 0 are the only π/ 2 radians ]-submodules for this T . F [ F [x]-submodules T . Proof. Note that the linear transformation T is expressed as the 2 × 2 matrix, i.e.
T =
0 −1 1
0
.
We first claim that both V and {0} are f [ ]-submodules. Then both V and {0} are subspaces f [x]-submodules. of V and T -inv ariant or T -stable T -invariant T -stable since T ( T (V ) V ) ⊆ V and
T ( T ({0}) = {0}.
Then we show that no other subspace W can be a submodule. So for every proper subspace of V ,, ax + by represents a straight line W = {(x, y) ∈ R2 |ax + by = 0 for some a and b ∈ R} of V on the xy-plane (0 , 0)T . Howev However er T ( xy-plane that passes through the origin (0, T (W ) W ) represents the line that is perpendicular to W and thus W is not T -invariant. T -invariant. 19. Let F = R, let V = R2 and let T be the linear transformation from V to V which is
projection onto the y-axis. -axis. Sho Show w that that V , V , 0, the x-axis and the y -axis are the only F [ F [x]submodules for this T . T . Proof. Observe that T =
0 0
. Then we we first claim that that V , V , 0, the x-axis and the y-axis are 0 1 ]-submodules for T . F [ F [x]-submodules T . It is immediate to show that V and 0 are submodules as discussed in problem problem #18. So let us consid consider er the x- and y-axes. -axes. Then observ observee that both the x-axis and y-axis are expressed as V x = {(x, 0)T ∈ R2 } and
(0, y)T ∈ R2 }, V y = {(0,
respectively, and both V x and V y are subspaces of V . ariant since V . Moreover they are T -inv T -invariant T ( T (V x ) = {0} ⊆ V x
and
T ( T (V y ) = V y .
Thus V x and V y are F [ ]-submodules. Now let us consider consider a proper subspace. subspace. W = {(x, y) ∈ F [x]-submodules. × 2 R |ax + by = 0 for some a and b ∈ R } distinct from both the x- and y -axes. Then (0, y)T ∈ R2 } T ( T (W ) W ) = {(0, Abstract Abstract Algebra Algebra by Dummit and Foote 4
Section 10.1.
Homework #1
Masaya Sato
and thus T ( T (W ) W ) represents the y-axis or each line ax + by = 0 is projected onto the y -axis. Therefore W is not T -invariant T -invariant and hence W is not a submodule. 20. Let F = R, let V = R2 and let T be the linear transformation from V to V which is rotation clockwise about the origin by π radian radians. s. Sho Show w that every subspace of V is an
F [ F [x]-submodule for this T . T . Proof. Again, it is obvious to show that both V and {0} are submodules. So let us think of a proper subspace W = {(x, y) ∈ R2 |ax + by = 0 for some a and b ∈ R}. Then T ( T (W ) W ) = W
since rotation of each line ax + by = 0 about the origin by π is invariant. invariant. Therefore W is ]-submodule of V . T -invariant T -invariant and hence every subspace of V is a F [ F [x]-submodule V . 21. Let n ∈ Z+ , n > 1 and let R be the ring of n × n matrices with entries from a field F . F .
Let M be the set of n × n matrices with arbitrary elements of F in the first column and zeros zeros elsewhere. elsewhere. Show that M is a submodule of R when R is considered as a left module over itself, but M is not a submodule of R when R is considered as a right R-module. Proof. Note that R = M n (F ) F ) and
a M = ...
11
an1
0 ··· 0 .. . . .. ’ . . . 0 ··· 0
(0.1)
where ai1 ∈ F for i = 1, . . . , n. that M is a submodule of the left module R n. We first claim that over itself. Since F is not empty, M is not empty. Then for all m, n ∈ M and for all r ∈ R rm + n = r [ai , 0, . . . , 0] + [ ai , 0, . . . , 0] = [rai + ai , 0, . . . , 0], where ai , ai , and 0 are column vectors in F n . Theref Therefore ore M is a submodule of R if R is considered considered as the left module over over itself. Howev However er if R is the right module over itself, then M is not a submodule. The following is a counterexample. Let m ∈ M be
1 m = ...
0 ··· 0 .. . . .. , . . . 1 0 ··· 0
where 1 ∈ F is the multiplicative identity, and let r ∈ R be
Then
1 1 · · · 1 r = ... ... . . . ... . 0 0 ··· 0 1 1 · · · 1 mr = ... ... . . . ... 1 1 ··· 1
and thus mr ∈ / M .
Abstract Abstract Algebra Algebra by Dummit and Foote 5
