hw04.2

8/11/2019 hw04.2

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Homework #04, due 2/10/10 = 10.1.1 , 10.1.8 , 10.1.9 , 10.1.10 .Additional problems recommended for study: 10.1.2 , 10.1.3 , 10.1.4 ,

10.1.5 , 10.1.6 , 10.1.7 , 10.1.11 , 10.1.12 , 10.1.1510.1.1 Assume R is a ring with 1 and M is a left R-module. Provethat 0 m = 0 and ( − 1)m = − m for all m ∈ M .

Let m ∈ M . Since M is a group with identity element 0, we have0 = 0+ 0 and 0+ 0 m = 0m. Then 0+0 m = 0m = (0+0) m = 0m +0 mby the module axiom ( r + s)m = rm + sm . Since M is a group, we cancancel 0m from both sides, leaving 0 = 0 m.

Since R is a ring with 1, we have 0 = 1+( − 1), so 0m = (1+( − 1))m =

1m + ( − 1)m, but 1m = m by one of the module axioms and 0m = 0by the rst part of this problem, so 0 = m + ( − 1)m. By adding − mto both sides we get − m = ( − 1)m.

10.1.8 Assume R is a ring with 1 and M is a left R-module. Anelement m of the R-module M is called a torsion element if rm = 0 forsome nonzero element r ∈ R. The set of torsion elements is denotedby

Tor (M ) := {m ∈ M |rm = 0 for some nonzero r ∈ R}

(a) Prove that if R is an integral domain then Tor (M ) is a submoduleof M (called the torsion submodule of M ).Note that, by hypothesis, R is commutative, has 1, and has no zero

divisors. Also, for every r ∈ R we have 0+ r 0 = r0 = r (0+0) = r0+ r 0,so, cancelling r0 from both sides leaves 0 = r0.

Let m, n ∈ Tor (M ). Then there are nonzero r, s ∈ R such thatrm = 0 = sn . Since R is an integral domain and neither r nor s iszero, we conclude that rs = 0. Then m + n ∈ Tor (M ) because

rs (m + n) = ( rs )m + ( rs )n= ( sr )m + ( rs )n R is commutative= s(rm ) + r(sn ) by a module axiom= s0 + r0 since rm = 0 = sn= 0 + 0 proved above= 0

8/11/2019 hw04.2


8/11/2019 hw04.2


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10.1.9 Assume R is a ring with 1 and M is a left R-module. If N is a submodule of M , the annihilator of N in R is dened to be

{r ∈ R |rn = 0 for all n ∈ N }. Prove that the annihilator of N in R isa 2-sided ideal of R.Let Ann R (N ) ⊆ R be the annihilator of N in R. Assume r ∈ Ann R (N )

and let s ∈ R. We wish to show sr ∈ Ann R (N ) and rs ∈ Ann R (N ).For every n ∈ N we have (sr )n = s(rn ) = s · 0 = 0, so sr ∈ Ann R (N ).Also, for every n ∈ N , sn ∈ N since n ∈ N and N is a submodule,so (rs )n = r(sn ) = 0 since r ∈ Ann R (N ). So far we have shownthat Ann R (N ) is closed under multiplication by elements of R, bothon the left and on the right. To complete the proof we need to showAnn R (N ) is closed under +. Let r, s ∈ Ann R (N ). Then for all n ∈ N ,we have rn = sn = 0, so (r + s)n = rn + sn = 0 + 0 = 0. Thereforer + s ∈ Ann R (N ).

10.1.10 Assume R is a ring with 1 and M is a left R-module. If I isan ideal of R, the annihilator of I in M is dened to be {m ∈ M |am =0 for all a ∈ I }. Prove that the annihilator of I in M is a submoduleof M .

Let Ann M (I ) be the annihilator of I in M . We will show Ann M (I )

is closed under − and closed under the action of elements in R. Letm, n ∈ Ann M (I ). Then, for all a ∈ I , am = an = 0, so a(m − n) =am − an = 0 − 0 = 0. This shows m − n ∈ Ann M (I ). Let r ∈ R. Thenrm ∈ Ann M (I ) because, for every a ∈ I , we have ar ∈ I since I is anideal, so (ar )m = 0 since m ∈ Ann M (I ), hence a(rm ) = 0 by a moduleaxiom.

hw04.2

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