Homework 1
Problem 1
Suppose
Suppose, towards a contradiction, that
Problem 2
Suppose
It is clear (always dangerous to say!) that
Since
Problem 3
Suppose
 Prove that if
is an integral domain then$R$ is a submodule of$\mathrm{Tor}(M)$ .$M$  Give an example of a ring
and$R$ module$R$ such that$M$ is not a submodule. (Hint: Consider torsion elements in the$\mathrm{Tor}(M)$ module$R$ for some specific ring$R$ .)$R$  Suppose
has a (nonzero) zero divisor and$R$ is nontrivial. Prove that$M$ has nonzero torsion elements.$M$  Suppose
is an$\mathrm{{\rm O}\x95}:M\beta \x86\x92N$ module morphism. Prove that$R$ .$\mathrm{{\rm O}\x95}(\mathrm{Tor}(M))\beta \x8a\x86\mathrm{Tor}(N)$
 First observe that
so${1}_{R}\beta \x8b\x850=0$ . Next suppose$0\beta \x88\x88\mathrm{Tor}(\mathrm{M})$ , and let$m,n\beta \x88\x88\mathrm{Tor}(\mathrm{M})$ be nonzero elements such that$r,s\beta \x88\x88R$ and$rm=0$ . We claim that$sn=0$ is nonzero and$rs$ . The first statement follows from the fact that we assumed$(rs)(m\beta \x88\x92n)=0$ is an integral domain; for the second, observe that$R$ Note that we used the commutativity of$(rs)(m\beta \x88\x92n)=(rs)m\beta \x88\x92(rs)n=(sr)m\beta \x88\x92(rs)n=s(rm)\beta \x88\x92r(sn)=s\beta \x8b\x850\beta \x88\x92r\beta \x8b\x850=0\beta \x88\x920=0.$ (the "integral" part of "integral domain") to switch from$R$ to$(rs)m$ . > 2. In light of part 1, we need to look at rings$(sr)m$ that are not integral domains. One such ring is$R$ . When considered as an module over itself, the elements${\mathbf{Z}}_{6}$ and$2$ are torsion elements, since$3$ and$3\beta \x8b\x852=0$ . On the other hand, the element$2\beta \x8b\x853=0$ is not torsion, as one can quickly verify that$2+3=5$ for each of the five nonzero elements of$r\beta \x8b\x855\beta \x890$ .${\mathbf{Z}}_{6}$  Let
be a nonzero zero divisor, so that$r\beta \x88\x88R$ for some nonzero$rs={0}_{R}$ . Let$s\beta \x88\x88R$ be any nonzero element, and consider the element$m\beta \x88\x88M$ . If$sm$ , then$sm=0$ is torsion; if$m$ , then$sm\beta \x890$ , and so$r(sm)=(rs)m={0}_{R}\beta \x8b\x85m=0$ is torsion. So in either case there exists a nonzero torsion element in$sm$ .$M$  Take any torsion element
and let$m\beta \x88\x88M$ be a nonzero element such that$r\beta \x88\x88R$ . Then observe that$rm={0}_{M}$ . This prove$r\beta \x8b\x85\mathrm{{\rm O}\x95}(m)=\mathrm{{\rm O}\x95}(rm)=\mathrm{{\rm O}\x95}({0}_{M})={0}_{N}$ is a torsion element in$\mathrm{{\rm O}\x95}(m)$ .$N$
Problem 4
Suppose
 For each submodule
on$N$ , the annihilator of$M$ in$N$ is defined to be the set of elements$R$ such that$r\beta \x88\x88R$ for all$rn=0$ . Prove that the annihilator of$n\beta \x88\x88N$ in$N$ is a 2sided ideal of$R$ .$R$  For each right ideal
of$I$ , the annihilator of$R$ in$I$ is defined to be the set of all elements$M$ such that$m\beta \x88\x88M$ for all$im=0$ . Prove that the annihilator of$i\beta \x88\x88I$ in$I$ is a submodule of$M$ .$M$  Consider the
module$\mathbf{Z}$ and ideal$M={\mathbf{Z}}_{24}\Gamma \x97{\mathbf{Z}}_{15}\Gamma \x97{\mathbf{Z}}_{50}$ . Determine the annihilator of$I=2\mathbf{Z}$ in$M$ and the annihilator of$\mathbf{Z}$ in$I$ .$M$

We first verify the annihilator of
in$N$ is an abelian subgroup. First recall that$R$ for every $n\in${0}_{R}\beta \x8b\x85n={0}_{M}$ . Take any$N$ in the annihilator of${r}_{1},{r}_{2}$ and observe that for all$N$ we have$n\beta \x88\x88N$ , and so$({r}_{1}\beta \x88\x92{r}_{2})n={r}_{1}n\beta \x88\x92{r}_{2}n={0}_{N}\beta \x88\x92{0}_{N}={0}_{N}$ is in the annihilator of${r}_{1}\beta \x88\x92{r}_{2}$ . By the Subgroup Criterion, we've thus proven the annihilator is a subgroup of$N$ under addition. Now fix any element$R$ of the annihilator and let$r$ be an arbitrary element. Then for every$s\beta \x88\x88R$ observe that$n\beta \x88\x88N$ , where for the last equality we once again used the fact that zero acts trivially. This proves$(sr)n=s(rn)=s\beta \x8b\x85{0}_{M}={0}_{M}$ is in the annihilator for every$sr$ and hence the annihilator is a left ideal of$s\beta \x88\x88R$ . Similarly, we have$R$ , where for the second equality we used the fact that$(rs)n=r(sn)=r\beta \x8b\x85{0}_{M}={0}_{M}$ is a submodule so$N$ and hence it is annihilated by$sn\beta \x88\x88N$ . This proves$r$ is in the annihilator for every$rs$ , and hence the annihilator is a right ideal of$s\beta \x88\x88R$ .$R$ 
This will prove very similar to the argument above. First note once again that
for every$i\beta \x8b\x85{0}_{M}={0}_{M}$ , and so$i\beta \x88\x88I$ is the annihilator of${0}_{M}$ in$I$ . Next suppose$M$ are in the annihilator of${m}_{1},{m}_{2}$ in$I$ . Then for every$M$ we have$i\beta \x88\x88I$ , and so$i({m}_{1}\beta \x88\x92{m}_{2})=i{m}_{1}\beta \x88\x92i{m}_{2}={0}_{M}\beta \x88\x92{0}_{M}={0}_{M}$ is in the annihilator of${m}_{1}\beta \x88\x92{m}_{2}$ in$I$ . We've thus proven the annihilator of$M$ in$I$ is a subgroup of$M$ . Finally, take any$M$ and$r\beta \x88\x88R$ in the annihilator of$m$ in$I$ . Then$M$ , since$i(rm)=(ir)m={i}^{\beta \x80\xb2}m={0}_{M}$ (as${i}^{\beta \x80\xb2}\beta \x88\x88I$ is a right ideal) and$I$ annihilates everything in$m$ . Thus$I$ is in the annihilator of$rm$ in$I$ , and hence that annihilator is a left$M$ submodule.$R$ 
By the definition, the annihilator of
in$M$ is the collection of all integers$\mathbf{Z}$ such that$n$ for every$nm=0$ . The module$m\beta \x88\x88M$ is a Cartesian product, which means that its operation is componentwise (addition) and that its zero element is the triple$M$ . If we denote the general element in$(0,0,0)$ as$M$ then$({m}_{1},{m}_{2},{m}_{3})$ , and so$n\beta \x8b\x85({m}_{1},{m}_{2},{m}_{3})=(n{m}_{1},n{m}_{2},n{m}_{3})$ annihilates that element exactly when$n$ in$n{m}_{1}=0$ ,${\mathbf{Z}}_{24}$ in$n{m}_{2}=0$ , and${\mathbf{Z}}_{15}$ in$n{m}_{3}=0$ . In other words, the desired annihilator is${\mathbf{Z}}_{50}$ , where each${A}_{1}\beta \x88\copyright {A}_{2}\beta \x88\copyright {A}_{3}$ is the annihilator in${A}_{i}$ of the corresponding factor group$\mathbf{Z}$ of${M}_{i}$ . By our basic knowledge of the cyclic groups$M$ , we know these annihilators are the ideals${\mathbf{Z}}_{k}$ ,${A}_{1}=24\mathbf{Z}$ , and${A}_{2}=15\mathbf{Z}$ . The intersection of those three ideals in${A}_{3}=50\mathbf{Z}$ is the ideal generated by their three generators, which (by definition!) is the least common multiple$\mathbf{Z}$ . So in summary, the annihilator of$\mathrm{lcm}(24,15,50)=600$ in$M$ is the ideal$\mathbf{Z}$ .$600\mathbf{Z}$ Turning things around, the annihilator of the ideal
in the given module$I=2\mathbf{Z}$ is the collection of all elements$M$ such that$({m}_{1},{m}_{2},{m}_{3})$ for every$(i{m}_{1},i{m}_{2},i{m}_{3})=(0,0,0)$ . Given the description of the ideal$i\beta \x88\x88I$ , the annihilator are those triples such that for every integer$I$ we have$k$ in$2k{m}_{1}=0$ ,${\mathbf{Z}}_{24}$ in$2k{m}_{2}=0$ , and${\mathbf{Z}}_{15}$ in$2k{m}_{3}=0$ . These conditions are satisfied exactly by${\mathbf{Z}}_{50}$ ,${m}_{1}\beta \x88\x8812{\mathbf{Z}}_{24}$ in${m}_{2}=0$ , and${\mathbf{Z}}_{15}$ . The annihilator of${m}_{3}\beta \x88\x8825{\mathbf{Z}}_{50}$ in$I$ is therefore the submodule$M$ , which is isomorphic as an abelian group to$(12{\mathbf{Z}}_{24})\Gamma \x970\Gamma \x97(25{\mathbf{Z}}_{50})$ .${\mathbf{Z}}_{2}\Gamma \x97{\mathbf{Z}}_{2}$
Problem 5
Give an example of a ring
There are many possible such examples. One familiar examples that of complex conjugation. Let
Problem 6
Suppose
Bonus challenge: Is your isomorphism natural in
First recall that
We will now define a set map