Homework 2

Suppose, towards a contradiction, that $r$ had a left inverse. Let $s\in R$ be a left inverse of $r$ in $R$, so that $sr=1_R$. On the one hand, we then have $s(rm)=s\cdot 0=0$; on the other hand we also have $s(rm)=(sr)m=1_R\cdot m=m$. Since $m$ was assumed to be nonzero, this is a contradiction.

It is clear (always dangerous to say!) that $IM$ is a subgroup of the abelian group $M$, so it only remains to verify $IM$ is closed under scaling, i.e., the action of $R$. To see that, suppose we have an element of $\sum _k{i}_k{m}_k$ of $IM$ and observe that $r\cdot \sum _k{i}_k{m}_k=\sum _{k}r(i_k{m}_k)=\sum _k(r{i}_k)m.$

Since $I$ is a left ideal of $R$, we have $r{i}_k\in R$ for every $k$ and so the final sum is once again another element of the set $IM$.

1. First observe that $1_R\cdot 0=0$ so $0\in \mathrm{Tor}(\mathrm{M})$. Next suppose $m,n\in \mathrm{Tor}(\mathrm{M})$, and let $r,s\in R$ be nonzero elements such that $rm=0$ and $sn=0$. We claim that $rs$ is nonzero and $(rs)(m-n)=0$. The first statement follows from the fact that we assumed $R$ is an integral domain; for the second, observe that $(rs)(m-n)=(rs)m-(rs)n=(sr)m-(rs)n=s(rm)-r(sn)=s\cdot 0-r\cdot 0=0-0=0.$ Note that we used the commutativity of $R$ (the "integral" part of "integral domain") to switch from $(rs)m$ to $(sr)m$. > 2. In light of part 1, we need to look at rings $R$ that are not integral domains. One such ring is ${\mathbf{Z}}_6$. When considered as an module over itself, the elements $2$ and $3$ are torsion elements, since $3\cdot 2=0$ and $2\cdot 3=0$. On the other hand, the element $2+3=5$ is not torsion, as one can quickly verify that $r\cdot 5\ne 0$ for each of the five nonzero elements of ${\mathbf{Z}}_6$.
2. Let $r\in R$ be a nonzero zero divisor, so that $rs=0_R$ for some nonzero $s\in R$. Let $m\in M$ be any nonzero element, and consider the element $sm$. If $sm=0$, then $m$ is torsion; if $sm\ne 0$, then $r(sm)=(rs)m=0_R\cdot m=0$, and so $sm$ is torsion. So in either case there exists a nonzero torsion element in $M$.
3. Take any torsion element $m\in M$ and let $r\in R$ be a nonzero element such that $rm=0_M$. Then observe that $r\cdot \varphi (m)=\varphi (rm)=\varphi (0_M)=0_N$. This prove $\varphi (m)$ is a torsion element in $N$.

1. Suppose $G$ is a finite abelian group, say of order $n$. By Lagrange's Theorem we then have $ng=0_G$ for every $g\in G$. But this exactly says that every $g\in G$ is torsion when $G$ is viewed as a $\mathbf{Z}$-module, and hence $G$ is torsion as a $\mathbf{Z}$-module.

2. One nice example is that of the group $U$ of roots of unity in $\mathbf{C}$, although that group is the rare abelian group that uses multiplicative notation. In that group we have (by definition!) that for each $\zeta \in U$ there is some positive integer $n$ with ${\zeta }^n=1$. Noting that the $\mathbf{Z}$-action on $U$ is by $m\star \zeta ={\zeta }^m$ and the "zero element" (i.e., identity) in $U$ is 1, we thus have that every $\zeta \in U$ is torsion when $U$ is viewed as a $\mathbf{Z}$-module.

   Here's another example. Suppose $F$ is an infinite field of positive characteristic; e.g., $F={\mathbf{Z}}_p(t)$. Considered as a $\mathbf{Z}$-module, every element in $F$ is killed by the characteristic of the field, and hence is torsion.

   A third example: the direct product of an infinite number of copies of a fixed finite abelian group, say $G={\displaystyle \prod _{i=1}^{\mathrm{\infty }}{\mathbf{Z}}_p}$.

1. We first verify the annihilator of $N$ in $R$ is an abelian subgroup. First recall that $0_R\cdot n=0_M$ for every $n\in $N$. Take any $r_1,r_2$ in the annihilator of $N$ and observe that for all $n\in N$ we have $(r_1-r_2)n=r_{1}n-r_{2}n=0_N-0_N=0_N$, and so $r_1-r_2$ is in the annihilator of $N$. By the Subgroup Criterion, we've thus proven the annihilator is a subgroup of $R$ under addition. Now fix any element $r$ of the annihilator and let $s\in R$ be an arbitrary element. Then for every $n\in N$ observe that $(sr)n=s(rn)=s\cdot 0_M=0_M$, where for the last equality we once again used the fact that zero acts trivially. This proves $sr$ is in the annihilator for every $s\in R$ and hence the annihilator is a left ideal of $R$. Similarly, we have $(rs)n=r(sn)=r\cdot 0_M=0_M$, where for the second equality we used the fact that $N$ is a submodule so $sn\in N$ and hence it is annihilated by $r$. This proves $rs$ is in the annihilator for every $s\in R$, and hence the annihilator is a right ideal of $R$.

2. This will prove very similar to the argument above. First note once again that $i\cdot 0_M=0_M$ for every $i\in I$, and so $0_M$ is the annihilator of $I$ in $M$. Next suppose $m_1,m_2$ are in the annihilator of $I$ in $M$. Then for every $i\in I$ we have $i(m_1-m_2)=i{m}_1-i{m}_2=0_M-0_M=0_M$, and so $m_1-m_2$ is in the annihilator of $I$ in $M$. We've thus proven the annihilator of $I$ in $M$ is a subgroup of $M$. Finally, take any $r\in R$ and $m$ in the annihilator of $I$ in $M$. Then $i(rm)=(ir)m=i^{\prime }m=0_M$, since $i^{\prime }\in I$ (as $I$ is a right ideal) and $m$ annihilates everything in $I$. Thus $rm$ is in the annihilator of $I$ in $M$, and hence that annihilator is a left $R$-submodule.

3. By the definition, the annihilator of $M$ in $\mathbf{Z}$ is the collection of all integers $n$ such that $nm=0$ for every $m\in M$. The module $M$ is a Cartesian product, which means that its operation is component-wise (addition) and that its zero element is the triple $(0,0,0)$. If we denote the general element in $M$ as $(m_1,m_2,m_3)$ then $n\cdot (m_1,m_2,m_3)=(n{m}_1,n{m}_2,n{m}_3)$, and so $n$ annihilates that element exactly when $n{m}_1=0$ in ${\mathbf{Z}}_{24}$, $n{m}_2=0$ in ${\mathbf{Z}}_{15}$, and $n{m}_3=0$ in ${\mathbf{Z}}_{50}$. In other words, the desired annihilator is $A_1\cap A_2\cap A_3$, where each $A_i$ is the annihilator in $\mathbf{Z}$ of the corresponding factor group $M_i$ of $M$. By our basic knowledge of the cyclic groups ${\mathbf{Z}}_k$, we know these annihilators are the ideals $A_1=24\mathbf{Z}$, $A_2=15\mathbf{Z}$, and $A_3=50\mathbf{Z}$. The intersection of those three ideals in $\mathbf{Z}$ is the ideal generated by their three generators, which (by definition!) is the least common multiple $\mathrm{lcm}(24,15,50)=600$. So in summary, the annihilator of $M$ in $\mathbf{Z}$ is the ideal $600\mathbf{Z}$.

   Turning things around, the annihilator of the ideal $I=2\mathbf{Z}$ in the given module $M$ is the collection of all elements $(m_1,m_2,m_3)$ such that $(i{m}_1,i{m}_2,i{m}_3)=(0,0,0)$ for every $i\in I$. Given the description of the ideal $I$, the annihilator are those triples such that for every integer $k$ we have $2k{m}_1=0$ in ${\mathbf{Z}}_{24}$, $2k{m}_2=0$ in ${\mathbf{Z}}_{15}$, and $2k{m}_3=0$ in ${\mathbf{Z}}_{50}$. These conditions are satisfied exactly by $m_1\in 12{\mathbf{Z}}_{24}$, $m_2=0$ in ${\mathbf{Z}}_{15}$, and $m_3\in 25{\mathbf{Z}}_{50}$. The annihilator of $I$ in $M$ is therefore the submodule $(12{\mathbf{Z}}_{24})\times 0\times (25{\mathbf{Z}}_{50})$, which is isomorphic as an abelian group to ${\mathbf{Z}}_2\times {\mathbf{Z}}_2$.

There are many possible such examples. One familiar examples that of complex conjugation. Let $R$ be the field of complex numbers, let $M$ and $N$ both be the abelian group of complex numbers (under addition), and let $f$ be complex conjugation. The familiar fact that complex conjugation is compatible with addition (i.e., $\overline{z+w}=\bar{z}+\bar{w}$) is really the statement that the map $f:\mathbf{C}\to \mathbf{C}$ is a morphism of abelian groups (i.e., $f(z+w)=f(z)+f(w)$.) But conjugation is not compatible with complex scaling. For instance, if $z=1+2i$ and $r=i$, then $\overline{rz}=\overline{i(1+2i)}=\overline{i-2}=-i-2$, while $r\bar{z}=i\cdot \overline{1+2i}=i(1-2i)=i+2$. In our notation, this exactly says that $f(rz)\ne rf(z)$, and so our map $f$ is not an $R$-module morphism.

First recall that $R$ is being viewed as a module over itself by way of left multiplication. Next, recall that the set ${\mathrm{Hom}}_R(R,M)$ always has the structure of an abelian group, and since $R$ is commutative here it also has the structure of an $R$-module.

We will now define a set map ${\mathrm{Hom}}_R(R,M)\to M$ and eventually prove it is a module isomorphism. How can we come up with a map? Suppose $f:R\to M$ is a module morphism. What do we know about $f$? At first glance, not much. By the group morphism property we must have $f(0_R)=0_M$, so that doesn't tell us much about the map $f$. But our ring $R$ has another distinguished element, its multiplicative identity $1_R$. That element gets mapped to some element $f(1_R)\in M$. We claim that this element (the image of $1_R$ in $M$) completely determines the map $f$ as a module morphism. To see this, suppose $r\in R$ is any element. Since $f$ is a module morphism, we must have $f(r)=f(r\cdot 1_R)=rf(1_R)$. In other words, the image of $r$ is determined by $f(1_R)$. We now have our candidate for a map from ${\mathrm{Hom}}_R(R,M)\to M$, which we will now name ${\mathrm{ev}}_{1_R}$ map ("evaluate at $1_R$"). We claim this map is an $R$-module isomorphism.

First we show it is group morphism, so suppose $f,g\in {\mathrm{Hom}}_R(R,M)$. Then ${\mathrm{ev}}_{1_R}(f+g)=(f+g)(1_R)=f(1_R)+g(1_R)={\mathrm{ev}}_{1_R}(f)+{\mathrm{ev}}_{1_R}(g)$, where the second equality holds by the definition of the group structure on ${\mathrm{Hom}}_R(R,M)$.

We next verify ${\mathrm{ev}}_{1_R}$ an $R$-module morphism. Take any $r\in R$ and $f\in {\mathrm{Hom}}_R(R,M)$, and observe that ${\mathrm{ev}}_{1_R}(rf)=(rf)(1_R)=r\cdot f(1_R)=r\cdot {\mathrm{ev}}_{1_R}(f)$, where the second equality holds by the definition of the $R$-action on ${\mathrm{Hom}}_R(R,M)$.

Finally, we verify ${\mathrm{ev}}_{1_R}$ is an isomorphism. We could check it is bijective, but I prefer to instead produce the inverse $R$-module morphism. To that end, define a map $g:M\to {\mathrm{Hom}}_R(R,M)$ as follows. For each element $m\in M$, let $f_m:R\to M$ be the map defined by $f_m(r)=rm$. Note that $f_m(1_R)=1_R\cdot m=m$, as one would hope. Also observe that $f_m$ is indeed an $R$-module morphism, since $f_m(r+s)=(r+s)m=rm+sm=f_m(r)+f_m(s)$ and $f_m(sr)=(sr)m=s(rm)=s{f}_m(r)$. We have therefore indeed defined a set map $g:M\to {\mathrm{Hom}}_R(R,M)$. This map is a group morphism since $f_{m_1+m_2}(r)=r(m_1+m_2)=r{m}_1+r{m}_2=f_{m_1}(r)+f_{m_2}(r)$, i.e., $g(m_1+m_2)=g(m_1)+g(m_2)$. This map is an $R$-module morphism since $f_{sm}(r)=r(sm)=(rs)m=(sr)m=s(rm)=s\cdot f_m(r)$, i.e., $g(sm)=s\cdot g(m)$. (Note that we used the commutativity of $R$ for the third equality.) Lastly, this $R$-module morphism $g$ is indeed inverse to ${\mathrm{ev}}_{1_R}$, since $({\mathrm{ev}}_{1_R}\circ g)(m)={\mathrm{ev}}_{1_R}\circ f_m=f_m(1_R)=1_R\cdot m=m$, i.e., ${\mathrm{ev}}_{1_R}\circ g$ is the identity module morphism on $R$. (You can also verify that $g\circ {\mathrm{ev}}_{1_R}$ is the identity module morphism on ${\mathrm{Hom}}_R(R,M)$).

Curious about the bonus challenge? Naturality in this case means that for every $R$-module morphism $g:M\to N$ we should have a commutative diagram as below:

Chasing the diagram around, suppose we have an element in the module on the top left, i.e., an $R$-module morphism $f:R\to M$. Going over and down gives $g(f(1_R))$, while going down and then over gives $(g\circ f)(1_R)$, which by the definition of function composition does indeed agree with $g(f(1_R))$. So we have indeed defined a natural isomorphism.

We know that when $R$ is a commutative ring and $M$ is an $R$-module, the set ${\mathrm{Hom}}_R(R,M)$ is an $R$-module and is isomorphic to $M$. So we already know that in the special case $M=R$ (when $R$ is a viewed as an $R$-module) the map ${\mathrm{ev}}_{1_R}:{\mathrm{Hom}}_R(R,R)\to R$ is an $R$-module isomorphism. We thus only need to verify it's actually a ring morphism. To see that, suppose $f_1,f_2\in {\mathrm{Hom}}_R(R,R)$ and recall that the product in the ring ${\mathrm{Hom}}_R(R,R)$ is given by composition (or $R$-module morphisms). Then note that
${\mathrm{ev}}_{1_R}(f_1\cdot f_2)=(f_1\cdot f_2)(1_R)=f_1(f_2(1_R)).$
Let $r^{\prime }=f_2(1_R)\in R$. Since $f_1$ is an $R$-module morphism, we must have $f_1(r^{\prime })=r^{\prime }\cdot f_1(1_R)=r^{\prime }\cdot {\mathrm{ev}}_{1_R}(f_1)$. Since $r^{\prime }=f_2(1_R)={\mathrm{ev}}_{1_R}(f_2)$, the above equality becomes
${\mathrm{ev}}_{1_R}(f_1\cdot f_2)={\mathrm{ev}}_{1_R}(f_2){\mathrm{ev}}_{1_R}(f_1).$
The right-hand side of the above equality is a product in the commutative ring $R$, so that commutativity allows us to conclude
${\mathrm{ev}}_{1_R}(f_1\cdot f_2)={\mathrm{ev}}_{1_R}(f_1){\mathrm{ev}}_{1_R}(f_2).$